Probability

How likely something is to happen.

Many events can't be predicted with total certainty. The best we can say is how likely they are to happen, using the idea of probability.

Tossing a Coin

When a coin is tossed, there are two possible outcomes:

Heads (H) or Tails (T)

  • the probability of the coin landing H is ½
  • the probability of the coin landing T is ½

Throwing Dice

When a single die is thrown, there are six possible outcomes: 1, 2, 3, 4, 5, 6 .

The probability of any one of them is 1 6

In general:

Probability of an event happening = Number of ways it can happen Total number of outcomes

Example: the chances of rolling a "4" with a die

Number of ways it can happen: 1 (there is only 1 face with a "4" on it)

Total number of outcomes: 6 (there are 6 faces altogether)

So the probability = 1 6

Example: there are 5 marbles in a bag: 4 are blue, and 1 is red. What is the probability that a blue marble gets picked?

Number of ways it can happen: 4 (there are 4 blues)

Total number of outcomes: 5 (there are 5 marbles in total)

So the probability = 4 5 = 0.8

Probability Line

We can show probability on a Probability Line :

Probability is always between 0 and 1

Probability is Just a Guide

Probability does not tell us exactly what will happen, it is just a guide

Example: toss a coin 100 times, how many Heads will come up?

Probability says that heads have a ½ chance, so we can expect 50 Heads .

But when we actually try it we might get 48 heads, or 55 heads ... or anything really, but in most cases it will be a number near 50.

Learn more at Probability Index .

Some words have special meaning in Probability:

Experiment : a repeatable procedure with a set of possible results.

Example: Throwing dice

We can throw the dice again and again, so it is repeatable.

The set of possible results from any single throw is {1, 2, 3, 4, 5, 6}

Outcome: A possible result.

Example: "6" is one of the outcomes of a throw of a die.

Trial: A single performance of an experiment.

Example: I conducted a coin toss experiment. After 4 trials I got these results:

Three trials had the outcome "Head", and one trial had the outcome "Tail"

Sample Space: all the possible outcomes of an experiment.

Example: choosing a card from a deck

There are 52 cards in a deck (not including Jokers)

So the Sample Space is all 52 possible cards : {Ace of Hearts, 2 of Hearts, etc... }

The Sample Space is made up of Sample Points:

Sample Point: just one of the possible outcomes

Example: Deck of Cards

  • the 5 of Clubs is a sample point
  • the King of Hearts is a sample point

"King" is not a sample point. There are 4 Kings, so that is 4 different sample points.

There are 6 different sample points in that sample space.

Event: one or more outcomes of an experiment

Example Events:

An event can be just one outcome:

  • Getting a Tail when tossing a coin
  • Rolling a "5"

An event can include more than one outcome:

  • Choosing a "King" from a deck of cards (any of the 4 Kings)
  • Rolling an "even number" (2, 4 or 6)

Hey, let's use those words, so you get used to them:

Example: Alex wants to see how many times a "double" comes up when throwing 2 dice.

The Sample Space is all possible Outcomes (36 Sample Points):

{1,1} {1,2} {1,3} {1,4} ... ... ... {6,3} {6,4} {6,5} {6,6}

The Event Alex is looking for is a "double", where both dice have the same number. It is made up of these 6 Sample Points :

{1,1} {2,2} {3,3} {4,4} {5,5} and {6,6}

These are Alex's Results:

 After 100 Trials , Alex has 19 "double" Events ... is that close to what you would expect?

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15 Probability Questions And Practice Problems for Middle and High School: Harder Exam Style Questions Included

Beki christian.

Probability questions and probability problems require students to work out how likely it is that something is to happen. Probabilities can be described using words or numbers. Probabilities range from 0 to 1 and can be written as fractions, decimals or percentages .

Here you’ll find a selection of probability questions of varying difficulty showing the variety you are likely to encounter in middle school and high school, including several harder exam style questions.

What are some real life examples of probability?

How to calculate probabilities, probability question: a worked example, 6th grade probability questions, 7th grade probability questions, 8th grade probability questions, 9th grade probability questions, 10th grade probability questions.

  • 11th & 12th grade probability questions

Looking for more middle school and high school probability math questions?

The more likely something is to happen, the higher its probability. We think about probabilities all the time.

For example, you may have seen that there is a 20% chance of rain on a certain day or thought about how likely you are to roll a 6 when playing a game, or to win in a raffle when you buy a ticket.

15 Probability Questions Worksheet

Want the 15 multiple choice probability questions from this blog in a handy downloadable format? Then look no further!

The probability of something happening is given by:

We can also use the following formula to help us calculate probabilities and solve problems:

  • Probability of something not occuring = 1 – probability of if occurring P(not\;A) = 1 - P(A)
  • For mutually exclusive events: Probability of event A OR event B occurring = Probability of event A + Probability of event B P(A\;or\;B) = P(A)+P(B)
  • For independent events: Probability of event A AND event B occurring = Probability of event A times probability of event B P(A\;and\;B) = P(A) × P(B)

Question: What is the probability of getting heads three times in a row when flipping a coin?

When flipping a coin, there are two possible outcomes – heads or tails. Each of these options has the same probability of occurring during each flip. The probability of either heads or tails on a single coin flip is ½.

Since there are only two possible outcomes and they have the same probability of occurring, this is called a binomial distribution.

Let’s look at the possible outcomes if we flipped a coin three times.

Let H=heads and T=tails.

The possible outcomes are: HHH, THH, THT, HTT, HHT, HTH, TTH, TTT

Each of these outcomes has a probability of ⅛.

Therefore, the probability of flipping a coin three times in a row and having it land on heads all three times is ⅛.

Middle school probability questions

In middle school, probability questions introduce the idea of the probability scale and the fact that probabilities sum to one. We look at theoretical and experimental probability as well as learning about sample space diagrams and venn diagrams.

1. Which number could be added to this spinner to make it more likely that the spinner will land on an odd number than a prime number?

GCSE Quiz False

Currently there are two odd numbers and two prime numbers so the chances of landing on an odd number or a prime number are the same. By adding 3, 5 or 11 you would be adding one prime number and one odd number so the chances would remain equal.

By adding 9 you would be adding an odd number but not a prime number. There would be three odd numbers and two prime numbers so the spinner would be more likely to land on an odd number than a prime number.

2. Ifan rolls a fair dice, with sides labeled A, B, C, D, E and F. What is the probability that the dice lands on a vowel?

A and E are vowels so there are 2 outcomes that are vowels out of 6 outcomes altogether.

Therefore the probability is   \frac{2}{6} which can be simplified to \frac{1}{3} .

3. Max tested a coin to see whether it was fair. The table shows the results of his coin toss experiment:

Heads          Tails

    26                  41

What is the relative frequency of the coin landing on heads?

Max tossed the coin 67 times and it landed on heads 26 times.

\text{Relative frequency (experimental probability) } = \frac{\text{number of successful trials}}{\text{total number of trials}} = \frac{26}{67}

4. Grace rolled two dice. She then did something with the two numbers shown. Here is a sample space diagram showing all the possible outcomes:

What did Grace do with the two numbers shown on the dice?

Add them together

Subtract the number on dice 2 from the number on dice 1

Multiply them

Subtract the smaller number from the bigger number

For each pair of numbers, Grace subtracted the smaller number from the bigger number.

For example, if she rolled a 2 and a 5, she did 5 − 2 = 3.

5. Alice has some red balls and some black balls in a bag. Altogether she has 25 balls. Alice picks one ball from the bag. The probability that Alice picks a red ball is x and the probability that Alice picks a black ball is 4x. Work out how many black balls are in the bag.

Since the probability of mutually exclusive events add to 1:  

\begin{aligned} x+4x&=1\\\\ 5x&=1\\\\ x&=\frac{1}{5} \end{aligned}

\frac{1}{5} of the balls are red and \frac{4}{5} of the balls are blue.

6. Arthur asked the students in his class whether they like math and whether they like science. He recorded his results in the venn diagram below.

How many students don’t like science?

We need to look at the numbers that are not in the ‘Like science’ circle. In this case it is 9 + 7 = 16.

High school probability questions

In high school, probability questions involve more problem solving to make predictions about the probability of an event. We also learn about probability tree diagrams, which can be used to represent multiple events, and conditional probability.

7. A restaurant offers the following options:

Starter – soup or salad

Main – chicken, fish or vegetarian

Dessert – ice cream or cake

How many possible different combinations of starter, main and dessert are there?

The number of different combinations is 2 × 3 × 2 = 12.

8. There are 18 girls and 12 boys in a class. \frac{2}{9} of the girls and \frac{1}{4} of the boys walk to school. One of the students who walks to school is chosen at random. Find the probability that the student is a boy. 

First we need to work out how many students walk to school:

\frac{2}{9} \text{ of } 18 = 4

\frac{1}{4} \text{ of } 12 = 3

7 students walk to school. 4 are girls and 3 are boys. So the probability the student is a boy is \frac{3}{7} .

9. Rachel flips a biased coin. The probability that she gets two heads is 0.16. What is the probability that she gets two tails?

We have been given the probability of getting two heads. We need to calculate the probability of getting a head on each flip.

Let’s call the probability of getting a head p.

The probability p, of getting a head AND getting another head is 0.16.

Therefore to find p:

The probability of getting a head is 0.4 so the probability of getting a tail is 0.6.

The probability of getting two tails is 0.6 × 0.6 = 0.36 .

10. I have a big tub of jelly beans. The probability of picking each different color of jelly bean is shown below:

If I were to pick 60 jelly beans from the tub, how many orange jelly beans would I expect to pick?

First we need to calculate the probability of picking an orange. Probabilities sum to 1 so 1 − (0.2 + 0.15 + 0.1 + 0.3) = 0.25.

The probability of picking an orange is 0.25.

The number of times I would expect to pick an orange jelly bean is 0.25 × 60 = 15 .

11. Dexter runs a game at a fair. To play the game, you must roll a dice and pick a card from a deck of cards.

To win the game you must roll an odd number and pick a picture card. The game can be represented by the tree diagram below.

Dexter charges players $1 to play and gives $3 to any winners. If 260 people play the game, how much profit would Dexter expect to make?

Completing the tree diagram:

Probability of winning is \frac{1}{2} \times \frac{4}{13} = \frac{4}{26}

If 260 play the game, Dexter would receive $260.

The expected number of winners would be \frac{4}{26} \times 260 = 40

Dexter would need to give away 40 × $3 = $120 .

Therefore Dexter’s profit would be $260 − $120 = $140.

12. A fair coin is tossed three times. Work out the probability of getting two heads and one tail.

There are three ways of getting two heads and one tail: HHT, HTH or THH.

The probability of each is \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8} 

Therefore the total probability is \frac{1}{8} +\frac{1}{8} + \frac{1}{8} = \frac{3}{8}

11th/12th grade probability questions

13. 200 people were asked about which athletic event they thought was the most exciting to watch. The results are shown in the table below.

A person is chosen at random. Given that that person chose 100m, what is the probability that the person was female?

Since we know that the person chose 100m, we need to include the people in that column only.

In total 88 people chose 100m so the probability the person was female is \frac{32}{88}   .

14.   Sam asked 50 people whether they like vegetable pizza or pepperoni pizza.

37 people like vegetable pizza. 

25 people like both. 

3 people like neither.

Sam picked one of the 50 people at random. Given that the person he chose likes pepperoni pizza, find the probability that they don’t like vegetable pizza.

We need to draw a venn diagram to work this out.

We start by putting the 25 who like both in the middle section. The 37 people who like vegetable pizza includes the 25 who like both, so 12 more people must like vegetable pizza. 3 don’t like either. We have 50 – 12 – 25 – 3 = 10 people left so this is the number that must like only pepperoni.

There are 35 people altogether who like pepperoni pizza. Of these, 10 do not like vegetable pizza. The probability is   \frac{10}{35} .

15. There are 12 marbles in a bag. There are n red marbles and the rest are blue marbles. Nico takes 2 marbles from the bag. Write an expression involving n for the probability that Nico takes one red marble and one blue marble.

We need to think about this using a tree diagram. If there are 12 marbles altogether and n are red then 12-n are blue.

To get one red and one blue, Nico could choose red then blue or blue then red so the probability is:

  • Ratio questions
  • Algebra questions
  • Trigonometry questions
  • Venn diagram questions
  • Long division questions
  • Pythagorean theorem questions

Do you have students who need extra support in math? Give your students more opportunities to consolidate learning and practice skills through personalized math tutoring with their own dedicated online math tutor. Each student receives differentiated instruction designed to close their individual learning gaps, and scaffolded learning ensures every student learns at the right pace. Lessons are aligned with your state’s standards and assessments, plus you’ll receive regular reports every step of the way. Personalized one-on-one math tutoring programs are available for: – 2nd grade tutoring – 3rd grade tutoring – 4th grade tutoring – 5th grade tutoring – 6th grade tutoring – 7th grade tutoring – 8th grade tutoring Why not learn more about how it works ?

The content in this article was originally written by secondary teacher Beki Christian and has since been revised and adapted for US schools by elementary math teacher Katie Keeton.

Pythagoras Theorem Questions [FREE]

Downloadable Pythagoras theorem worksheet containing 15 multiple choice questions with a mix of worded problems and deeper problem solving questions.

Includes an answer key and follows variation theory with plenty of opportunities for students to work independently at their own level.

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Probability Questions with Solutions

Tutorial on finding the probability of an event. In what follows, S is the sample space of the experiment in question and E is the event of interest. n(S) is the number of elements in the sample space S and n(E) is the number of elements in the event E.

Questions and their Solutions

Answers to the above exercises, more references and links.

Teach yourself statistics

How to Solve Probability Problems

You can solve many simple probability problems just by knowing two simple rules:

  • The probability of any sample point can range from 0 to 1.
  • The sum of probabilities of all sample points in a sample space is equal to 1.

The following sample problems show how to apply these rules to find (1) the probability of a sample point and (2) the probability of an event.

Probability of a Sample Point

The probability of a sample point is a measure of the likelihood that the sample point will occur.

Example 1 Suppose we conduct a simple statistical experiment . We flip a coin one time. The coin flip can have one of two equally-likely outcomes - heads or tails. Together, these outcomes represent the sample space of our experiment. Individually, each outcome represents a sample point in the sample space. What is the probability of each sample point?

Solution: The sum of probabilities of all the sample points must equal 1. And the probability of getting a head is equal to the probability of getting a tail. Therefore, the probability of each sample point (heads or tails) must be equal to 1/2.

Example 2 Let's repeat the experiment of Example 1, with a die instead of a coin. If we toss a fair die, what is the probability of each sample point?

Solution: For this experiment, the sample space consists of six sample points: {1, 2, 3, 4, 5, 6}. Each sample point has equal probability. And the sum of probabilities of all the sample points must equal 1. Therefore, the probability of each sample point must be equal to 1/6.

Probability of an Event

The probability of an event is a measure of the likelihood that the event will occur. By convention, statisticians have agreed on the following rules.

  • The probability of any event can range from 0 to 1.
  • The probability of event A is the sum of the probabilities of all the sample points in event A.
  • The probability of event A is denoted by P(A).

Thus, if event A were very unlikely to occur, then P(A) would be close to 0. And if event A were very likely to occur, then P(A) would be close to 1.

Example 1 Suppose we draw a card from a deck of playing cards. What is the probability that we draw a spade?

Solution: The sample space of this experiment consists of 52 cards, and the probability of each sample point is 1/52. Since there are 13 spades in the deck, the probability of drawing a spade is

P(Spade) = (13)(1/52) = 1/4

Example 2 Suppose a coin is flipped 3 times. What is the probability of getting two tails and one head?

Solution: For this experiment, the sample space consists of 8 sample points.

S = {TTT, TTH, THT, THH, HTT, HTH, HHT, HHH}

Each sample point is equally likely to occur, so the probability of getting any particular sample point is 1/8. The event "getting two tails and one head" consists of the following subset of the sample space.

A = {TTH, THT, HTT}

The probability of Event A is the sum of the probabilities of the sample points in A. Therefore,

P(A) = 1/8 + 1/8 + 1/8 = 3/8

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Probability - Problem Solving

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  • Geoff Pilling
  • Sandeep Bhardwaj

To solve problems on this page, you should be familiar with

  • Uniform Probability
  • Probability - By Outcomes
  • Probability - Rule of Sum
  • Probability - Rule of Product
  • Probability - By Complement
  • Probability - Independent Events
  • Conditional Probability

Problem Solving - Basic

Problem solving - intermediate, problem solving - difficult.

If I throw 2 standard 5-sided dice, what is the probability that the sum of their top faces equals to 10? Assume both throws are independent to each other. Solution : The only way to obtain a sum of 10 from two 5-sided dice is that both die shows 5 face up. Therefore, the probability is simply \( \frac15 \times \frac15 = \frac1{25} = .04\)

If from each of the three boxes containing \(3\) white and \(1\) black, \(2\) white and \(2\) black, \(1\) white and \(3\) black balls, one ball is drawn at random. Then the probability that \(2\) white and \(1\) black balls will be drawn is?

2 fair 6-sided dice are rolled. What is the probability that the sum of these dice is \(10\)? Solution : The event for which I obtain a sum of 10 is \(\{(4,6),(6,4),(5,5) \}\). And there is a total of \(6^2 = 36\) possible outcomes. Thus the probability is simply \( \frac3{36} = \frac1{12} \approx 0.0833\)

If a fair 6-sided dice is rolled 3 times, what is the probability that we will get at least 1 even number and at least 1 odd number?

Three fair cubical dice are thrown. If the probability that the product of the scores on the three dice is \(90\) is \(\dfrac{a}{b}\), where \(a,b\) are positive coprime integers, then find the value of \((b-a)\).

You can try my other Probability problems by clicking here

Suppose a jar contains 15 red marbles, 20 blue marbles, 5 green marbles, and 16 yellow marbles. If you randomly select one marble from the jar, what is the probability that you will have a red or green marble? First, we can solve this by thinking in terms of outcomes. You could draw a red, blue, green, or yellow marble. The probability that you will draw a green or a red marble is \(\frac{5 + 15}{5+15+16+20}\). We can also solve this problem by thinking in terms of probability by complement. We know that the marble we draw must be blue, red, green, or yellow. In other words, there is a probability of 1 that we will draw a blue, red, green, or yellow marble. We want to know the probability that we will draw a green or red marble. The probability that the marble is blue or yellow is \(\frac{16 + 20}{5+15+16+20}\). , Using the following formula \(P(\text{red or green}) = 1 - P(\text{blue or yellow})\), we can determine that \(P(\text{red or green}) = 1 - \frac{16 + 20}{5+15+16+20} = \frac{5 + 15}{5+15+16+20}\).

Two players, Nihar and I, are playing a game in which we alternate tossing a fair coin and the first player to get a head wins. Given that I toss first, the probability that Nihar wins the game is \(\dfrac{\alpha}{\beta}\), where \(\alpha\) and \(\beta\) are coprime positive integers.

Find \(\alpha + \beta\).

If I throw 3 fair 5-sided dice, what is the probability that the sum of their top faces equals 10? Solution : We want to find the total integer solution for which \(a +b+c=10 \) with integers \(1\leq a,b,c \leq5 \). Without loss of generality, let \(a\leq b \leq c\). We list out the integer solutions: \[ (1,4,5),(2,3,5), (2,4,4), (3,3,4) \] When relaxing the constraint of \(a\leq b \leq c\), we have a total of \(3! + 3! + \frac{3!}{2!} + \frac{3!}{2!} = 18 \) solutions. Because there's a total of \(5^3 = 125\) possible combinations, the probability is \( \frac{18}{125} = 14.4\%. \ \square\)

Suppose you and 5 of your friends each brought a hat to a party. The hats are then put into a large box for a random-hat-draw. What is the probability that nobody selects his or her own hat?

How many ways are there to choose exactly two pets from a store with 8 dogs and 12 cats? Since we haven't specified what kind of pets we pick, we can choose any animal for our first pick, which gives us \( 8+12=20\) options. For our second choice, we have 19 animals left to choose from. Thus, by the rule of product, there are \( 20 \times 19 = 380 \) possible ways to choose exactly two pets. However, we have counted every pet combination twice. For example, (A,B) and (B,A) are counted as two different choices even when we have selected the same two pets. Therefore, the correct number of possible ways are \( {380 \over 2} = 190 \)

A bag contains blue and green marbles. If 5 green marbles are removed from the bag, the probability of drawing a green marble from the remaining marbles would be 75/83 . If instead 7 blue marbles are added to the bag, the probability of drawing a blue marble would be 3/19 . What was the number of blue marbles in the bag before any changes were made?

Bob wants to keep a good-streak on Brilliant, so he logs in each day to Brilliant in the month of June. But he doesn't have much time, so he selects the first problem he sees, answers it randomly and logs out, despite whether it is correct or incorrect.

Assume that Bob answers all problems with \(\frac{7}{13}\) probability of being correct. He gets only 10 problems correct, surprisingly in a row, out of the 30 he solves. If the probability that happens is \(\frac{p}{q}\), where \(p\) and \(q\) are coprime positive integers, find the last \(3\) digits of \(p+q\).

Out of 10001 tickets numbered consecutively, 3 are drawn at random .

Find the chance that the numbers on them are in Arithmetic Progression .

The answer is of the form \( \frac{l}{k} \) .

Find \( k - l \) where \(k\) and \(l\) are co-prime integers.

HINT : You might consider solving for \(2n + 1\) tickets .

You can try more of my Questions here .

A bag contains a blue ball, some red balls, and some green balls. You reach into​ the bag and pull out three balls at random. The probability you pull out one of each color is exactly 3%. How many balls were initially in the bag?

More probability questions

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Amanda decides to practice shooting hoops from the free throw line. She decides to take 100 shots before dinner.

Her first shot has a 50% chance of going in.

But for Amanda, every time she makes a shot, it builds her confidence, so the probability of making the next shot goes up, But every time she misses, she gets discouraged so the probability of her making her next shot goes down.

In fact, after \(n\) shots, the probability of her making her next shot is given by \(P = \dfrac{b+1}{n+2}\), where \(b\) is the number of shots she has made so far (as opposed to ones she has missed).

So, after she has completed 100 shots, if the probability she has made exactly 83 of them is \(\dfrac ab\), where \(a\) and \(b\) are coprime positive integers, what is \(a+b\)?

Photo credit: http://polymathprogrammer.com/

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Probability Word Problems

In these lessons, we will learn how to solve a variety of probability problems.

Related Pages Probability Tree Diagrams Probability Without Replacement Theoretical vs. Experimental Probability More Lessons On Probability

Here we shall be looking into solving probability word problems involving:

  • Probability and Sample Space
  • Probability and Frequency Table
  • Probability and Area
  • Probability of Simple Events
  • Probability and Permutations
  • Probability and Combinations
  • Probability of Independent Events

We will now look at some examples of probability problems.

Example: At a car park there are 100 vehicles, 60 of which are cars, 30 are vans and the remainder are lorries. If every vehicle is equally likely to leave, find the probability of: a) a van leaving first. b) a lorry leaving first. c) a car leaving second if either a lorry or van had left first.

Solution: a) Let S be the sample space and A be the event of a van leaving first. n(S) = 100 n(A) = 30

c) If either a lorry or van had left first, then there would be 99 vehicles remaining, 60 of which are cars. Let T be the sample space and C be the event of a car leaving. n(T) = 99 n(C) = 60

Example: A survey was taken on 30 classes at a school to find the total number of left-handed students in each class. The table below shows the results:

A class was selected at random. a) Find the probability that the class has 2 left-handed students. b) What is the probability that the class has at least 3 left-handed students? c) Given that the total number of students in the 30 classes is 960, find the probability that a student randomly chosen from these 30 classes is left-handed.

a) Let S be the sample space and A be the event of a class having 2 left-handed students. n(S) = 30 n(A) = 5

b) Let B be the event of a class having at least 3 left-handed students. n(B) = 12 + 8 + 2 = 22

c) First find the total number of left-handed students:

Total no. of left-handed students = 2 + 10 + 36 + 32 + 10 = 90

Here, the sample space is the total number of students in the 30 classes, which was given as 960.

Let T be the sample space and C be the event that a student is left-handed. n(T) = 960 n(C) = 90

Probability And Area

Example: ABCD is a square. M is the midpoint of BC and N is the midpoint of CD. A point is selected at random in the square. Calculate the probability that it lies in the triangle MCN.

Area of square = 2x × 2x = 4x 2

example problems for probability

This video shows some examples of probability based on area.

Probability Of Simple Events

The following video shows some examples of probability problems. A few examples of calculating the probability of simple events.

  • What is the probability of the next person you meeting having a phone number that ends in 5?
  • What is the probability of getting all heads if you flip 3 coins?
  • What is the probability that the person you meet next has a birthday in February? (Non-leap year)

This video introduces probability and gives many examples to determine the probability of basic events.

A bag contains 8 marbles numbered 1 to 8 a. What is the probability of selecting a 2 from the bag? b. What is the probability of selecting an odd number? c. What is the probability of selecting a number greater than 6?

Using a standard deck of cards, determine each probability. a. P(face card) b. P(5) c. P(non face card)

Using Permutations To Solve Probability Problems

This video shows how to evaluate factorials, how to use permutations to solve probability problems, and how to determine the number of permutations with indistinguishable items.

A permutation is an arrangement or ordering. For a permutation, the order matters.

  • If a class has 28 students, how many different arrangements can 5 students give a presentation to the class?
  • How many ways can the letters of the word PHEONIX be arranged?
  • How many ways can you order 3 blue marbles, 4 red marbles and 5 green marbles? Marbles of the same color look identical.

Using Combinations To Solve Probability Problems

This video shows how to evaluate combinations and how to use combinations to solve probability problems.

A combination is a grouping or subset of items. For a combination, the order does not matter.

  • The soccer team has 20 players. There are always 11 players on the field. How many different groups of players can be in the field at the same time?
  • A student needs 8 more classes to complete her degree. If she has met the prerequisites for all the courses, how many ways can she take 4 class next semester?
  • There are 4 men and 5 women in a small office. The customer wants a site visit from a group of 2 men and 2 women. How many different groups van be formed from the office?

How To Find The Probability Of Different Events?

This video explains how to determine the probability of different events. This can be found that can be found using combinations and basic probability.

  • The probability of drawing 2 cards that are both face cards.
  • The probability of drawing 2 cards that are both aces.
  • The probability of drawing 4 cards all from the same suite.

A group of 10 students made up of 6 females and 4 males form a committee of 4. What is the probability the committee is all male? What is the probability that the committee is all female? What is the probability the committee is made up of 2 females and 2 males?

How To Find The Probability Of Multiple Independent Events?

This video explains the counting principle and how to determine the number of ways multiple independent events can occur.

  • How many ways can students answer a 3-question true of false quiz?
  • How many passwords using 6 digits where the first digit must be letters and the last four digits must be numbers?
  • A restaurant offers a dinner special in which you get to pick 1 item from 4 different categories. How many different meals are possible?
  • A door lock on a classroom requires entry of 4 digits. All digits must be numbers, but the digits can not be repeated. How many unique codes are possible?

How To Find The Probability Of A Union Of Two Events?

This video shows how to determine the probability of a union of two events.

  • If you roll 2 dice at the same time, what is the probability the sum is 6 or a pair of odd numbers?
  • What is the probability of selecting 1 card that is red or a face card?

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Probability

Probability defines the likelihood of occurrence of an event. There are many real-life situations in which we may have to predict the outcome of an event. We may be sure or not sure of the results of an event. In such cases, we say that there is a probability of this event to occur or not occur. Probability generally has great applications in games, in business to make predictions, and also it has extensive applications in this new area of artificial intelligence.

The probability of an event can be calculated by the probability formula by simply dividing the favourable number of outcomes by the total number of possible outcomes. The value of the probability of an event happening can lie between 0 and 1 because the favourable number of outcomes can never be more than the total number of outcomes. Also, the favorable number of outcomes cannot be negative. Let us discuss the basics of probability in detail in the following sections.

What is Probability?

Probability can be defined as the ratio of the number of favorable outcomes to the total number of outcomes of an event. For an experiment having 'n' number of outcomes, the number of favorable outcomes can be denoted by x. The formula to calculate the probability of an event is as follows.

Probability(Event) = Favorable Outcomes/Total Outcomes = x/n

Probability is used to predict the outcomes for the tossing of coins, rolling of dice, or drawing a card from a pack of playing cards. The probability is classified into two types:

  • Theoretical probability
  • Experimental probability

To understand each of these types, click on the respective links.

Terminology of Probability Theory

The following terms in probability theorey help in a better understanding of the concepts of probability.

Experiment: A trial or an operation conducted to produce an outcome is called an experiment.

Sample Space: All the possible outcomes of an experiment together constitute a sample space . For example, the sample space of tossing a coin is {head, tail}.

Favorable Outcome: An event that has produced the desired result or expected event is called a favorable outcome. For example, when we roll two dice, the possible/favorable outcomes of getting the sum of numbers on the two dice as 4 are (1,3), (2,2), and (3,1).

Trial: A trial denotes doing a random experiment.

Random Experiment: An experiment that has a well-defined set of outcomes is called a random experiment . For example, when we toss a coin, we know that we would get ahead or tail, but we are not sure which one will appear.

Event: The total number of outcomes of a random experiment is called an event .

Equally Likely Events: Events that have the same chances or probability of occurring are called equally likely events. The outcome of one event is independent of the other. For example, when we toss a coin, there are equal chances of getting a head or a tail.

Exhaustive Events: When the set of all outcomes of an event is equal to the sample space, we call it an exhaustive event .

Mutually Exclusive Events: Events that cannot happen simultaneously are called mutually exclusive events . For example, the climate can be either hot or cold. We cannot experience the same weather simultaneously.

Events in Probability

In probability theory, an event is a set of outcomes of an experiment or a subset of the sample space. If P(E) represents the probability of an event E, then, we have,

  • P(E) = 0 if and only if E is an impossible event.
  • P(E) = 1 if and only if E is a certain event.
  • 0 ≤ P(E) ≤ 1.

Suppose, we are given two events, "A" and "B", then the probability of event A, P(A) > P(B) if and only if event "A" is more likely to occur than the event "B". Sample space(S) is the set of all of the possible outcomes of an experiment and n(S) represents the number of outcomes in the sample space.

P(E) = n(E)/n(S)

P(E’) = (n(S) - n(E))/n(S) = 1 - (n(E)/n(S))

E’ represents that the event will not occur.

Therefore, now we can also conclude that, P(E) + P(E’) = 1

Probability Formula

The probability equation defines the likelihood of the happening of an event. It is the ratio of favorable outcomes to the total favorable outcomes. The probability formula can be expressed as,

probability formula says p of a equals number of favorable outcomes over total number of outcomes.

i.e., P(A) = n(A)/n(S)

  • P(A) is the probability of an event 'B'.
  • n(A) is the number of favorable outcomes of an event 'B'.
  • n(S) is the total number of events occurring in a sample space.

Different Probability Formulas

Probability formula with addition rule : Whenever an event is the union of two other events, say A and B, then P(A or B) = P(A) + P(B) - P(A∩B) P(A ∪ B) = P(A) + P(B) - P(A∩B)

Probability formula with the complementary rule: Whenever an event is the complement of another event, specifically, if A is an event, then P(not A) = 1 - P(A) or P(A') = 1 - P(A). P(A) + P(A′) = 1.

Probability formula with the conditional rule : When event A is already known to have occurred, the probability of event B is known as conditional probability and is given by: P(B∣A) = P(A∩B)/P(A)

Probability formula with multiplication rule : Whenever an event is the intersection of two other events, that is, events A and B need to occur simultaneously. Then

  • P(A ∩ B) = P(A)⋅P(B) (in case of independent events )
  • P(A∩B) = P(A)⋅P(B∣A) (in case of dependent events )

Calculating Probability

In an experiment, the probability of an event is the possibility of that event occurring. The probability of any event is a value between (and including) "0" and "1". Follow the steps below for calculating probability of an event A:

  • Step 1: Find the sample space of the experiment and count the elements. Denote it by n(S).
  • Step 2: Find the number of favorable outcomes and denote it by n(A).
  • Step 3: To find probability, divide n(A) by n(S). i.e., P(A) = n(A)/n(S).

Here are some examples that well describe the process of finding probability.

Example 1 : Find the probability of getting a number less than 5 when a dice is rolled by using the probability formula.

To find: Probability of getting a number less than 5 Given: Sample space, S = {1,2,3,4,5,6} Therefore, n(S) = 6

Let A be the event of getting a number less than 5. Then A = {1,2,3,4} So, n(A) = 4

Using the probability equation, P(A) = (n(A))/(n(s)) p(A) = 4/6 m = 2/3

Answer: The probability of getting a number less than 5 is 2/3.

Example 2: What is the probability of getting a sum of 9 when two dice are thrown?

There is a total of 36 possibilities when we throw two dice. To get the desired outcome i.e., 9, we can have the following favorable outcomes. (4,5),(5,4),(6,3)(3,6). There are 4 favorable outcomes. Probability of an event P(E) = (Number of favorable outcomes) ÷ (Total outcomes in a sample space) Probability of getting number 9 = 4 ÷ 36 = 1/9

Answer: Therefore the probability of getting a sum of 9 is 1/9.

Probability Tree Diagram

A tree diagram in probability is a visual representation that helps in finding the possible outcomes or the probability of any event occurring or not occurring. The tree diagram for the toss of a coin given below helps in understanding the possible outcomes when a coin is tossed. Each branch of the tree is associated with the respective probability (just like how 0.5 is written on each brack in the figure below). Remember that the sum of probabilities of all branches that start from the same point is always 1 (here, 0.5 + 0.5 = 1).

probability tree diagram when a coin is tossed has two outcomes head and tail each with probability of 0 point 5

Types of Probability

There can be different perspectives or types of probabilities based on the nature of the outcome or the approach followed while finding probability of an event happening. The four types of probabilities are,

Classical Probability

Empirical probability, subjective probability, axiomatic probability.

Classical probability, often referred to as the "priori" or "theoretical probability", states that in an experiment where there are B equally likely outcomes, and event X has exactly A of these outcomes, then the probability of X is A/B, or P(X) = A/B. For example, when a fair die is rolled, there are six possible outcomes that are equally likely. That means, there is a 1/6 probability of rolling each number on the die.

The empirical probability or the experimental perspective evaluates probability through thought experiments. For example, if a weighted die is rolled, such that we don't know which side has the weight, then we can get an idea for the probability of each outcome by rolling the die number of times and calculating the proportion of times the die gives that outcome and thus find the probability of that outcome.

Subjective probability considers an individual's own belief of an event occurring. For example, the probability of a particular team winning a football match on a fan's opinion is more dependent upon their own belief and feeling and not on a formal mathematical calculation.

In axiomatic probability, a set of rules or axioms by Kolmogorov are applied to all the types. The chances of occurrence or non-occurrence of any event can be quantified by the applications of these axioms, given as,

  • The smallest possible probability is zero, and the largest is one.
  • An event that is certain has a probability equal to one.
  • Any two mutually exclusive events cannot occur simultaneously, while the union of events says only one of them can occur.

Coin Toss Probability

Let us now look into the probability of tossing a coin . Quite often in games like cricket, for making a decision as to who would bowl or bat first, we sometimes use the tossing of a coin and decide based on the outcome of the toss. Let us check how we can use the concept of probability in the tossing of a single coin. Further, we shall also look into the tossing of two and three coins.

Tossing a Coin

A single coin on tossing has two outcomes, a head, and a tail. The concept of probability which is the ratio of favorable outcomes to the total number of outcomes can be used in finding probability of getting the head and the probability of getting a tail.

Total number of possible outcomes = 2; Sample Space = {H, T}; H: Head, T: Tail

  • P(H) = Number of heads/Total outcomes = 1/2
  • P(T)= Number of Tails/ Total outcomes = 1/2

Tossing Two Coins

In the process of tossing two coins, we have a total of four (= 2 2 ) outcomes. The probability formula can be used to find the probability of two heads, one head, no head, and a similar probability can be calculated for the number of tails. The probability calculations for the two heads are as follows.

Total number of outcomes = 4; Sample Space = {(H, H), (H, T), (T, H), (T, T)}

  • P(2H) = P(0 T) = Number of outcome with two heads/Total Outcomes = 1/4
  • P(1H) = P(1T) = Number of outcomes with only one head/Total Outcomes = 2/4 = 1/2
  • P(0H) = (2T) = Number of outcome with two heads/Total Outcomes = 1/4

Tossing Three Coins

The number of total outcomes on tossing three coins simultaneously is equal to 2 3 = 8. For these outcomes, we can find the probability of getting one head, two heads, three heads, and no head. A similar probability can also be calculated for the number of tails.

Total number of outcomes = 2 3 = 8 Sample Space = {(H, H, H), (H, H, T), (H, T, H), (T, H, H), (T, T, H), (T, H, T), (H, T, T), (T, T, T)}

  • P(0H) = P(3T) = Number of outcomes with no heads/Total Outcomes = 1/8
  • P(1H) = P(2T) = Number of Outcomes with one head/Total Outcomes = 3/8
  • P(2H) = P(1T) = Number of outcomes with two heads /Total Outcomes = 3/8
  • P(3H) = P(0T) = Number of outcomes with three heads/Total Outcomes = 1/8

Dice Roll Probability

Many games use dice to decide the moves of players across the games. A dice has six possible outcomes and the outcomes of a dice is a game of chance and can be obtained by using the concepts of probability. Some games also use two dice, and there are numerous probabilities that can be calculated for outcomes using two dice. Let us now check the outcomes, their probabilities for one dice and two dice respectively.

Rolling One Dice

The total number of outcomes on rolling a die is 6, and the sample space is {1, 2, 3, 4, 5, 6}. Here we shall compute the following few probabilities to help in better understanding the concept of probability on rolling one dice.

  • P( Even Number ) = Number of even number outcomes/Total Outcomes = 3/6 = 1/2
  • P( Odd Number ) = Number of odd number outcomes/Total Outcomes = 3/6 = 1/2
  • P( Prime Number ) = Number of prime number outcomes/Total Outcomes = 3/6 = 1/2

Rolling Two Dice

The total number of outcomes on rolling two dice is 6 2 = 36. The following image shows the sample space of 36 outcomes on rolling two dice.

Sample Space of Outcomes of Two Dice helps in finding Probabilities and it has 36 elements

Let us check a few probabilities of the outcomes from two dice. The probabilities are as follows.

  • Probability of getting a doublet(Same number) = 6/36 = 1/6
  • Probability of getting a number 3 on at least one dice = 11/36
  • Probability of getting a sum of 7 = 6/36 = 1/6

As we see, when we roll a single die, there are 6 possibilities. When we roll two dice, there are 36 (= 6 2 ) possibilities. When we roll 3 dice we get 216 (= 6 3 ) possibilities. So a general formula to represent the number of outcomes on rolling 'n' dice is 6 n .

Probability of Drawing Cards

A deck containing 52 cards is grouped into four suits of clubs, diamonds, hearts, and spades. Each of the clubs, diamonds, hearts, and spades have 13 cards each, which sum up to 52. Now let us discuss the probability of drawing cards from a pack. The symbols on the cards are shown below. Spades and clubs are black cards. Hearts and diamonds are red cards.

Sample Space of deck of cards contains 52 cards out of which 26 are black and 26 are red.

The 13 cards in each suit are ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, jack, queen, king. In these, the jack, the queen, and the king are called face cards. We can understand the card probability from the following examples.

  • The probability of drawing a black card is P(Black card) = 26/52 = 1/2
  • The probability of drawing a hearts card is P(Hearts) = 13/52 = 1/4
  • The probability of drawing a face card is P(Face card) = 12/52 = 3/13
  • The probability of drawing a card numbered 4 is P(4) = 4/52 = 1/13
  • The probability of drawing a red card numbered 4 is P(4 Red) = 2/52 = 1/26

Probability Theorems

The following theorems of probability are helpful to understand the applications of probability and also perform the numerous calculations involving probability.

Theorem 1: The sum of the probability of happening of an event and not happening of an event is equal to 1. P(A) + P(A') = 1.

Theorem 2: The probability of an impossible event or the probability of an event not happening is always equal to 0. P(ϕ) = 0.

Theorem 3: The probability of a sure event is always equal to 1. P(A) = 1

Theorem 4: The probability of happening of any event always lies between 0 and 1. 0 < P(A) < 1

Theorem 5: If there are two events A and B, we can apply the formula of the union of two sets and we can derive the formula for the probability of happening of event A or event B as follows.

P(A∪B) = P(A) + P(B) - P(A∩B)

Also for two mutually exclusive events A and B, we have P( A U B) = P(A) + P(B)

Bayes' Theorem on Conditional Probability

Bayes' theorem describes the probability of an event based on the condition of occurrence of other events. It is also called conditional probability . It helps in calculating the probability of happening of one event based on the condition of happening of another event.

For example, let us assume that there are three bags with each bag containing some blue, green, and yellow balls. What is the probability of picking a yellow ball from the third bag? Since there are blue and green colored balls also, we can arrive at the probability based on these conditions also. Such a probability is called conditional probability.

The formula for Bayes' theorem is \(\begin{align}P(A|B) = \dfrac{ P(B|A)·P(A)} {P(B)}\end{align}\)

where, \(\begin{align}P(A|B) \end{align}\) denotes how often event A happens on a condition that B happens.

where, \(\begin{align}P(B|A) \end{align}\) denotes how often event B happens on a condition that A happens.

\(\begin{align}P(A) \end{align}\) the likelihood of occurrence of event A.

\(\begin{align}P(B) \end{align}\) the likelihood of occurrence of event B.

Law of Total Probability

If there are n number of events in an experiment, then the sum of the probabilities of those n events is always equal to 1.

P(A 1 ) + P(A 2 ) + P(A 3 ) + … + P(A n ) = 1

Important Notes on Probability:

  • Probability is a measure of how likely an event is to happen.
  • Probability is represented as a fraction and always lies between 0 and 1.
  • An event can be defined as a subset of sample space.
  • The sample of throwing a coin is {head, tail} and the sample space of throwing dice is {1, 2, 3, 4, 5, 6}.
  • A random experiment cannot predict the exact outcomes but only some probable outcomes.

☛ Related Articles:

  • Event Probability Calculator
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Probability Examples

Example 1: What is the probability of getting a sum of 10 when two dice are thrown?

There are 36 possibilities when we throw two dice.

The desired outcome is 10. To get 10, we can have three favorable outcomes.

{(4,6),(6,4),(5,5)}

Probability of an event = number of favorable outcomes/ sample space

Probability of getting number 10 = 3/36 =1/12

Answer: Therefore the probability of getting a sum of 10 is 1/12.

Example 2: In a bag, there are 6 blue balls and 8 yellow balls. One ball is selected randomly from the bag. Find the probability of getting a blue ball.

Let us assume the probability of drawing a blue ball to be P(B)

Number of favorable outcomes to get a blue ball = 6

Total number of balls in the bag = 14

P(B) = Number of favorable outcomes/Total number of outcomes = 6/14 = 3/7

Answer: Therefore the probability of drawing a blue ball is 3/7.

Example 3: There are 5 cards numbered: 2, 3, 4, 5, 6. Find the probability of picking a prime number, and putting it back, you pick a composite number.

The two events are independent. Thus we use the product of the probability of the events.

P(getting a prime) = n(favorable events)/ n(sample space) = {2, 3, 5}/{2, 3, 4, 5, 6} = 3/5

p(getting a composite) = n(favorable events)/ n(sample space) = {4, 6}/{2, 3, 4, 5, 6}= 2/5

Thus the total probability of the two independent events = P(prime) × P(composite)

= 3/5 × (2/5)

Answer: Therefore the probability of picking a prime number and a prime number again is 6/25.

Example 4: Find the probability of getting a face card from a standard deck of cards using the probability equation.

Solution: To find: Probability of getting a face card Given: Total number of cards = 52 Number of face cards = Favorable outcomes = 12 Using the probability formula, Probability = (Favorable Outcomes)÷(Total Favourable Outcomes) P(face card) = 12/52 m = 3/13

Answer: The probability of getting a face card is 3/13

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Practice Questions on Probability

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FAQs on Probability

What is the meaning of probability in statistics.

Probability is a branch of math which deals with finding out the likelihood of the occurrence of an event. Probability measures the chance of an event happening and is equal to the number of favorable events divided by the total number of events. The value of probability ranges between 0 and 1, where 0 denotes uncertainty and 1 denotes certainty.

How to Find Probability?

The probability can be found by first knowing the sample space of the outcomes of an experiment. A probability is generally calculated for an event (x) within the sample space. The probability of an event happening is obtained by dividing the number of outcomes of an event by the total number of possible outcomes or sample space.

What are the Three Types of Probability?

The three types of probabilities are theoretical probability, experimental probability, and axiomatic probability. The theoretical probability calculates the probability based on formulas and input values. The experimental probability gives a realistic value and is based on the experimental values for calculation. Quite often the theoretical and experimental probability differ in their results. And the axiomatic probability is based on the axioms which govern the concepts of probability.

How To Calculate Probability?

The probability of any event depends upon the number of favorable outcomes and the total outcomes. Finding probability is finding the ratio of the number of favorable outcomes to the total outcomes in that sample space. It is expressed as, Probability of an event P(E) = (Number of favorable outcomes) ÷ (Number of Elements in Sample space).

What is Conditional Probability?

The conditional probability predicts the happening of one event based on the happening of another event. If there are two events A and B, conditional probability is a chance of occurrence of event B provided the event A has already occurred. The formula for the conditional probability of happening of event B, given that event A, has happened is P(B/A) = P(A ∩ B)/P(A).

What is Experimental Probability?

The experimental probability is based on the results and the values obtained from the probability experiments. Experimental probability is defined as the ratio of the total number of times an event has occurred to the total number of trials conducted. The results of the experimental probability are based on real-life instances and may differ in values from theoretical probability.

What is a Probability Distribution?

The two important probability distributions are binomial distribution and Poisson distribution. The binomial distribution is defined for events with two probability outcomes and for events with a multiple number of times of such events. The Poisson distribution is based on the numerous probability outcomes in a limited space of time, distance, sample space. An example of the binomial distribution is the tossing of a coin with two outcomes, and for conducting such a tossing experiment with n number of coins. A Poisson distribution is for events such as antigen detection in a plasma sample, where the probabilities are numerous.

How are Probability and Statistics Related?

The probability calculates the happening of an experiment and it calculates the happening of a particular event with respect to the entire set of events. For simple events of a few numbers of events, it is easy to calculate the probability. But for calculating probabilities involving numerous events and to manage huge data relating to those events we need the help of statistics . Statistics helps in rightly analyzing

How Probability is Used in Real Life?

Probability has huge applications in games and analysis. Also in real life and industry areas where it is about prediction we make use of probability. The prediction of the price of a stock, or the performance of a team in cricket requires the use of probability concepts. Further, the new technology field of artificial intelligence is extensively based on probability.

Where Do We Use the Probability Formula In Our Real Life?

The following activities in our real-life tend to follow the probability equation:

  • Weather forecasting
  • Playing cards
  • Voting strategy in politics
  • Rolling a dice.
  • Pulling out the exact matching socks of the same color
  • Chances of winning or losing in any sports.

How was Probability Discovered?

The use of the word "probable" started first in the seventeenth century when it was referred to actions or opinions which were held by sensible people. Further, the word probable in the legal content was referred to a proposition that had tangible proof. The field of permutations and combinations, statistical inference, cryptoanalysis, frequency analysis have altogether contributed to this current field of probability.

What is the Conditional Probability Formula?

The conditional probability depends upon the happening of one event based on the happening of another event. The conditional probability formula of happening of event B, given that event A, has already happened is expressed as P(B/A) = P(A ∩ B)/P(A).

Test Prep Review

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Probability Practice Problems

1. on a six-sided die, each side has a number between 1 and 6. what is the probability of throwing a 3 or a 4, 2. three coins are tossed up in the air, one at a time. what is the probability that two of them will land heads up and one will land tails up, 3. a two-digit number is chosen at random. what is the probability that the chosen number is a multiple of 7, 4. a bag contains 14 blue, 6 red, 12 green, and 8 purple buttons. 25 buttons are removed from the bag randomly. how many of the removed buttons were red if the chance of drawing a red button from the bag is now 1/3, 5. there are 6 blue marbles, 3 red marbles, and 5 yellow marbles in a bag. what is the probability of selecting a blue or red marble on the first draw, 6. using a six-sided die, carlin has rolled a six on each of 4 successive tosses. what is the probability of carlin rolling a six on the next toss, 7. a regular deck of cards has 52 cards. assuming that you do not replace the card you had drawn before the next draw, what is the probability of drawing three aces in a row.

  • 1 in 132600

8. An MP3 player is set to play songs at random from the fifteen songs it contains in memory. Any song can be played at any time, even if it is repeated. There are 5 songs by Band A, 3 songs by Band B, 2 by Band C, and 5 by Band D. If the player has just played two songs in a row by Band D, what is the probability that the next song will also be by Band D?

  • Not enough data to determine.

9. Referring again to the MP3 player described in Question 8, what is the probability that the next two songs will both be by Band B?

10. if a bag of balloons consists of 47 white balloons, 5 yellow balloons, and 10 black balloons, what is the approximate likelihood that a balloon chosen randomly from the bag will be black, 11. in a lottery game, there are 2 winners for every 100 tickets sold on average. if a man buys 10 tickets, what is the probability that he is a winner, answers and explanations.

1.  B:  On a six-sided die, the probability of throwing any number is 1 in 6. The probability of throwing a 3 or a 4 is double that, or 2 in 6. This can be simplified by dividing both 2 and 6 by 2.

Therefore, the probability of throwing either a 3 or 4 is 1 in 3.

2.  D:  Shown below is the sample space of possible outcomes for tossing three coins, one at a time. Since there is a possibility of two outcomes (heads or tails) for each coin, there is a total of 2*2*2=8 possible outcomes for the three coins altogether. Note that H represents heads and T represents tails:

HHH HHT HTT HTH TTT TTH THT THH

Notice that out of the 8 possible outcomes, only 3 of them (HHT, HTH, and THH) meet the desired condition that two coins land heads up and one coin lands tails up. Probability, by definition, is the number of desired outcomes divided by the number of possible outcomes. Therefore, the probability of two heads and one tail is 3/8, Choice D.

3.  E:  There are 90 two-digit numbers (all integers from 10 to 99). Of those, there are 13 multiples of 7: 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98.

4.  B:  Add the 14 blue, 6 red, 12 green, and 8 purple buttons to get a total of 40 buttons. If 25 buttons are removed, there are 15 buttons remaining in the bag. If the chance of drawing a red button is now 1/3, then 5 of the 15 buttons remaining must be red. The original total of red buttons was 6. So, one red button was removed.

5.  D:  Use this ratio for probability:

Probability = Number of Desired Outcomes

Number of Possible Outcomes

There are 6 blue marbles and 3 red marbles for a total of 9 desired outcomes. Add the total number of marbles to get the total number of possible outcomes, 14. The probability that a red or blue marble will be selected is 9/14.

6.  C:  The outcomes of previous rolls do not affect the outcomes of future rolls. There is one desired outcome and six possible outcomes. The probability of rolling a six on the fifth roll is 1/6, the same as the probability of rolling a six on any given individual roll.

7.  D:  The probability of getting three aces in a row is the product of the probabilities for each draw. For the first ace, that is 4 in 52 or 1 in 13; for the second, it is 3 in 51 or 1 in 27; and for the third, it is 2 in 50 or 1 in 25. So the overall probability,  P , is P=1/13*1/17*1/25=1/5,525

8.  B:  The probability of playing a song by a particular band is proportional to the number of songs by that band divided by the total number of songs, or 5/15=1/3 for B and D. The probability of playing any particular song is not affected by what has been played previously, since the choice is random and songs may be repeated.

9.  A:  Since 3 of the 15 songs are by Band B, the probability that any one song will be by that band is 3/15=1/5. The probability that the next two songs are by Band B is equal to the product of two probabilities, where each probability is that the next song is by Band B: 1/5*1/5=1/25 The same probability of 1/5 may be multiplied twice because whether or not the first song is by Band B has no impact on whether the second song is by Band B. They are independent events.

10.  B:  First, calculate the total number of balloons in the bag: 47 + 5 + 10 = 62.

Ten of these are black, so divide this number by 62. Then, multiply by 100 to express the probability as a percentage:

10 / 62 = 0.16

0.16 100 = 16%

11. C: First, simplify the winning rate. If there are 2 winners for every 100 tickets, there is 1 winner for every 50 tickets sold. This can be expressed as a probability of 1/50 or 0.02. In order to account for the (unlikely) scenarios of more than a single winning ticket, calculate the probability that none of the tickets win and then subtract that from 1. There is a probability of 49/50 that a given ticket will not win. For all ten to lose that would be (49/50)^(10) ≈ 0.817. Therefore, the probability that at least one ticket wins is 1 − 0.817 = 0.183 or about 18.3%

  • Math Article

Probability Questions

The probability questions , with answers, are provided here for students to make them understand the concept in an easy way. The chapter Probability has been included in Class 9, 10, 11 and 12. Therefore, it is a very important chapter. The questions here will be provided, as per NCERT guidelines. Get Probability For Class 10 at BYJU’S.

The application of probability can be seen in Maths as well as in day to day life. It is necessary to learn the basics of this concept. The questions here will cover the basics as well as the hard level problems for all levels of students. Thus, students will be confident in solving problems based on it. Also, solving these probability problems will help them to participate in competitive exams, going further.

Definition: Probability is nothing but the possibility of an event occurring. For example, when a test is conducted, then the student can either get a pass or fail. It is a state of probability.

Also read: Probability

The probability of happening of an event E is a number P(E) such that:

0 ≤ P(E) ≤ 1

Probability Formula: If an event E occurs, then the empirical probability of an event to happen is:

P(E) = Number of trials in which Event happened/Total number of trials

The theoretical probability of an event E, P(E), is defined as:

P(E) = (Number of outcomes favourable to E)/(Number of all possible outcomes of the experiment)

Impossible event: The probability of an occurrence/event impossible to happen is 0. Such an event is called an impossible event.

Sure event: The probability of an event that is sure to occur is 1. Such an event is known as a sure event or a certain event.

Probability Questions & Answers

1. Two coins are tossed 500 times, and we get:

Two heads: 105 times

One head: 275 times

No head: 120 times

Find the probability of each event to occur.

Solution: Let us say the events of getting two heads, one head and no head by E 1 , E 2 and E 3 , respectively.

P(E 1 ) = 105/500 = 0.21

P(E 2 ) = 275/500 = 0.55

P(E 3 ) = 120/500 = 0.24

The Sum of probabilities of all elementary events of a random experiment is 1.

P(E 1 )+P(E 2 )+P(E 3 ) = 0.21+0.55+0.24 = 1

2. A tyre manufacturing company kept a record of the distance covered before a tyre needed to be replaced. The table shows the results of 1000 cases.

If a tyre is bought from this company, what is the probability that :

(i) it has to be substituted before 4000 km is covered?

(ii) it will last more than 9000 km?

(iii) it has to be replaced after 4000 km and 14000 km is covered by it?

Solution: (i) Total number of trials = 1000.

The frequency of a tyre required to be replaced before covering 4000 km = 20

So, P(E 1 ) = 20/1000 = 0.02

(ii) The frequency that tyre will last more than 9000 km = 325 + 445 = 770

So, P(E 2 ) = 770/1000 = 0.77

(iii) The frequency that tyre requires replacement between 4000 km and 14000 km = 210 + 325 = 535.

So, P(E 3 ) = 535/1000 = 0.535

3. The percentage of marks obtained by a student in the monthly tests are given below:

Based on the above table, find the probability of students getting more than 70% marks in a test.

Solution: The total number of tests conducted is 5.

The number of tests when students obtained more than 70% marks = 3.

So, P(scoring more than 70% marks) = ⅗ = 0.6

4. One card is drawn from a deck of 52 cards, well-shuffled. Calculate the probability that the card will

(i) be an ace,

(ii) not be an ace.

Solution: Well-shuffling ensures equally likely outcomes.

(i) There are 4 aces in a deck.

Let E be the event the card drawn is ace.

The number of favourable outcomes to the event E = 4

The number of possible outcomes = 52

Therefore, P(E) = 4/52 = 1/13

(ii) Let F is the event of ‘card is not an ace’

The number of favourable outcomes to F = 52 – 4 = 48

Therefore, P(F) = 48/52 = 12/13

5. Two players, Sangeet and Rashmi, play a tennis match. The probability of Sangeet winning the match is 0.62. What is the probability that Rashmi will win the match?

Solution: Let S and R denote the events that Sangeeta wins the match and Reshma wins the match, respectively.

The probability of Sangeet to win = P(S) = 0.62

The probability of Rashmi to win = P(R) = 1 – P(S)

= 1 – 0.62 = 0.38

6. Two coins (a one rupee coin and a two rupee coin) are tossed once. Find a sample space.

Solution: Either Head(H) or Tail(T) can be the outcomes.

Heads on both coins = (H,H) = HH

Head on 1st coin and Tail on the 2nd coin = (H,T) = HT

Tail on 1st coin and Head on the 2nd coin = (T,H) = TH

Tail on both coins = (T,T) = TT

Therefore, the sample space is S = {HH, HT, TH, TT}

7. Consider the experiment in which a coin is tossed repeatedly until a head comes up. Describe the sample space.

Solution: In the random experiment where the head can appear on the 1st toss, or the 2nd toss, or the 3rd toss and so on till we get the head of the coin. Hence, the required sample space is :

S= {H, TH, TTH, TTTH, TTTTH,…}

8. Consider the experiment of rolling a die. Let A be the event ‘getting a prime number’, B be the event ‘getting an odd number’. Write the sets representing the events

(ii) A and B

(iii) A but not B

(iv) ‘not A’.

Solution: S = {1, 2, 3, 4, 5, 6}, A = {2, 3, 5} and B = {1, 3, 5}

(i) A or B = A ∪ B = {1, 2, 3, 5}

(ii) A and B = A ∩ B = {3,5}

(iii) A but not B = A – B = {2}

(iv) not A = A′ = {1,4,6}

9. A coin is tossed three times, consider the following events.

P: ‘No head appears’,

Q: ‘Exactly one head appears’ and

R: ‘At Least two heads appear’.

Check whether they form a set of mutually exclusive and exhaustive events.

Solution: The sample space of the experiment is:

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} and

Q = {HTT, THT, TTH},

R = {HHT, HTH, THH, HHH}

P ∪ Q ∪ R = {TTT, HTT, THT, TTH, HHT, HTH, THH, HHH} = S

Therefore, P, Q and R are exhaustive events.

P ∩ R = φ and

Therefore, the events are mutually exclusive.

Hence, P, Q and R form a set of mutually exclusive and exhaustive events.

10. If P(A) = 7/13, P(B) = 9/13 and P(A∩B) = 4/13, evaluate P(A|B).

Solution: P(A|B) = P(A∩B)/P(B) = (4/13)/(9/13) = 4/9.

Video Lesson

Probability important topics.

example problems for probability

Probability Important Questions

example problems for probability

Related Links

  • Important Questions Class 9 Maths Chapter 15 Probability
  • Important Questions Class 10 Maths Chapter 15 Probability
  • Important Questions Class 11 Maths Chapter 16 Probability
  • Important Questions Class 12 Maths Chapter 13 Probability

Practice Questions

Solve the following probability questions.

  • Write the sample space for rolling two dice.
  • If two coins are tossed simultaneously, what is the probability of getting exactly two heads?
  • From a well-shuffled deck of 52 cards, what is the probability of getting a king?
  • In a bag, there are 5 red balls and 7 black balls. What is the probability of getting a black ball?
  • If the probability of an event happening is 0.7, then what is the probability of an event that will not happen?

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  • Probability Distribution | Formula, Types, & Examples

Probability Distribution | Formula, Types, & Examples

Published on June 9, 2022 by Shaun Turney . Revised on June 21, 2023.

A probability distribution is a mathematical function that describes the probability of different possible values of a variable . Probability distributions are often depicted using graphs or probability tables.

Common probability distributions include the binomial distribution, Poisson distribution , and uniform distribution. Certain types of probability distributions are used in hypothesis testing, including the standard normal distribution , the  F distribution, and Student’s t distribution .

Table of contents

What is a probability distribution, discrete probability distributions, continuous probability distributions, how to find the expected value and standard deviation, how to test hypotheses using null distributions, probability distribution formulas, other interesting articles, frequently asked questions about probability distributions.

A probability distribution is an idealized frequency distribution .

A frequency distribution describes a specific sample or dataset. It’s the number of times each possible value of a variable occurs in the dataset.

The number of times a value occurs in a sample is determined by its probability of occurrence. Probability is a number between 0 and 1 that says how likely something is to occur:

  • 0 means it’s impossible.
  • 1 means it’s certain.

The higher the probability of a value, the higher its frequency in a sample.

More specifically, the probability of a value is its relative frequency in an infinitely large sample.

Infinitely large samples are impossible in real life, so probability distributions are theoretical. They’re idealized versions of frequency distributions that aim to describe the population the sample was drawn from.

Probability distributions are used to describe the populations of real-life variables, like coin tosses or the weight of chicken eggs. They’re also used in hypothesis testing to determine p values .

The farmer weighs 100 random eggs and describes their frequency distribution using a histogram:

frequency_distribution_example_egg_weight

She can get a rough idea of the probability of different egg sizes directly from this frequency distribution. For example, she can see that there’s a high probability of an egg being around 1.9 oz., and there’s a low probability of an egg being bigger than 2.1 oz.

Suppose the farmer wants more precise probability estimates. One option is to improve her estimates by weighing many more eggs.

A better option is to recognize that egg size appears to follow a common probability distribution called a normal distribution . The farmer can make an idealized version of the egg weight distribution by assuming the weights are normally distributed:

normal_distribution_example_egg_weight

Variables that follow a probability distribution are called random variables . There’s special notation you can use to say that a random variable follows a specific distribution:

  • Random variables are usually denoted by X.
  • The ~ (tilde) symbol means “follows the distribution.”
  • The distribution is denoted by a capital letter (usually the first letter of the distribution’s name), followed by brackets that contain the distribution’s parameters .

For example, the following notation means “the random variable X follows a normal distribution with a mean of µ and a variance of σ 2 .”

X \sim N(\mu,\sigma^2)

There are two types of probability distributions:

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A discrete probability distribution is a probability distribution of a categorical or discrete variable .

Discrete probability distributions only include the probabilities of values that are possible. In other words, a discrete probability distribution doesn’t include any values with a probability of zero. For example, a probability distribution of dice rolls doesn’t include 2.5 since it’s not a possible outcome of dice rolls.

The probability of all possible values in a discrete probability distribution add up to one. It’s certain (i.e., a probability of one) that an observation will have one of the possible values.

Probability tables

A probability table represents the discrete probability distribution of a categorical variable . Probability tables can also represent a discrete variable with only a few possible values or a continuous variable that’s been grouped into class intervals .

A probability table is composed of two columns:

  • The values or class intervals
  • Their probabilities

Probability mass functions

A probability mass function (PMF) is a mathematical function that describes a discrete probability distribution. It gives the probability of every possible value of a variable.

A probability mass function can be represented as an equation or as a graph.

The probability mass function of the distribution is given by the formula:

P(X = k) = \dfrac{e^{-\lambda} \lambda^k}{k!}

This probability mass function can also be represented as a graph:

probability_example_egg_weight

Notice that the variable can only have certain values, which are represented by closed circles. You can have two sweaters or 10 sweaters, but you can’t have 3.8 sweaters.

Common discrete probability distributions

A continuous probability distribution is the probability distribution of a continuous variable .

A continuous variable can have any value between its lowest and highest values. Therefore, continuous probability distributions include every number in the variable’s range .

The probability that a continuous variable will have any specific value is so infinitesimally small that it’s considered to have a probability of zero. However, the probability that a value will fall within a certain interval of values within its range is greater than zero.

Probability density functions

A probability density function (PDF) is a mathematical function that describes a continuous probability distribution. It provides the probability density of each value of a variable, which can be greater than one.

A probability density function can be represented as an equation or as a graph.

In graph form, a probability density function is a curve. You can determine the probability that a value will fall within a certain interval by calculating the area under the curve within that interval. You can use reference tables or software to calculate the area.

The area under the whole curve is always exactly one because it’s certain (i.e., a probability of one) that an observation will fall somewhere in the variable’s range.

A cumulative distribution function is another type of function that describes a continuous probability distribution.

f(x) = \dfrac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\dfrac{x-\mu}{\sigma}\right)^2}

The probability of an egg being exactly 2 oz. is zero. Although an egg can weigh very close to 2 oz., it is extremely improbable that it will weigh exactly 2 oz. Even if a regular scale measured an egg’s weight as being 2 oz., an infinitely precise scale would find a tiny difference between the egg’s weight and 2 oz.

The probability that an egg is within a certain weight interval, such as 1.98 and 2.04 oz., is greater than zero and can be represented in the graph of the probability density function as a shaded region:

shaded_area_example

Common continuous probability distributions

You can find the expected value and standard deviation of a probability distribution if you have a formula, sample, or probability table of the distribution.

The expected value is another name for the mean of a distribution. It’s often written as E ( x ) or µ. If you take a random sample of the distribution, you should expect the mean of the sample to be approximately equal to the expected value.

  • If you have a formula describing the distribution, such as a probability density function, the expected value is usually given by the µ parameter. If there’s no µ parameter, the expected value can be calculated from the other parameters using equations that are specific to each distribution.
  • If you have a sample , then the mean of the sample is an estimate of the expected value of the population’s probability distribution. The larger the sample size, the better the estimate will be.
  • If you have a probability table , you can calculate the expected value by multiplying each possible outcome by its probability, and then summing these values.

What is the expected value of robin eggs per nest?

Multiply each possible outcome by its probability:

Sum the values:

E ( x ) = 0.4 + 1.5 + 1.2

The standard deviation of a distribution is a measure of its variability. It’s often written as σ.

  • If you have a formula describing the distribution, such as a probability density function, the standard deviation is sometimes given by the σ parameter. If there’s no σ parameter, the standard deviation can often be calculated from other parameters using formulas that are specific to each distribution.
  • If you have a sample , the standard deviation of the sample is an estimate of the standard deviation of the population’s probability distribution. The larger the sample size, the better the estimate will be.
  • If you have a probability table , you can calculate the standard deviation by calculating the deviation between each value and the expected value, squaring it, multiplying it by its probability, and then summing the values and taking the square root.

Square the values and multiply them by their probability:

Sum the values and take the square root:

σ = √ (0.242 + 0.005 + 0.243)

σ = √ (0.49)

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Null distributions are an important tool in hypothesis testing . A null distribution is the probability distribution of a test statistic when the null hypothesis of the test is true.

All hypothesis tests involve a test statistic . Some common examples are z , t , F , and chi-square. A test statistic summarizes the sample in a single number, which you then compare to the null distribution to calculate a p value .

The p value is the probability of obtaining a value equal to or more extreme than the sample’s test statistic, assuming that the null hypothesis is true. In practical terms, it’s the area under the null distribution’s probability density function curve that’s equal to or more extreme than the sample’s test statistic.

t_distribution_example_egg_weight

If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.

  • Student’s t table
  • Student’s t distribution
  • Descriptive statistics
  • Measures of central tendency
  • Correlation coefficient

Methodology

  • Cluster sampling
  • Stratified sampling
  • Types of interviews
  • Cohort study
  • Thematic analysis

Research bias

  • Implicit bias
  • Cognitive bias
  • Survivorship bias
  • Availability heuristic
  • Nonresponse bias
  • Regression to the mean

Probability is the relative frequency over an infinite number of trials.

For example, the probability of a coin landing on heads is .5, meaning that if you flip the coin an infinite number of times, it will land on heads half the time.

Since doing something an infinite number of times is impossible, relative frequency is often used as an estimate of probability. If you flip a coin 1000 times and get 507 heads, the relative frequency, .507, is a good estimate of the probability.

In a normal distribution , data are symmetrically distributed with no skew. Most values cluster around a central region, with values tapering off as they go further away from the center.

The measures of central tendency (mean, mode, and median) are exactly the same in a normal distribution.

Normal distribution

Probability distributions belong to two broad categories: discrete probability distributions and continuous probability distributions . Within each category, there are many types of probability distributions.

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Mathematics LibreTexts

5.3: Probability Rules- “And” and “Or”

  • Last updated
  • Save as PDF
  • Page ID 91507

  • Maxie Inigo, Jennifer Jameson, Kathryn Kozak, Maya Lanzetta, & Kim Sonier
  • Coconino Community College

Learning Objectives

Students will be able to:

  • Determine if two events are mutually exclusive and/or independent.
  • Apply the "Or" rule to calculate the probability that either of two events occurs.
  • Apply the "And" rule to calculate the probability that both of two events occurs.

Many probabilities in real life involve more than one event. If we draw a single card from a deck we might want to know the probability that it is either red or a jack. If we look at a group of students, we might want to know the probability that a single student has brown hair and blue eyes. When we combine two events we make a single event called a compound event . To create a compound event, we can use the word “and” or the word “or” to combine events. It is very important in probability to pay attention to the words “and” and “or” if they appear in a problem. The word “and” restricts the field of possible outcomes to only those outcomes that simultaneously describe all events. The word “or” broadens the field of possible outcomes to those that describe one or more events.

Example \(\PageIndex{1}\): Counting Students

Suppose a teacher wants to know the probability that a single student in her class of 30 students is taking either Art or English. She asks the class to raise their hands if they are taking Art and counts 13 hands. Then she asks the class to raise their hands if they are taking English and counts 21 hands. The teacher then calculates

\[P(\text{Art or English}) = \dfrac{13+21}{30} = \dfrac{33}{30} \nonumber\]

The teacher knows that this is wrong because probabilities must be between zero and one, inclusive. After thinking about it she remembers that nine students are taking both Art and English. These students raised their hands each time she counted, so the teacher counted them twice. When we calculate probabilities we have to be careful to count each outcome only once.

http://media.townhall.com/townhall/reu/ha/2013/190/b00cc532-24d8-4028-9beb-877e2c63baf7.jpg

Mutually Exclusive Events

An experiment consists of drawing one card from a well shuffled deck of 52 cards. Consider the events E : the card is red, F : the card is a five, and G : the card is a spade. It is possible for a card to be both red and a five at the same time but it is not possible for a card to be both red and a spade at the same time. It would be easy to accidentally count a red five twice by mistake. It is not possible to count a red spade twice.

Definition: Mutually Exclusive

Two events are mutually exclusive if they have no outcomes in common.

Example \(\PageIndex{2}\): Mutually Exclusive with Dice

Two fair dice are tossed and different events are recorded. Let the events E , F and G be as follows:

  • E = {the sum is five} = {(1, 4), (2, 3), (3, 2), (4, 1)}
  • F = {both numbers are even} = {(2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6)}
  • G = {both numbers are less than five} = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 3), (3, 4), (4,1), (4, 2), (4, 3), (4,4)}
  • Are events E and F mutually exclusive?

Yes. E and F are mutually exclusive because they have no outcomes in common. It is not possible to add two even numbers to get a sum of five.

  • Are events E and G mutually exclusive?

No. E and G are not mutually exclusive because they have some outcomes in common. The pairs (1, 4), (2, 3), (3, 2) and (4, 1) all have sums of 5 and both numbers are less than five.

  • Are events F and G mutually exclusive?

No. F and G are not mutually exclusive because they have some outcomes in common. The pairs (2, 2), (2, 4), (4, 2) and (4, 4) all have two even numbers that are less than five.

Addition Rule for “Or” Probabilities

The addition rule for probabilities is used when the events are connected by the word “or”. Remember our teacher in Example \(\PageIndex{1}\) at the beginning of the section? She wanted to know the probability that her students were taking either art or English. Her problem was that she counted some students twice. She needed to add the number of students taking art to the number of students taking English and then subtract the number of students she counted twice. After dividing the result by the total number of students she will find the desired probability. The calculation is as follows:

\[ \begin{align*} P(\text{art or English}) &= \dfrac{\# \text{ taking art + } \# \text{ taking English - } \# \text{ taking both}}{\text{total number of students}} \\[4pt] &= \dfrac{13+21-9}{30} \\[4pt] &= \dfrac{25}{30} \approx {0.833} \end{align*}\]

The probability that a student is taking art or English is 0.833 or 83.3%.

When we calculate the probability for compound events connected by the word “or” we need to be careful not to count the same thing twice. If we want the probability of drawing a red card or a five we cannot count the red fives twice. If we want the probability a person is blonde-haired or blue-eyed we cannot count the blue-eyed blondes twice. The addition rule for probabilities adds the number of blonde-haired people to the number of blue-eyed people then subtracts the number of people we counted twice.

If A and B are any events then

\[P(A\, \text{or}\, B) = P(A) + P(B) – P(A \,\text{and}\, B).\]

If A and B are mutually exclusive events then \(P(A \,\text{and}\, B) = 0\), so then

\[P(A \, \text{or}\, B) = P(A) + P(B).\]

Example \(\PageIndex{3}\): Additional Rule for Drawing Cards

A single card is drawn from a well shuffled deck of 52 cards. Find the probability that the card is a club or a face card.

There are 13 cards that are clubs, 12 face cards (J, Q, K in each suit) and 3 face cards that are clubs.

\[ \begin{align*} P(\text{club or face card}) &= P(\text{club}) + P(\text{face card}) - P(\text{club and face card}) \\[4pt] &= \dfrac{13}{52} + \dfrac{12}{52} - \dfrac{3}{52} \\[4pt] &= \dfrac{22}{52} = \dfrac{11}{26} \approx {0.423} \end{align*}\]

The probability that the card is a club or a face card is approximately 0.423 or 42.3%.

A simple way to check this answer is to take the 52 card deck and count the number of physical cards that are either clubs or face cards. If you were to set aside all of the clubs and face cards in the deck, you would end up with the following:

{2 Clubs, 3 Clubs, 4 Clubs, 5 Clubs, 6 Clubs, 7 Clubs, 8 Clubs, 9 Clubs, 10 Clubs, J Clubs, Q Clubs, K Clubs, A Clubs, J Hearts, Q Hearts, K Hearts, J Spades, Q Spades, K Spades, J Diamonds, Q Diamonds, K Diamonds}

That is 22 cards out of the 52 card deck, which gives us a probably of: \[ \begin{align*} \dfrac{22}{52} = \dfrac{11}{26} \approx {0.423} \end{align*}\]

This confirms our earlier answer using the formal Addition Rule.

Example \(\PageIndex{4}\): Addition Rule for Tossing a Coin and Rolling a Die

An experiment consists of tossing a coin then rolling a die. Find the probability that the coin lands heads up or the number is five.

Let H represent heads up and T represent tails up. The sample space for this experiment is S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}.

  • There are six ways the coin can land heads up, {H1, H2, H3, H4, H5, H6}.
  • There are two ways the die can land on five, {H5, T5}.
  • There is one way for the coin to land heads up and the die to land on five, {H5}.

\[ \begin{align*} P(\text{heads or five}) &= P(\text{heads}) + P(\text{five}) - P(\text{both heads and five}) \\[4pt] &= \dfrac{6}{12} + \dfrac{2}{12} - \dfrac{1}{12} \\[4pt] &= \dfrac{7}{12} = \approx {0.583} \end{align*}\]

The probability that the coin lands heads up or the number is five is approximately 0.583 or 58.3%.

Example \(\PageIndex{5}\): Addition Rule for Satisfaction of Car Buyers

Two hundred fifty people who recently purchased a car were questioned and the results are summarized in the following table.

Find the probability that a person bought a new car or was not satisfied.

\[\begin{align*} P(\text{new car or not satisfied}) &= P(\text{new car}) + P(\text{not satisfied}) - P(\text{new car and not satisfied}) \\[4pt] &= \dfrac{120}{250} + \dfrac{75}{250} - \dfrac{28}{250} = \dfrac{167}{250} \approx 0.668 \end{align*}\]

The probability that a person bought a new car or was not satisfied is approximately 0.668 or 66.8%.

Independent Events

Sometimes we need to calculate probabilities for compound events that are connected by the word “and.” Tossing a coin multiple times or rolling dice are independent events. Each time you toss a fair coin the probability of getting heads is ½. It does not matter what happened the last time you tossed the coin. It’s similar for dice. If you rolled double sixes last time that does not change the probability that you will roll double sixes this time. Drawing two cards without replacement is not an independent event. When you draw the first card and set it aside, the probability for the second card is now out of 51 cards not 52 cards.

Definition: Independent Events

Two events are independent events if the occurrence of one event has no effect on the probability of the occurrence of the other event.

Example \(\PageIndex{6}\): Determining When Events are Independent

Are these events independent?

a) A fair coin is tossed two times. The two events are (1) first toss is a head and (2) second toss is a head.

b) The two events (1) “It will rain tomorrow in Houston” and (2) “It will rain tomorrow in Galveston” (a city near Houston).

c) You draw a card from a deck, then draw a second card without replacing the first.

a) The probability that a head comes up on the second toss is \(\frac{1}{2}\) regardless of whether or not a head came up on the first toss, so these events are independent .

b) These events are not independent because it is more likely that it will rain in Galveston on days it rains in Houston than on days it does not.

c) The probability of the second card being red depends on whether the first card is red or not, so these events are not independent .

Multiplication Rule for “And” Probabilities: Independent Events

If events A and B are independent events, then \( P(\text{A and B}) = P(A) \cdot P(B)\).

Example \(\PageIndex{7}\): Independent Events for Tossing Coins

Suppose a fair coin is tossed four times. What is the probability that all four tosses land heads up?

The tosses of the coins are independent events. Knowing a head was tossed on the first trial does not change the probability of tossing a head on the second trial.

\(P(\text{four heads in a row}) = P(\text{1st heads and 2nd heads and 3rd heads and 4th heads})\)

\( = P(\text{1st heads}) \cdot P(\text{2nd heads}) \cdot P(\text{3rd heads}) \cdot P(\text{4th heads})\)

\( = \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2} \cdot \dfrac{1}{2}\)

\( = \dfrac{1}{16}\)

The probability that all four tosses land heads up is \(\dfrac{1}{16}\).

Example \(\PageIndex{8}\): Independent Events for Drawing Marbles

A bag contains five red and four white marbles. A marble is drawn from the bag, its color recorded and the marble is returned to the bag. A second marble is then drawn. What is the probability that the first marble is red and the second marble is white?

Since the first marble is put back in the bag before the second marble is drawn these are independent events.

\[\begin{align*} P(\text{1st red and 2nd white}) &= P(\text{1st red}) \cdot P(\text{2nd white}) \\[4pt] &= \dfrac{5}{9} \cdot \dfrac{4}{9} = \dfrac{20}{81}\end{align*}\]

The probability that the first marble is red and the second marble is white is \(\dfrac{20}{81}\).

Example \(\PageIndex{9}\): Independent Events for Faulty Alarm Clocks

Abby has an important meeting in the morning. She sets three battery-powered alarm clocks just to be safe. If each alarm clock has a 0.03 probability of malfunctioning, what is the probability that all three alarm clocks fail at the same time?

Since the clocks are battery powered we can assume that one failing will have no effect on the operation of the other two clocks. The functioning of the clocks is independent.

\[\begin{align*} P(\text{all three fail}) &= P(\text{first fails}) \cdot P(\text{second fails})\cdot P(\text{third fails}) \\[4pt] &= (0.03)(0.03)(0.03) \\[4pt] &= 2.7 \times 10^{-5} \end{align*}\]

The probability that all three clocks will fail is approximately 0.000027 or 0.0027%. It is very unlikely that all three alarm clocks will fail.

At Least Once Rule for Independent Events

Many times we need to calculate the probability that an event will happen at least once in many trials. The calculation can get quite complicated if there are more than a couple of trials. Using the complement to calculate the probability can simplify the problem considerably. The following example will help you understand the formula.

Example \(\PageIndex{10}\): At Least Once Rule

The probability that a child forgets her homework on a given day is 0.15. What is the probability that she will forget her homework at least once in the next five days?

Assume that whether she forgets or not one day has no effect on whether she forgets or not the second day.

If P (forgets) = 0.15, then P (not forgets) = 0.85.

\[\begin{align*} P(\text{forgets at least once in 5 tries}) &= P(\text{forgets 1, 2, 3, 4 or 5 times in 5 tries}) \\[4pt] & = 1 - P(\text{forgets 0 times in 5 tries}) \\[4pt] &= 1 - P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \cdot P(\text{not forget}) \\[4pt] &= 1 - (0.85)(0.85)(0.85)(0.85)(0.85) \\[4pt] & = 1 - (0.85)^{5} = 0.556 \end{align*}\]

The probability that the child will forget her homework at least one day in the next five days is 0.556 or 55.6%

The idea in Example \(\PageIndex{9}\) can be generalized to get the At Least Once Rule.

Definition: At Least Once Rule

If an experiment is repeated n times, the n trials are independent and the probability of event A occurring one time is P(A) then the probability that A occurs at least one time is: \(P(\text{A occurs at least once in n trials}) = 1 - P(\overline{A})^{n}\)

Example \(\PageIndex{11}\): At Least Once Rule for Bird Watching

The probability of seeing a falcon near the lake during a day of bird watching is 0.21. What is the probability that a birdwatcher will see a falcon at least once in eight trips to the lake?

Let A be the event that he sees a falcon so P(A) = 0.21. Then, \(P(\overline{A}) = 1 - 0.21 = 0.79\).

\(P(\text{at least once in eight tries}) = 1 - P(\overline{A})^{8}\)

\( = 1 - (0.79)^{8}\)

\( = 1 - (0.152) = 0.848\)

The probability of seeing a falcon at least once in eight trips to the lake is approximately 0.848 or 84.8%.

Example \(\PageIndex{12}\): At Least Once Rule for Guessing on Multiple Choice Tests

A multiple choice test consists of six questions. Each question has four choices for answers, only one of which is correct. A student guesses on all six questions. What is the probability that he gets at least one answer correct?

Let A be the event that the answer to a question is correct. Since each question has four choices and only one correct choice, \(P(\text{correct}) = \dfrac{1}{4}\).

That means \(P(\text{not correct}) =1 - \dfrac{1}{4} = \dfrac{3}{4}\).

\[ \begin{align*} P(\text{at least one correct in six trials}) &= 1 - P(\text{not correct})^{6} \\[4pt] &= 1 - \left(\dfrac{3}{4}\right)^{6} \\[4pt] &= 1 - (0.178) = 0.822 \end{align*}\]

The probability that he gets at least one answer correct is 0.822 or 82.2%.

Probabilities from Two-Way Tables

Two-way tables can be used to define events and find their probabilities using two different approaches: intuitively or using the probability rules. We can calculate “and” and "or" probabilities by combining the data in relevant cells.

Example \(\PageIndex{13}\): Probabilities from a Two-Way Table

Continuation of Example \(\PageIndex{5}\):

A person is chosen at random. Find the probability that the person:

  • bought a new car

\[\begin{align*} P(\text{new car}) &= \dfrac{\text{number of new car}}{\text{number of people}} \\[4pt] &= \dfrac{120}{250} = 0.480 = 48.0 \% \end{align*} \]

  • was satisfied

\[\begin{align*} P(\text{satisfied}) &= \dfrac{\text{number of satisfied}}{\text{number of people}} \\[4pt] &= \dfrac{175}{250} = 0.700 = 70.0 \% \end{align*} \]

  • bought a new car and was satisfied

\[\begin{align*} P(\text{new car and satisfied}) &= \dfrac{\text{number of new car and satisfied}}{\text{number of people}} \\[4pt] &= \dfrac{92}{250} = 0.368 = 36.8 \% \end{align*} \]

  • bought a new car or was satisfied

\[\begin{align*} P(\text{new car or satisfied}) &= \dfrac{\text{number of new car + number of satisfied - number of new car and satisfied}}{\text{number of people}} \\[4pt] &= \dfrac{120 + 175 - 92}{250} = \dfrac{203}{250} = 0.812 = 81.2 \% \end{align*} \]

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Statistics LibreTexts

4.3: The Addition and Multiplication Rules of Probability

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When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not.

The Multiplication Rule

If A and B are two events defined on a sample space, then:

\[P(A \text{ AND } B) = P(B)P(A|B) \label{eq1}\]

This rule may also be written as:

\[P(A|B) = \dfrac{P(A \text{ AND } B)}{P(B)} \nonumber\]

(The probability of \(A\) given \(B\) equals the probability of \(A\) and \(B\) divided by the probability of \(B\).)

If \(A\) and \(B\) are independent , then

\[P(A|B) = P(A). \nonumber\]

and Equation \ref{eq1} becomes

\[P(A \text{ AND } B) = P(A)P(B). \nonumber\]

The Addition Rule

If A and B are defined on a sample space, then:

\[P(A \text{ OR } B) = P(A) + P(B) - P(A \text{ AND } B) \label{eq5}\]

If A and B are mutually exclusive , then

\[P(A \text{ AND } B) = 0. \nonumber\]

and Equation \ref{eq5} becomes

\[P(A \text{ OR } B) = P(A) + P(B). \nonumber\]

Example \(\PageIndex{1}\)

Klaus is trying to choose where to go on vacation. His two choices are: \(\text{A} = \text{New Zealand}\) and \(\text{B} = \text{Alaska}\).

  • Klaus can only afford one vacation. The probability that he chooses \(\text{A}\) is \(P(\text{A}) = 0.6\) and the probability that he chooses \(\text{B}\) is \(P(\text{B}) = 0.35\).
  • \(P(\text{A AND B}) = 0\) because Klaus can only afford to take one vacation
  • Therefore, the probability that he chooses either New Zealand or Alaska is \(P(\text{A OR B}) = P(\text{A}) + P(\text{B}) = 0.6 + 0.35 = 0.95\). Note that the probability that he does not choose to go anywhere on vacation must be 0.05.

Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. \(\text{A} =\) the event Carlos is successful on his first attempt. \(P(\text{A}) = 0.65\). \(\text{B} =\) the event Carlos is successful on his second attempt. \(P(\text{B}) = 0.65\). Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.

  • What is the probability that he makes both goals?
  • What is the probability that Carlos makes either the first goal or the second goal?
  • Are \(\text{A}\) and \(\text{B}\) independent?
  • Are \(\text{A}\) and \(\text{B}\) mutually exclusive?

a. The problem is asking you to find \(P(\text{A AND B}) = P(\text{B AND A})\). Since \(P(\text{B|A}) = 0.90: P(\text{B AND A}) = P(\text{B|A}) P(\text{A}) = (0.90)(0.65) = 0.585\)

Carlos makes the first and second goals with probability 0.585.

b. The problem is asking you to find \(P(\text{A OR B})\).

\[P(\text{A OR B}) = P(\text{A}) + P(\text{B}) - P(\text{A AND B}) = 0.65 + 0.65 - 0.585 = 0.715\]

Carlos makes either the first goal or the second goal with probability 0.715.

c. No, they are not, because \(P(\text{B AND A}) = 0.585\).

\[P(\text{B})P(\text{A}) = (0.65)(0.65) = 0.423\]

\[0.423 \neq 0.585 = P(\text{B AND A})\]

So, \(P(\text{B AND A})\) is not equal to \(P(\text{B})P(\text{A})\).

d. No, they are not because \(P(\text{A and B}) = 0.585\).

To be mutually exclusive, \(P(\text{A AND B})\) must equal zero.

Exercise \(\PageIndex{1}\)

Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. \(\text{C} =\) the event that Helen makes the first shot. \(P(\text{C}) = 0.75\). \(\text{D} =\) the event Helen makes the second shot. \(P(\text{D}) = 0.75\). The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws?

\[P(\text{D|C}) = 0.85\]

\[P(\text{C AND D}) = P(\text{D AND C})\]

\[P(\text{D AND C}) = P(\text{D|C})P(\text{C}) = (0.85)(0.75) = 0.6375\]

Helen makes the first and second free throws with probability 0.6375.

Example \(\PageIndex{2}\)

A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.

  • What is the probability that the member is a novice swimmer?
  • What is the probability that the member practices four times a week?
  • What is the probability that the member is an advanced swimmer and practices four times a week?
  • What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not?
  • Are being a novice swimmer and practicing four times a week independent events? Why or why not?
  • \(\dfrac{28}{150}\)
  • \(\dfrac{80}{150}\)
  • \(\dfrac{40}{150}\)
  • \(P(\text{advanced AND intermediate}) = 0\), so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.
  • No, these are not independent events. \[P(\text{novice AND practices four times per week}) = 0.0667\]\[P(\text{novice})P(\text{practices four times per week}) = 0.0996\] \[0.0667 \neq 0.0996\]

Exercise \(\PageIndex{2}\)

A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year?

\[P = \dfrac{200-140-40}{200} = \dfrac{20}{200} = 0.1\]

Example \(\PageIndex{3}\)

Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25.

Let: \(\text{M} =\) math class, \(\text{S} =\) speech class, \(\text{M|S} =\) math given speech

  • What is the probability that Felicity enrolls in math and speech? Find \(P(\text{M AND S}) = P(\text{M|S})P(\text{S})\).
  • What is the probability that Felicity enrolls in math or speech classes? Find \(P(\text{M OR S}) = P(\text{M}) + P(\text{S}) - P(\text{M AND S})\).
  • Are \(\text{M}\) and \(\text{S}\) independent? Is \(P(\text{M|S}) = P(\text{M})\)?
  • Are \(\text{M}\) and \(\text{S}\) mutually exclusive? Is \(P(\text{M AND S}) = 0\)?

a. 0.1625, b. 0.6875, c. No, d. No

Exercise \(\PageIndex{3}\)

A student goes to the library. Let events \(\text{B} =\) the student checks out a book and \(\text{D} =\) the student check out a DVD. Suppose that \(P(\text{B}) = 0.40, P(\text{D}) = 0.30\) and \(P(\text{D|B}) = 0.5\).

  • Find \(P(\text{B AND D})\).
  • Find \(P(\text{B OR D})\).
  • \(P(\text{B AND D}) = P(\text{D|B})P(\text{B}) = (0.5)(0.4) = 0.20\).
  • \(P(\text{B OR D}) = P(\text{B}) + P(\text{D}) − P(\text{B AND D}) = 0.40 + 0.30 − 0.20 = 0.50\)

Example \(\PageIndex{4}\)

Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let \(\text{B} =\) woman develops breast cancer and let \(\text{N} =\) tests negative. Suppose one woman is selected at random.

  • What is the probability that the woman develops breast cancer? What is the probability that woman tests negative?
  • Given that the woman has breast cancer, what is the probability that she tests negative?
  • What is the probability that the woman has breast cancer AND tests negative?
  • What is the probability that the woman has breast cancer or tests negative?
  • Are having breast cancer and testing negative independent events?
  • Are having breast cancer and testing negative mutually exclusive?
  • \(P(\text{B}) = 0.143; P(\text{N}) = 0.85\)
  • \(P(\text{N|B}) = 0.02\)
  • \(P(\text{B AND N}) = P(\text{B})P(\text{N|B}) = (0.143)(0.02) = 0.0029\)
  • \(P(\text{B OR N}) = P(\text{B}) + P(\text{N}) - P(\text{B AND N}) = 0.143 + 0.85 - 0.0029 = 0.9901\)
  • No. \(P(\text{N}) = 0.85; P(\text{N|B}) = 0.02\). So, \(P(\text{N|B})\) does not equal \(P(\text{N})\).
  • No. \(P(\text{B AND N}) = 0.0029\). For \(\text{B}\) and \(\text{N}\) to be mutually exclusive, \(P(\text{B AND N})\) must be zero

Exercise \(\PageIndex{4}\)

A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports?

Let \(\text{A} =\) student is a senior going to college.

Let \(\text{B} =\) student plays sports.

\(P(\text{B}) = \dfrac{140}{200}\)

\(P(\text{B|A}) = \dfrac{50}{140}\)

\(P(\text{A AND B}) = P(\text{B|A})P(\text{A})\)

\(P(\text{A AND B}) = (\dfrac{140}{200}\))(\(\dfrac{50}{140}) = \dfrac{1}{4}\)

Example \(\PageIndex{5}\)

Refer to the information in Example \(\PageIndex{4}\). \(\text{P} =\) tests positive.

  • Given that a woman develops breast cancer, what is the probability that she tests positive. Find \(P(\text{P|B}) = 1 - P(\text{N|B})\).
  • What is the probability that a woman develops breast cancer and tests positive. Find \(P(\text{B AND P}) = P(\text{P|B})P(\text{B})\).
  • What is the probability that a woman does not develop breast cancer. Find \(P(\text{B′}) = 1 - P(\text{B})\).
  • What is the probability that a woman tests positive for breast cancer. Find \(P(\text{P}) = 1 - P(\text{N})\).

a. 0.98; b. 0.1401; c. 0.857; d. 0.15

Exercise \(\PageIndex{5}\)

A student goes to the library. Let events \(\text{B} =\) the student checks out a book and \(\text{D} =\) the student checks out a DVD. Suppose that \(P(\text{B}) = 0.40, P(\text{D}) = 0.30\) and \(P(\text{D|B}) = 0.5\).

  • Find \(P(\text{B′})\).
  • Find \(P(\text{D AND B})\).
  • Find \(P(\text{B|D})\).
  • Find \(P(\text{D AND B′})\).
  • Find \(P(\text{D|B′})\).
  • \(P(\text{B′}) = 0.60\)
  • \(P(\text{D AND B}) = P(\text{D|B})P(\text{B}) = 0.20\)
  • \(P(\text{B|D}) = \dfrac{P(\text{B AND D})}{P(\text{D})} = \dfrac{(0.20)}{(0.30)} = 0.66\)
  • \(P(\text{D AND B′}) = P(\text{D}) - P(\text{D AND B}) = 0.30 - 0.20 = 0.10\)
  • \(P(\text{D|B′}) = P(\text{D AND B′})P(\text{B′}) = (P(\text{D}) - P(\text{D AND B}))(0.60) = (0.10)(0.60) = 0.06\)
  • DiCamillo, Mark, Mervin Field. “The File Poll.” Field Research Corporation. Available online at www.field.com/fieldpollonline...rs/Rls2443.pdf (accessed May 2, 2013).
  • Rider, David, “Ford support plummeting, poll suggests,” The Star, September 14, 2011. Available online at www.thestar.com/news/gta/2011..._suggests.html (accessed May 2, 2013).
  • “Mayor’s Approval Down.” News Release by Forum Research Inc. Available online at www.forumresearch.com/forms/News Archives/News Releases/74209_TO_Issues_-_Mayoral_Approval_%28Forum_Research%29%2820130320%29.pdf (accessed May 2, 2013).
  • “Roulette.” Wikipedia. Available online at http://en.Wikipedia.org/wiki/Roulette (accessed May 2, 2013).
  • Shin, Hyon B., Robert A. Kominski. “Language Use in the United States: 2007.” United States Census Bureau. Available online at www.census.gov/hhes/socdemo/l...acs/ACS-12.pdf (accessed May 2, 2013).
  • Data from the Baseball-Almanac, 2013. Available online at www.baseball-almanac.com (accessed May 2, 2013).
  • Data from U.S. Census Bureau.
  • Data from the Wall Street Journal.
  • Data from The Roper Center: Public Opinion Archives at the University of Connecticut. Available online at www.ropercenter.uconn.edu/ (accessed May 2, 2013).
  • Data from Field Research Corporation. Available online at www.field.com/fieldpollonline (accessed May 2,2 013).

The multiplication rule and the addition rule are used for computing the probability of \(\text{A}\) and \(\text{B}\), as well as the probability of \(\text{A}\) or \(\text{B}\) for two given events \(\text{A}\), \(\text{B}\) defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events \(\text{A}\) and \(\text{B}\) are mutually exclusive events when they do not have any outcomes in common.

Formula Review

The multiplication rule: \(P(\text{A AND B}) = P(\text{A|B})P(\text{B})\)

The addition rule: \(P(\text{A OR B}) = P(\text{A}) + P(\text{B}) - P(\text{A AND B})\)

Use the following information to answer the next ten exercises. Forty-eight percent of all Californians registered voters prefer life in prison without parole over the death penalty for a person convicted of first degree murder. Among Latino California registered voters, 55% prefer life in prison without parole over the death penalty for a person convicted of first degree murder. 37.6% of all Californians are Latino.

In this problem, let:

  • \(\text{C} =\) Californians (registered voters) preferring life in prison without parole over the death penalty for a person convicted of first degree murder.
  • \(\text{L} =\) Latino Californians

Suppose that one Californian is randomly selected.

Find \(P(\text{C})\).

Exercise \(\PageIndex{6}\)

Find \(P(\text{L})\).

Exercise \(\PageIndex{7}\)

Find \(P(\text{C|L})\).

Exercise \(\PageIndex{8}\)

In words, what is \(\text{C|L}\)?

\(\text{C|L}\) means, given the person chosen is a Latino Californian, the person is a registered voter who prefers life in prison without parole for a person convicted of first degree murder.

Exercise \(\PageIndex{9}\)

Find \(P(\text{L AND C})\)

Exercise \(\PageIndex{10}\)

In words, what is \(\text{L AND C}\)?

\(\text{L AND C}\) is the event that the person chosen is a Latino California registered voter who prefers life without parole over the death penalty for a person convicted of first degree murder.

Exercise \(\PageIndex{11}\)

Are \(\text{L}\) and \(\text{C}\) independent events? Show why or why not.

Exercise \(\PageIndex{12}\)

Find \(P(\text{L OR C})\).

Exercise \(\PageIndex{13}\)

In words, what is \(\text{L OR C}\)?

Exercise \(\PageIndex{14}\)

Are \(\text{L}\) and \(\text{C}\) mutually exclusive events? Show why or why not.

No, because \(P(\text{L AND C})\) does not equal 0.

  • \(P(\text{A|B}) = P(\text{A})\)
  • \(P(\text{B|A}) = P(\text{B})\)
  • \(P(\text{A AND B}) = P(\text{A})P(\text{B})\)

Probability | Theory, solved examples and practice questions

When MS and MBA applicants ask us – ‘ What are my chances of getting into Harvard? ‘ or ‘ What’s my probability of getting scholarships from Oxford? ‘ we get tongue-tied. There are so many variables at play, it’s difficult to give an accurate answer.

But when you get probability questions in your GRE and GMAT exam syllabus , you don’t have to get flummoxed. Understanding the basic rules and formulas of probability will help you score high in the entrance exams.  

Meaning and definition of Probability

As the Oxford dictionary states it, Probability means ‘The extent to which something is probable; the likelihood of something happening or being the case’.

In mathematics too, probability indicates the same – the likelihood of the occurrence of an event.

Examples of events can be :

  • Tossing a coin with the head up
  • Drawing a red pen from a pack of different coloured pens
  • Drawing a card from a deck of 52 cards etc.

Either an event will occur for sure, or not occur at all. Or there are possibilities to different degrees the event may occur.

An event that occurs for sure is called a Certain event and its probability is 1.

An event that doesn’t occur at all is called an impossible event and its probability is 0.

This means that all other possibilities of an event occurrence lie between 0 and 1.

This is depicted as follows:

0 <= P(A) <= 1

where A is an event and P(A) is the probability of the occurrence of the event.

This also means that a probability value can never be negative.

Every event will have a set of possible outcomes. It is called the ‘sample space’.

Consider the example of tossing a coin.

When a coin is tossed, the possible outcomes are Head and Tail. So, the sample space is represented as {H, T}.

Similarly when two coins are tossed, the sample space is {(H,H), (H,T), (T,H), (T,T)}.

The probability of head each time you toss the coin is 1/2. So is the probability of tail.

Basic formula of probability

As you might know from the list of GMAT maths formulas , the Probability of the occurrence of an event A is defined as:

P(A) = (No. of ways A can occur)/(Total no. of possible outcomes)

Another example is the rolling of dice. When a single die is rolled, the sample space is {1,2,3,4,5,6}.

What is the probability of rolling a 5 when a die is rolled?

No. of ways it can occur = 1

Total no. of possible outcomes = 6

So the probability of rolling a particular number when a die is rolled = 1/6.

Compound probability

Compound probability is when the problem statement asks for the likelihood of the occurrence of more than one outcome.  

Formula for compound probability

  • P(A or B) = P(A) + P(B) – P(A and B)

where A and B are any two events.

P(A or B) is the probability of the occurrence of atleast one of the events.

P(A and B) is the probability of the occurrence of both A and B at the same time.

Mutually exclusive events:

Mutually exclusive events are those where the occurrence of one indicates the non-occurrence of the other

When two events cannot occur at the same time, they are considered mutually exclusive.

Note: For a mutually exclusive event, P(A and B) = 0.

Example 1: What is the probability of getting a 2 or a 5 when a die is rolled?

Taking the individual probabilities of each number, getting a 2 is 1/6 and so is getting a 5.

Applying the formula of compound probability,

Probability of getting a 2 or a 5,

P(2 or 5) = P(2) + P(5) – P(2 and 5)

==>      1/6 + 1/6 – 0

==>      2/6 = 1/3.

Example 2: Consider the example of finding the probability of selecting a black card or a 6 from a deck of 52 cards.

We need to find out P(B or 6)

Probability of selecting a black card  = 26/52

Probability of selecting a 6                 = 4/52

Probability of selecting both a black card and a 6 = 2/52

P(B or 6)          = P(B) + P(6) – P(B and 6)

= 26/52 + 4/52 – 2/52

Independent and Dependent Events

Independent event.

When multiple events occur, if the outcome of one event DOES NOT affect the outcome of the other events, they are called independent events.

Say, a die is rolled twice. The outcome of the first roll doesn’t affect the second outcome. These two are independent events.

Example 1: Say, a coin is tossed twice. What is the probability of getting two consecutive tails ?

Probability of getting a tail in one toss = 1/2

The coin is tossed twice. So 1/2 * 1/2 = 1/4 is the answer.

Here’s the verification of the above answer with the help of sample space.

When a coin is tossed twice, the sample space is {(H,H), (H,T), (T,H), (T,T)}.

Our desired event is (T,T) whose occurrence is only once out of four possible outcomes and hence, our answer is 1/4.

Example 2: Consider another example where a pack contains 4 blue, 2 red and 3 black pens. If a pen is drawn at random from the pack, replaced and the process repeated 2 more times, What is the probability of drawing 2 blue pens and 1 black pen?

Here, total number of pens = 9

Probability of drawing 1 blue pen = 4/9 Probability of drawing another blue pen = 4/9 Probability of drawing 1 black pen = 3/9 Probability of drawing 2 blue pens and 1 black pen = 4/9 * 4/9 * 3/9 = 48/729 = 16/243

Dependent Events

When two events occur, if the outcome of one event affects the outcome of the other, they are called dependent events.

Consider the aforementioned example of drawing a pen from a pack, with a slight difference.

Example 1: A pack contains 4 blue, 2 red and 3 black pens. If 2 pens are drawn at random from the pack, NOT replaced and then another pen is drawn. What is the probability of drawing 2 blue pens and 1 black pen?

Probability of drawing 1 blue pen = 4/9 Probability of drawing another blue pen = 3/8 Probability of drawing 1 black pen = 3/7 Probability of drawing 2 blue pens and 1 black pen = 4/9 * 3/8 * 3/7 = 1/14

Let’s consider another example:

Example 2: What is the probability of drawing a king and a queen consecutively from a deck of 52 cards, without replacement.

Probability of drawing a king = 4/52 = 1/13

After drawing one card, the number of cards are 51.

Probability of drawing a queen = 4/51.

Now, the probability of drawing a king and queen consecutively is 1/13 * 4/51 = 4/663

Conditional probability

Conditional probability is calculating the probability of an event given that another event has already occured .

The formula for conditional probability P(A|B), read as P(A given B) is

P(A|B) = P (A and B) / P(B)

Consider the following example:

Example: In a class, 40% of the students study math and science. 60% of the students study math. What is the probability of a student studying science given he/she is already studying math?

P(M and S) = 0.40

P(M) = 0.60

P(S|M) = P(M and S)/P(S) = 0.40/0.60 = 2/3 = 0.67

Complement of an event

A complement of an event A can be stated as that which does NOT contain the occurrence of A.

A complement of an event is denoted as P(A c ) or P(A’).

P(A c ) = 1 – P(A)

or it can be stated, P(A)+P(A c ) = 1

For example,

if A is the event of getting a head in coin toss, A c is not getting a head i.e., getting a tail.

if A is the event of getting an even number in a die roll, A c is the event of NOT getting an even number i.e., getting an odd number.

if A is the event of randomly choosing a number in the range of -3 to 3, A c is the event of choosing every number that is NOT negative i.e., 0,1,2 & 3 (0 is neither positive or negative).

Example: A single coin is tossed 5 times. What is the probability of getting at least one head?

Consider solving this using complement.

Probability of getting no head = P(all tails) = 1/32

P(at least one head) = 1 – P(all tails) = 1 – 1/32 = 31/32.

Sample Probability questions with solutions

Probability example 1.

What is the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled.

Let the event of the occurrence of a number that is odd be ‘A’ and the event of the occurrence of a number that is less than 5 be ‘B’. We need to find P(A or B).

P(A) = 3/6 (odd numbers = 1,3 and 5)

P(B) = 4/6 (numbers less than 5 = 1,2,3 and 4)

P(A and B) = 2/6 (numbers that are both odd and less than 5 = 1 and 3)

Now, P(A or B)            = P(A) + P(B) – P(A or B)

= 3/6 + 4/6 – 2/6

P(A or B) = 5/6.

Probability Example 2

A box contains 4 chocobars and 4 ice creams. Tom eats 3 of them one after another. What is the probability of sequentially choosing 2 chocobars and 1 icecream?

Probability of choosing 1 chocobar = 4/8 = 1/2

After taking out 1 chocobar, the total number is 7.

Probability of choosing 2nd chocobar = 3/7

Probability of choosing 1 icecream out of a total of 6 = 4/6 = 2/3

So the final probability of choosing 2 chocobars and 1 icecream = 1/2 * 3/7 * 2/3 = 1/7

Probability Example 3

When two dice are rolled, find the probability of getting a greater number on the first die than the one on the second, given that the sum should equal 8.

Let the event of getting a greater number on the first die be G.

There are 5 ways to get a sum of 8 when two dice are rolled = {(2,6),(3,5),(4,4), (5,3),(6,2)}.

And there are two ways where the number on the first die is greater than the one on the second given that the sum should equal 8, G = {(5,3), (6,2)}.

Therefore, P(Sum equals 8) = 5/36 and P(G) = 2/36.

Now, P(G|sum equals 8)         = P(G and sum equals 8)/P(sum equals 8)

= (2/36)/(5/36)

Probability Quiz: Sample probability questions for practice

Problem 1: Click here

A bag contains blue and red balls. Two balls are drawn randomly without replacement. The probability of selecting a blue and then a red ball is 0.2. The probability of selecting a blue ball in the first draw is 0.5. What is the probability of drawing a red ball, given that the first ball drawn was blue? a) 0.4 b) 0.2 c) 0.1 d) 0.5

Answer 1: Click here

Problem 2: Click here

A die is rolled thrice. What is the probability that the sum of the rolls is atleast 5. a) 1/216 b) 1/6 c) 3/216 d) 212/216

Answer 2: Click here

  Learn how to solve: – Simple and compound interest problems – Speed, distance and time problems – Ratio and proportion – List of Maths Formulas

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43 thoughts on “Probability | Theory, solved examples and practice questions”

a number of people gave a hat check girl one hat. suppose all the tickets got misplaced, so all the hat were given back randomly. a) if its 2 people determine the probability at least one person got their hat returned. b) if its 3 people determine the probability at least one person got their hat returned. c) if its 4 people determine the probability at least one person got their hat returned. d) if its 5 people determine the probability at least one person got their hat returned.

Hi I’m Algia and I need help in solving this problem, can you help me please.

a) 1-(0.5)=0.5 b) 1-(0.667*0.5)=0.667 c) 1-(0.75*0.667*0.5)=0.75 d) 1-(0.8*0.75*0.667*0.5)=0.8

Really nice sequence.

P(h’)=1-P(h) etc. At least one hat is correctly returned is compliment that no hat is returned correctly.

The math here is totally wrong

Feel free to enlighten!

you make the assumption the events are independent, they are not, if you there are only 2 people, it is impossible for just 1 person to be given the right hat, it must be both people.

You make the assumption of independence, which does not hold in this question. instead we have to use the inclusion-exclusion principle.

For a) the probability that at least 1 person gets the got their hat returned. A is the event the 1st person gets their hat and B is the event the 2nd person gets their hat. We want to find P(A U B) = ?

P(A U B) = P(A) + P(B) – P(A n B)

P(A) = 1/2 P(B) = 1/2 P(A n B) = 1/4

P(A U B) = 1/2 + 1/2 – 1/4 = 3/4

Think it in this way. Representation here: Got Hat – G, Didn’t Get Hat – D. If there are 2 people, the total possible outcomes would be = 2^2 = 4 with them being {(G,G),(D,D),(G,D),(D,G)}. Now, probability that at least one person got hat returned = 1 – P(no one gets hat) = 1 – 1/4 = 3/4.

Similarly, in 3 people case, probability that at least one person gets hat = 1 – P(no one gets hat) = 1 – 1/8 = 7/8.

and so on…

General Formula derived for ‘n’ people case : P(At least one person gets hat returned) = 1 – P(no one gets hat) = (1 – 1/2^n)

Hi Last question must be 212/216 right ?

Tell me the way u did that sol. Plzz…

yes its must be 212/216

I think it should be 212/216 Bcz we have 4 number of event to getting a number of sum less than 5 {(1,1,1),(1,1,2),(1,2,1),(2,1,1)} It means p(e) = 4/216 Nd getting a number of sum at least 5 Is 1-4/216=212/216

you are telling a wrong way so stop telling me your bloody wrong answer it’s a correct answer it’s should not be 212/216..

@Emmanuel, I don’t think you can isolate the two variables like that since, as you mentioned, these are dependent events. The total number of ways of returning the hat is n! and the total number of ways of returning the hats such that no one gets their own one is (n-1)!. Therefore, the probability that at least one person gets their own hat is P(X>0) = 1-P(X=0) = 1 – (n-1)!/n! = 1 – 1/n.

A woman bought 5basket of tomatos each costing 1250naira,in her discovery she observe that 90% of the tomatos where damage resulting to a loss of 510naira.(a)what is the probability of obtaining an average of 50 if the cost per bag is 50 above the cost?(b)what will be the actual price for selling the tomato at cost plus(+) 25%?

Simple way: Q: A die is rolled thrice. What is the probability that the sum of the roll is at least 5?

A: At least 5 is equal or greater 5 ===> which is P(x> or =5)= 1-P(x<5) Then: P(x<5)= 4/216. . .taken thrice rolls =6*6*6=216

Therefore: 1-P(x 53/54

This is incorrect. Correct answer is: P(x<5) = 3 Therefore Answer = 1 – 3/216 = 213/216.

the personal director of a company wishes to select applicant for advanced training without regard to sex. let “W” denotes women and “M” denotes men and the pattern of arrival be M WWW MMM WW M WWW MMMM W M W MM WWW MM W MMMM WW M WW MMMM WW M WWWW MM WW M W WW. will you conclude that the applicants have arrived in a random fashion?

pavement.before any 250 m length of a pavement is accepted by the state highway department,the thickness of a30 m s mointored by an altrasonic to verify compliance to specification .each section is rejected if a measurment thickness less than 10cm;otherwise the all section is accepted .from past experment ,the stat highway engineer know the 85%of all section constructed by the contructor comply with specification . however the relability of altrusonic thickness testing is only 75 ,so that there is a 25 percent chane of errorneous concolusion based on the determenation of thickness with ultrasonic . what is the probablity that a poorly constructed section is accepted on the base of ultrasonic test?

The chance or probability of getting accepted is 0.85; the chance of getting accepted even when bad is 0.25. So therefore the chance of being bad and getting selected can be solved using the conditional probability theorem given by:

P(A/B)= P(AnB)/P(B). Going by this the answer is: 0.25 x 0.85= 0.2125

solution the possible out come of rolling die is =6 here in this case since it is rolled 3 our sample space is 6×6×6=216 we have asked to solve the probability of sum which will be atleast 5 this means 5 and more is possible. so that we have to search the possibilities of less than five to easy our work this will be like[111][112][121] = 3 out comes onlywso p(s`)=3/216 when p(s`) is probability of sum less than five or probability of sum greater than equal to five. since the sum of p(s) and p(s`)=1 p(s)=1-p(s~) 1-3/216=213/216

what about 2,1,1?

Two cards are drawn at random from an ordinary deck of 52 card. Find the probability P that (a) Both are spade (b) One is a spade and one is heart

Ans: (a) Probability of getting spade 1st time is 13/52 and Probability of getting spade 2nd time is is 12/51 Total Probability is 13*12/(52*51) = 156/2652 (b) Probability of getting spade is 13/52 and Probability of getting Heart is 12/51 Total probability is 13*13/(52*51) = 169/2652

Question ‘b’ says that one is a spade and one is a heart. Therefore the possibilities are ‘heart-spade’ and ‘spade-heart’. The answer should be: ((13*13/52*51) + (13*13/52*51)) = 169/2652 + 169/2652 =338/2652 = 13/102

copying the solution offerred by @ diriba

The above solution is good but a little faulty because it considered only the possibility of obtaining a ‘1’ on the first die, it omitted the possibility of getting a ‘2’ on the first die i.e (using the same notation) [211], this is the fourth possible outcome. Hence P(s)= 1- P(s’) = 1-4/216 =212/216 =53/54

A bag contains blue and red balls. Two balls are drawn randomly without replacement. The probability of selecting a blue and then a red ball is 0.2. The probability of selecting a blue ball in the first draw is 0.5. What is the probability of drawing a red ball, given that the first ball drawn was blue? Solution please

Lets assume probability of picking a red ball is X. The probability of selecting a blue ball and then a red ball, P(B)*P(R)=.2 .5*X=.2 x=.5/.2 x=.4

P(R|B)=P(R and B)/P(B) =0.2/0.5=0.4

The probability of snow tomorrow is 0.6. And the probability that it will bi colder is 0.7. The probability that it will not snow and not bi colder is 0.1 .What is probability that it will not snow if it is colder tomorrow? please solve it … and tell me answer.. thanks …

A= event for it will snow tomorrow B= event for it will be cold a= event for it will not snow b= event for it will not be cold P(A) = 0.6; P(B)= 0.7; P(a n b)=0.1 P(A)+P(B)-P(AnB)+P(a n b)=1 Inserting values, you’ll have P(AnB)=0.4. P(Bna)= P(B)-P(BnA)

This will give you P(Bna)= 0.3.

By conditional theorem,

P(a/B)=P(anB)/P(B)

This will give you 3/7 as the answer.

pl first draw a tree diagram: P(S)=0.6, P(N.S.)=0.4 P(Cold)=0.7, P(Not Cold)=0.3 P(No S + Not Cold)= 0.4*0.3=0.12=0.1 P(No S + Cold)=0.4*0.7=0.28

in a class 10 boys and 5 girls .three students are selected random one after the other.find the probability that

1)first two are boys and third is girl 2)first and third is of same gender and third is of opposite gender

please help me in solving this

A) 10/15*9/14*5/13 B) 1st case: 1st & 2nd are boys & 3rd is girl 10/15*9/14*5/13 2nd case: 1st & 2nd are girls & 3rd is boy 5/15*4/14*10/13

n(b)=10 n(g)=5 n(t)=15 (a) p(first two are boys and third is a girl) p(B,B,G) 10/15*9/14*5/13= 5/1*3/7*1/13=15/91 (b) p(first and third is of the same gender and second is opposite) P(B,G,B) or p(G,B,G) (10/15*5/14*9/13)+(5/15*10/14*4/13)= (450/2730)+(200/2730)= 450+200/2730= 650/2730= 5/21

There are three boxes, one of which contains a prize. A contestant is given two chances, such that if he chooses the wrong box in the first round, that box is removed from the selection and he then chooses between the two remaining boxes. 1. What is the probability that the contestant wins? 2. Does the contestant’s probability of winning increases on the second round?

It’s a Monty Hall problem. You can google it.

As for your question, As the first box chosen if found empty is removed and you HAVE/Switch to pick from other two, the P(W) = 2/3.

Above answer can be explained as Prob. of winning on first box + Prob. of choosing wrong * Prob. of Choosing right between the two => 1/3+2/3*1/2 => 2/3

The answer to the second: Yes probability increases as its a 50% chance to win as 1 wrong box is eliminated.

1) 10C2*5C1/15C3? 2) (10C1*5C1*9C1/15C3) + (5C1*10C1*4C1/15C3)?

Plz solve it

XYZ company wants to start a food outlet in pakistan. There is a 40% and 60% chance of stating in hyderabad and karachi respectively. If he start the outlet in hyderabad there is 30% chance that it will be in saddar and 70% chance that it will be in defence area. If they start the outlet in karachi there is 50% chance that it will be in defence, 30% in clifton and 20% in pechs. Determine probability of starting the outlet in: (a) saddar (b) defence area of any city (c) clifton given that the outlet is started in karachi

a) P(H,S) = 40% x 30% =0.4 x 0.3 = 0.12 = 12% b) P(H, D) + P(K, D) = 40% x 70% + 60% x 50% = 0.4 x 0.7 + 0.6 x 0.5 = 0.28 + 0.3 = 0.58 = 58% c) P(C|K) = 30%

In maternity clinic the probability of new born was females is 55%=0.55

So,the probabilitt of the next three deliveries are females is 0.55×0.55×0.55=0.166 or 16.6%

Nah, that’s not correct. Births are independent events. It doesn’t matter what the gender of previous births were. The odds of the next birth being female is still 50%. The probability of three females in a row is simply (0.5)^3 = 1/8 = 0.125

a) 15*(0.2)^4*(0.8)^2 = 0.01536 b) 0.01536 + 6*(0.2)^5*(0.8)^1 + 1*(0.2)^6*(0.8)^0 = 0.01696 c) (0.8)^6 = 0.262144

hopefully you haven’t been waiting over a year for a response 🙂 Here you go though…

A: (2/3)^ 3 = 8/27 B: 1 * 2/3 * 1/3 = 2/9 C: 1 * 1/3 * 1/3 = 1/9

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Statology

Statistics Made Easy

10 Examples of Using Probability in Real Life

Probability refers to how likely an event is to occur.

Probability is used in all types of areas in real life including weather forecasting, sports betting, investing, and more.

The following examples share how probability is used in 10 real-life situations on a regular basis.

Example 1: Weather Forecasting

Perhaps the most common real life example of using probability is weather forecasting .

Probability is used by weather forecasters to assess how likely it is that there will be rain, snow, clouds, etc. on a given day in a certain area.

Forecasters will regularly say things like “there is an 80% chance of rain today between 2PM and 5PM” to indicate that there’s a high likelihood of rain during certain hours.

Example 2: Sports Betting

Probability is heavily used by sports betting companies to determine the odds they should set for certain teams to win certain games.

For example, a sports betting company may look at the current record of two teams and determine that team A has a 90% probability of winning while team B has just a 10% probability of winning.

Based on these probabilities, the company would offer a higher payout for people who bet on team B to win since it’s highly unlikely that team B will actually win.

Example 3: Politics

Political forecasters use probability to predict the chances that certain candidates will win various elections.

For example, a forecaster might say that candidate A has a 60% chance of winning, candidate B has a 20% chance of winning, candidate C has a 10% chance of winning, etc. to give voters an idea of how likely it is that each candidate will win.

Note : A real-life example of a site that uses probability to perform political forecasting is FiveThirtyEight .

Example 4: Sales Forecasting

Many retail companies use probability to predict the chances that they’ll sell a certain amount of goods in a given day, week, or month.

This allows the companies to predict how much inventory they’ll need. For example, a company might use a forecasting model that tells them the probability of selling at least 100 products on a certain day is 90%.

This means they’ll need to make sure they have at least 100 products on hand to sell (or preferably more) so they don’t run out.

Example 5: Health Insurance

Health insurance companies often use probability to determine how likely it is that certain individuals will spend a certain amount on healthcare each year.

For example, a company might use factors like age, existing medical conditions, current health status, etc. to determine that there’s a 90% probability that a certain individual will spend $10,000 or more on healthcare in a given year.

Individuals who are likely to spend more on healthcare will be charged higher premiums because the insurance company knows that they’ll be more expensive to insure.

Example 6: Grocery Store Staffing

Grocery stores often use probability to determine how many workers they should schedule to work on a given day.

For example, a grocery store may use a model that tells them there is a 75% chance that they’ll have more than 800 customers come into the store on a given day.

Based on this probability, they’ll schedule a certain amount of workers to be at the store on that day to handle that many customers.

Example 7: Natural Disasters

The environmental departments of countries often use probability to determine how likely it is that a natural disaster like a hurricane, tornado, earthquake, etc. will strike the country in a given year.

If the probability is quite high, then the department will make decisions about housing, resource allocation, etc. that will minimize the effects done by the natural disaster.

Example 8: Traffic

Ordinary people use probability every day when they decide to drive somewhere.

Based on the time of day, location in the city, weather conditions, etc. we all tend to make probability predictions about how bad traffic will be during a certain time.

For example, if you think there’s a 90% probability that traffic will be heavy from 4PM to 5:30PM in your area then you may decide to wait to drive somewhere during that time.

Example 9: Investing

Investors use probability to assess how likely it is that a certain investment will pay off.

For example, a given investor might determine that there is a 1% chance that the stock of company A will increase 100x during the upcoming year.

Based on this probability, the investor will decide how much of their net worth to invest in the stock.

Example 10: Card Games

Probability is routinely used by anyone who plays card games on a regular basis.

For example, professional poker players use probability to determine how likely it is that a certain hand of cards will win and this informs them on how much they should bet.

If a player knows that there is a high probability that they will win a certain hand based on their cards, they will be more likely to bet more money.

Conversely, if they think the probability that they’ll win is low then they may bet significantly less money.

Additional Resources

The following tutorials provide additional information about probability:

Examples of Using Conditional Probability in Real Life Probability vs. Proportion: What’s the Difference? Probability vs. Likelihood: What’s the Difference? Law of Total Probability: Definition & Examples

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Please provide examples of the question.    . 4. Submit Table 4...

Please provide examples of the question.

Answer & Explanation

To clarify the improvements based on the "just right" scales for each attribute, let's formulate examples of how to interpret and address the feedback for improvement:

 Color - Example Question: "For the Tostitos sample, a significant number of panelists found the color to be too dark. What adjustment can be made to improve the color perception towards the 'just right' category?" - Example Improvemen: Lighten the color of the Tostitos sample slightly to align with the majority preference for a lighter appearance.

 Flavor -Example Question: "If the Western Family sample received feedback indicating the flavor is too bland, what specific flavor enhancement would move it towards the 'just right' category?" - Example Improvement: Enhance the flavor profile of the Western Family sample by increasing its seasoning intensity to address the blandness feedback.

 Aftertaste -Example Question: "Given feedback that the Tostitos Multigrain has a too strong aftertaste, what formulation adjustment could balance the aftertaste?" -Example Improvement : Modify the seasoning blend for the Tostitos Multigrain to reduce components that may contribute to a lingering or overpowering aftertaste.

 Saltiness - Example Question: "For the Western Family Lightly Salted chips, with feedback indicating they are too salty, how can the recipe be adjusted?" - Example Improvement: Reduce the amount of salt used in the Western Family Lightly Salted chips to decrease the overall saltiness level.

 Oiliness in Mouth - Example Question: "If consumers find the oiliness in mouth feel for the 867 sample too excessive, what change in cooking process or ingredients could mitigate this?" -Example Improvement: Adjust the frying time or temperature for the 867 sample, or explore using an oil with a lighter mouthfeel to reduce perceived oiliness.

 Crunchiness - Example Question: "Considering that the crunchiness of the 736 sample is not meeting the 'just right' expectation, being too crunchy, what adjustment should be made?" - Example Improvement: Alter the thickness of the 736 chip or adjust the frying time to achieve a less intense crunch that aligns with consumer expectations.

These examples demonstrate how to use specific feedback from the "just right" scales to identify and implement product improvements. The goal is to align the product attributes more closely with consumer preferences by making informed adjustments based on panel feedback.

Improving product attributes based on consumer feedback is an essential aspect of product development, particularly in the food industry where sensory experiences play a significant role in consumer satisfaction. When feedback indicates that a product's attributes—such as color, flavor, aftertaste, saltiness, oiliness, and crunchiness—fall outside the "just right" category, manufacturers must carefully consider adjustments to align more closely with consumer preferences.

For color, if a product like Tostitos is found to be too dark by a significant number of panelists, the improvement strategy involves adjusting the cooking process or ingredient proportions to lighten the product's color. This could mean reducing cooking time or temperature to prevent over-browning, ensuring the visual appeal meets consumer expectations.

Flavor adjustments are particularly nuanced, as they involve balancing spices, seasonings, and core ingredients to hit the right notes of taste. For instance, if the Western Family chips are deemed too bland, enhancing the flavor profile might involve increasing the intensity of specific seasonings or introducing new flavor components to create a more robust and satisfying taste experience.

Aftertaste is another critical attribute that can affect consumer satisfaction. A strong or lingering aftertaste, as might be reported for Tostitos Multigrain, requires formulation adjustments. This could include reducing or substituting ingredients that contribute to the aftertaste, ensuring the product leaves a pleasant and balanced flavor in the mouth post-consumption.

Saltiness is a common concern in snack foods. Feedback indicating a product is too salty, such as with Western Family Lightly Salted chips, necessitates a reduction in salt content. This adjustment helps in meeting health-conscious consumer demands while ensuring the taste remains appealing without the overpowering presence of salt.

Oiliness affects the mouthfeel and can turn consumers away if perceived as too high. Adjusting the oiliness, especially for products like Tostitos Multigrain that might be reported as too oily, involves changing the type of oil used, modifying the frying process, or even altering the product's formulation to achieve a more desirable mouthfeel that balances richness without being greasy.

Crunchiness is crucial for the sensory experience of snack foods. Feedback suggesting a product is either too crunchy or not crunchy enough requires adjustments in the product's texture. This might involve altering the thickness of the chip, the frying time, or the ingredients used to achieve a crunch that satisfies consumer expectations for a pleasant eating experience.

In summary, responding to consumer feedback on sensory attributes involves a delicate balance of art and science. Manufacturers must understand the nuances of each attribute and how they contribute to the overall product experience. Adjusting these attributes based on feedback ensures that the product not only meets but exceeds consumer expectations, thereby enhancing satisfaction and loyalty.

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  3. Probability Problems and Independent Events

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  5. How to Calculate Probability (with Cheat Sheets)

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  6. Finding Probabilities Using Combinations in One Step

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VIDEO

  1. Probability,important terms of probability

  2. Probability example

  3. Basic example problems on Probability

  4. Probability distribution

  5. PROBLEM SOLUTION OF PROBABILITY

  6. A Simple Probability Problem

COMMENTS

  1. Probability

    How likely something is to happen. Many events can't be predicted with total certainty. The best we can say is how likely they are to happen, using the idea of probability. Tossing a Coin When a coin is tossed, there are two possible outcomes: Heads (H) or Tails (T) Also: the probability of the coin landing H is ½

  2. 15 Probability Questions And Practice Problems

    Probability questions and probability problems require students to work out how likely it is that something is to happen. Probabilities can be described using words or numbers. Probabilities range from 0 to 1 and can be written as fractions, decimals or percentages.

  3. Probability: the basics (article)

    The best example for understanding probability is flipping a coin: There are two possible outcomes—heads or tails. What's the probability of the coin landing on Heads? We can find out using the equation P ( H) =? . You might intuitively know that the likelihood is half/half, or 50%. But how do we work that out? Probability = In this case:

  4. Probability Examples with Questions and Answers

    Cases = 7 P (A) = 7/8 OR P (of getting at least one head) = 1 - P (no head)⇒ 1 - (1/8) = 7/8 Example 2: Find the probability of getting a numbered card when a card is drawn from the pack of 52 cards. Sol: Total Cards = 52. Numbered Cards = (2, 3, 4, 5, 6, 7, 8, 9, 10) 9 from each suit 4 × 9 = 36 P (E) = 36/52 = 9/13

  5. Probability Questions with Solutions

    Solution to Question 1 Let us first write the sample space S of the experiment. Let E be the event "an even number is obtained" and write it down. classical probability. Two coins are tossed, find the probability that two heads are obtained. Each coin has two possible outcomes H (heads) and T (Tails).

  6. Probability

    Quiz Unit test About this unit Probability tells us how often some event will happen after many repeated trials. You've experienced probability when you've flipped a coin, rolled some dice, or looked at a weather forecast.

  7. 7.6: Basic Concepts of Probability

    The probability of an event is a number between 0 and 1 (inclusive). If the probability of an event is 0, then the event is impossible. On the other hand, an event with probability 1 is certain to occur. In general, the higher the probability of an event, the more likely it is that the event will occur.

  8. How to Solve Probability Problems

    The following sample problems show how to apply these rules to find (1) the probability of a sample point and (2) the probability of an event. Probability of a Sample Point. The probability of a sample point is a measure of the likelihood that the sample point will occur. Example 1 Suppose we conduct a simple statistical experiment. We flip a ...

  9. PDF Twenty problems in probability

    Twenty problems in probability This section is a selection of famous probability puzzles, job interview questions (most high- ... that i is the last new point. For example, p0 = 0. 14. [R. Stanley] Choose X1,...,Xn from [0,1]. Let pn be the probability that Xi + X ... probability that the elevator is exactly on the 13th floor when Smith ...

  10. Probability

    2 fair 6-sided dice are rolled. What is the probability that the sum of these dice is 10 10? Solution: The event for which I obtain a sum of 10 is \ { (4,6), (6,4), (5,5) \} {(4,6),(6,4),(5,5)}. And there is a total of 6^2 = 36 62 = 36 possible outcomes. Thus the probability is simply \frac3 {36} = \frac1 {12} \approx 0.0833 363 = 121 ≈ 0.0833

  11. Probability Problems (video lessons, examples and solutions)

    We will now look at some examples of probability problems. Example: At a car park there are 100 vehicles, 60 of which are cars, 30 are vans and the remainder are lorries. If every vehicle is equally likely to leave, find the probability of: a) a van leaving first. b) a lorry leaving first.

  12. Simple probability (practice)

    Lesson 1: Basic probability Math > 7th grade > Statistics and probability > Basic probability Simple probability Google Classroom You might need: Calculator Jake is going to call one person from his contacts at random. He has 30 total contacts. 16 of those contacts are people he met at school. What is P (call a person from school) ?

  13. Probability

    The probability of any event is a value between (and including) "0" and "1". Follow the steps below for calculating probability of an event A: Step 1: Find the sample space of the experiment and count the elements. Denote it by n (S). Step 2: Find the number of favorable outcomes and denote it by n (A).

  14. Probability in Maths

    Math Article Probability Probability Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one. Probability has been introduced in Maths to predict how likely events are to happen.

  15. Probability Practice Problems

    1. On a six-sided die, each side has a number between 1 and 6. What is the probability of throwing a 3 or a 4? 1 in 6 1 in 3 1 in 2 1 in 4 2. Three coins are tossed up in the air, one at a time. What is the probability that two of them will land heads up and one will land tails up? 0 1/8 1/4 3/8 3. A two-digit number is chosen at random.

  16. Probabilities: Problems with Solutions

    Throw 2 dices simultaneously. What is the probability that the summation of the numbers is multiply of 4?

  17. Probability Questions

    1. Two coins are tossed 500 times, and we get: Two heads: 105 times One head: 275 times No head: 120 times Find the probability of each event to occur. Solution: Let us say the events of getting two heads, one head and no head by E 1, E 2 and E 3, respectively. P (E 1) = 105/500 = 0.21 P (E 2) = 275/500 = 0.55 P (E 3) = 120/500 = 0.24

  18. Probability Distribution

    For example, the following notation means "the random variable X follows a normal distribution with a mean of µ and a variance of σ 2 .". There are two types of probability distributions: Discrete probability distributions. Continuous probability distributions.

  19. 5.3: Probability Rules- "And" and "Or"

    Example \(\PageIndex{3}\): Additional Rule for Drawing Cards. A single card is drawn from a well shuffled deck of 52 cards. Find the probability that the card is a club or a face card. ... Using the complement to calculate the probability can simplify the problem considerably. The following example will help you understand the formula.

  20. 4.3: The Addition and Multiplication Rules of Probability

    The multiplication rule and the addition rule are used for computing the probability of A and B, and the probability of A or B for two given events A, B. ... Carlos makes the first and second goals with probability 0.585. b. The problem is asking you to find \(P(\text{A OR B})\). ... Refer to the information in Example \(\PageIndex{4 ...

  21. Statistics and probability

    This introduction to probability and statistics explores probability models, sample spaces, compound events, random samples, and a whole lot more. Basic probability. Learn. Statistics and probability FAQ ... Comparing distributions with dot plots (example problem) (Opens a modal) Practice. Valid claims Get 3 of 4 questions to level up!

  22. Probability

    Probability of getting no head = P(all tails) = 1/32. P(at least one head) = 1 - P(all tails) = 1 - 1/32 = 31/32. Sample Probability questions with solutions. Probability Example 1. What is the probability of the occurrence of a number that is odd or less than 5 when a fair die is rolled. Solution

  23. 10 Examples of Using Probability in Real Life

    Perhaps the most common real life example of using probability is weather forecasting. Probability is used by weather forecasters to assess how likely it is that there will be rain, snow, clouds, etc. on a given day in a certain area.

  24. Please provide examples of the question. . 4. Submit Table 4

    -Example Improvement: Adjust the frying time or temperature for the 867 sample, or explore using an oil with a lighter mouthfeel to reduce perceived oiliness. Crunchiness - Example Question: "Considering that the crunchiness of the 736 sample is not meeting the 'just right' expectation, being too crunchy, what adjustment should be made?"