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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

Photo credit: www.figurerealm.com

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 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.

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Unit 7: Probability

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. Go deeper with your understanding of probability as you learn about theoretical, experimental, and compound probability, and investigate permutations, combinations, and more!

Basic theoretical probability

• Intro to theoretical probability (Opens a modal)
• Probability: the basics (Opens a modal)
• Simple probability: yellow marble (Opens a modal)
• Simple probability: non-blue marble (Opens a modal)
• Intuitive sense of probabilities (Opens a modal)
• The Monty Hall problem (Opens a modal)
• Simple probability Get 5 of 7 questions to level up!
• Comparing probabilities Get 5 of 7 questions to level up!

Probability using sample spaces

• Probability with counting outcomes (Opens a modal)
• Example: All the ways you can flip a coin (Opens a modal)
• Die rolling probability (Opens a modal)
• Subsets of sample spaces (Opens a modal)
• Subsets of sample spaces Get 3 of 4 questions to level up!

Basic set operations

• Intersection and union of sets (Opens a modal)
• Relative complement or difference between sets (Opens a modal)
• Universal set and absolute complement (Opens a modal)
• Subset, strict subset, and superset (Opens a modal)
• Bringing the set operations together (Opens a modal)
• Basic set notation Get 5 of 7 questions to level up!

Experimental probability

• Experimental probability (Opens a modal)
• Theoretical and experimental probabilities (Opens a modal)
• Making predictions with probability (Opens a modal)
• Simulation and randomness: Random digit tables (Opens a modal)
• Experimental probability Get 5 of 7 questions to level up!
• Making predictions with probability Get 5 of 7 questions to level up!

Randomness, probability, and simulation

• Experimental versus theoretical probability simulation (Opens a modal)
• Theoretical and experimental probability: Coin flips and die rolls (Opens a modal)
• Random number list to run experiment (Opens a modal)
• Random numbers for experimental probability (Opens a modal)
• Statistical significance of experiment (Opens a modal)
• Interpret results of simulations Get 3 of 4 questions to level up!

Addition rule

• Probability with Venn diagrams (Opens a modal)
• Addition rule for probability (Opens a modal)
• Addition rule for probability (basic) (Opens a modal)
• Adding probabilities Get 3 of 4 questions to level up!
• Two-way tables, Venn diagrams, and probability Get 3 of 4 questions to level up!

Multiplication rule for independent events

• Sample spaces for compound events (Opens a modal)
• Compound probability of independent events (Opens a modal)
• Probability of a compound event (Opens a modal)
• "At least one" probability with coin flipping (Opens a modal)
• Free-throw probability (Opens a modal)
• Three-pointer vs free-throw probability (Opens a modal)
• Probability without equally likely events (Opens a modal)
• Independent events example: test taking (Opens a modal)
• Die rolling probability with independent events (Opens a modal)
• Probabilities involving "at least one" success (Opens a modal)
• Sample spaces for compound events Get 3 of 4 questions to level up!
• Independent probability Get 3 of 4 questions to level up!
• Probabilities of compound events Get 3 of 4 questions to level up!
• Probability of "at least one" success Get 3 of 4 questions to level up!

Multiplication rule for dependent events

• Dependent probability introduction (Opens a modal)
• Dependent probability: coins (Opens a modal)
• Dependent probability example (Opens a modal)
• Independent & dependent probability (Opens a modal)
• The general multiplication rule (Opens a modal)
• Dependent probability (Opens a modal)
• Dependent probability Get 3 of 4 questions to level up!

Conditional probability and independence

• Calculating conditional probability (Opens a modal)
• Conditional probability explained visually (Opens a modal)
• Conditional probability using two-way tables (Opens a modal)
• Conditional probability tree diagram example (Opens a modal)
• Tree diagrams and conditional probability (Opens a modal)
• Conditional probability and independence (Opens a modal)
• Analyzing event probability for independence (Opens a modal)
• Calculate conditional probability Get 3 of 4 questions to level up!
• Dependent and independent events Get 3 of 4 questions to level up!

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?

Solved Probability Problems

Solved probability problems and solutions are given here for a concept with clear understanding.

Students can get a fair idea on the probability questions which are provided with the detailed step-by-step answers to every question.

Solved probability problems with solutions :

The graphic above shows a container with 4 blue triangles, 5 green squares and 7 red circles. A single object is drawn at random from the container.

Match the following events with the corresponding probabilities:

Number of blue triangles in a container = 4

Number of green squares = 5

Number of red circles = 7

Total number of objects = 4 + 5 + 7 = 16

(i) The objects is not a circle:

P(the object is a circle)

= Number of circles/Total number of objects

P(the object is not a circle)

= 1 - P(the object is a circle)

= (16 - 7)/16

(ii) The objects is a triangle:

P(the object is a triangle)

= Number of triangle/Total number of objects

(iii) The objects is not a triangle:

= Number of triangles/Total number of objects

P(the object is not a triangle)

= 1 - P(the object is a triangle)

= (16 - 4)/16

(iv) The objects is not a square:

P(the object is a square)

= Number of squares/Total number of objects

P(the object is not a square)

= 1 - P(the object is a square)

= (16 - 5)/16

(v) The objects is a circle:

(vi) The objects is a square:

Match the following events with the corresponding probabilities are shown below:

2. A single card is drawn at random from a standard deck of 52 playing cards.

Match each event with its probability.

Note: fractional probabilities have been reduced to lowest terms. Consider the ace as the highest card.

Total number of playing cards = 52

(i) The card is a diamond:

Number of diamonds in a deck of 52 cards = 13

P(the card is a diamond)

= Number of diamonds/Total number of playing cards

(ii) The card is a red king:

Number of red king in a deck of 52 cards = 2

P(the card is a red king)

= Number of red kings/Total number of playing cards

(iii) The card is a king or queen:

Number of kings in a deck of 52 cards = 4

Number of queens in a deck of 52 cards = 4

Total number of king or queen in a deck of 52 cards = 4 + 4 = 8

P(the card is a king or queen)

= Number of king or queen/Total number of playing cards

(iv) The card is either a red card or an ace:

Total number of red card or an ace in a deck of 52 cards = 28

P(the card is either a red card or an ace)

= Number of cards which is either a red card or an ace/Total number of playing cards

(v) The card is not a king:

P(the card is a king)

= Number of kings/Total number of playing cards

P(the card is not a king)

= 1 - P(the card is a king)

= (13 - 1)/13

(vi) The card is a five or lower:

Number of cards is a five or lower = 16

P(the card is a five or lower)

= Number of card is a five or lower/Total number of playing cards

(vii) The card is a king:

(viii) The card is black:

Number of black cards in a deck of 52 cards = 26

P(the card is black)

= Number of black cards/Total number of playing cards

3. A bag contains 3 red balls and 4 black balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is 

(ii) not black.

(i) Total number of possible outcomes = 3 + 4 = 7.

Number of favourable outcomes for the event E.

                              = Number of black balls = 4.

So, P(E) = \(\frac{\textrm{Number of Favourable Outcomes for the Event E}}{\textrm{Total Number of Possible Outcomes}}\)

             = \(\frac{4}{7}\).

(ii) The event of the ball being not black = \(\bar{E}\).

Hence, required probability = P(\(\bar{E}\))

                                        = 1 - P(E)

                                        = 1 - \(\frac{4}{7}\)

                                        = \(\frac{3}{7}\).

4. If the probability of Serena Williams a particular tennis match is 0.86, what is the probability of her losing the match?

Let E = the event of Serena Williams winning.

From the question, P(E) = 0.86.

Clearly, \(\bar{E}\) = the event of Serena Williams losing.

So, P(\(\bar{E}\)) = 1 - P(E) 

                            = 1 - 0.86

                            = 0.14

                            = \(\frac{14}{100}\)

                            = \(\frac{7}{50}\).

5. Find the probability of getting 53 Sunday in a leap year.

A leap year has 366 days. So, it has 52 weeks and 2 days.

So, 52 Sundays are assured. For 53 Sundays, one of the two remaining days must be a Sunday. 

For the remaining 2 days we can have

(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday).

So, total number of possible outcomes = 7.

Number of favourable outcomes for the event E = 2,   [namely, (Sunday, Monday), (Saturday, Sunday)].

So, by definition: P(E) = \(\frac{2}{7}\).

6. A lot of 24 bulbs contains 25% defective bulbs. A bulb is drawn at random from the lot. It is found to be not defective and it is not put back. Now, one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

25% of 24 = \(\frac{25}{100}\) × 24 = 6.

So, there are 6 defective bulbs and 18 bulbs are not defective. 

After the first draw, the lot is left with 6 defective bulbs and 17 non-defective bulbs.

So, when the second bulb is drwn, the total number of possible outcomes = 23  (= 6+ 17).

Number of favourable outcomes for the event E = number of non-defective bulbs = 17.

So, the required probability = P(E) = (\frac{17}{23}\).

The examples can help the students to practice more questions on probability by following the concept provided in the solved probability problems.

• Probability

Random Experiments

Experimental Probability

Events in Probability

Empirical Probability

Coin Toss Probability

Probability of Tossing Two Coins

Probability of Tossing Three Coins

Complimentary Events

Mutually Exclusive Events

Mutually Non-Exclusive Events

Conditional Probability

Theoretical Probability

Odds and Probability

Playing Cards Probability

Probability and Playing Cards

Probability for Rolling Two Dice

Probability for Rolling Three Dice

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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

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

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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Browse Course Material

Course info, instructors.

• Dr. Jeremy Orloff
• Dr. Jennifer French Kamrin

Departments

• Mathematics

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• Discrete Mathematics
• Probability and Statistics

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Introduction to probability and statistics, problem sets with solutions.

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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,

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).

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.

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.

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
• Probability and Statistics
• 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).

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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?

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

Get this 15 probability questions for your middle and high school students. Allows students tackle problems at their own pace and includes an answer key!

How to calculate probabilities

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)

Probability question: A worked example

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.

6th grade probability questions

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?

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} .

7th grade probability questions

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.

8th grade probability questions

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.

9th grade probability questions

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 .

10th grade probability questions

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:

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

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• Pythagorean theorem questions

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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.

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Mathematics > Probability

Title: fast mixing in sparse random ising models.

Abstract: Motivated by the community detection problem in Bayesian inference, as well as the recent explosion of interest in spin glasses from statistical physics, we study the classical Glauber dynamics for sampling from Ising models with sparse random interactions. It is now well-known that when the interaction matrix has spectral diameter less than $1$, Glauber dynamics mixes in $O(n\log n)$ steps. Unfortunately, such criteria fail dramatically for interactions supported on arguably the most well-studied sparse random graph: the Erdős--Rényi random graph $G(n,d/n)$, due to the presence of almost linearly many outlier eigenvalues of unbounded magnitude. We prove that for the \emph{Viana--Bray spin glass}, where the interactions are supported on $G(n,d/n)$ and randomly assigned $\pm\beta$, Glauber dynamics mixes in $n^{1+o(1)}$ time with high probability as long as $\beta \le O(1/\sqrt{d})$, independent of $n$. We further extend our results to random graphs drawn according to the $2$-community stochastic block model, as well as when the interactions are given by a "centered" version of the adjacency matrix. The latter setting is particularly relevant for the inference problem in community detection. Indeed, we build on this result to demonstrate that Glauber dynamics succeeds at recovering communities in the stochastic block model in an upcoming paper. The primary technical ingredient in our proof is showing that with high probability, a sparse random graph can be decomposed into two parts --- a \emph{bulk} which behaves like a graph with bounded maximum degree and a well-behaved spectrum, and a \emph{near-forest} with favorable pseudorandom properties. We then use this decomposition to design a localization procedure that interpolates to simpler Ising models supported only on the near-forest, and then execute a pathwise analysis to establish a modified log-Sobolev inequality.

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