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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%

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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Course info, instructors.

  • Dr. Jeremy Orloff
  • Dr. Jennifer French Kamrin

Departments

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  • Probability and Statistics

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Introduction to probability and statistics, exams with solutions.

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  • Practice Test Questions

Basic Probability and Statistics Quick Review and Practice Questions

  • Posted by Brian Stocker
  • Date August 15, 2017
  • Comments 5 comments

Simple Probability and Statistics – A Quick Review

The probability of an event is given by –

The Number Of Ways Event A Can Occur The total number Of Possible Outcomes

So for example if there are 4 red balls and 3 yellow balls in a bag, the probability of choosing a red ball will be  4/7

Another example:

In a certain game, players toss a coin and roll a dice. A player wins if the coin comes up heads, or the dice with a number greater than 4. In 20 games, how many times will a player win?

a. 13 b. 8 c. 11 d. 15

Correct Answer: A

First determine the possible number of outcomes, the sample space of this event will be:

S = { (H,1),(H,2),(H,3),(H,4),(H,5),(H,6) (T,1),(T,2),(T,3),(T,4),(T,5),(T,6) }

So there are a total of 12 outcomes and 8 winning outcomes.  The probability of a win in a single event is P (W)

= 8/12 = 2/3. In 20 games the probability of a win = 2/3 × 20 = 13

practice problems for probability

Basic Probability and Statistics Practice Questions

1. There are 3 blue, 1 white and 4 red identical balls inside a bag. If it is aimed to take two balls out of the bag consecutively, what is the probability to have 1 blue and 1 white ball?

a. 3/28 b. 1/12 c. 1/7 d. 3/7

2. A boy has 4 red, 5 green and 2 yellow balls. He  chooses two balls  randomly for play. What is the probability  that one is red and other is green?

a. 2/11 b. 19/22 c. 20/121 d. 9/11

3. There are 5 blue, 5 green and 5 red books on a shelf.  Two books are selected randomly. What is the probability  of choosing two books of different colors?

a. 1/3 b. 2/5 c. 4/7 d. 5/7

4. How many different ways can a reader choose 3 books out of 4, ignoring the order of selection?

a. 3 b. 4 c. 9 d. 12

5. There is a die and a coin. The dice is rolled and the coin is flipped according to the number the die is rolled. If the die is rolled only once, what is the probability of 4 successive heads?

a. 3/64 b. 1/16 c. 3/16 d. 1/4

6. Smith and Simon are playing a card game. Smith will win if the drawn card form the deck of 52 is either 7 or a diamond, and Simon will win if the drawn card is an even number. Which statement is more likely to be correct?

a. Smith will win more games. b. Simon will win more games. c. They have same winning probability. d. Decision could not be made from the provided data.

7. A box contains 30 red, green and blue balls. The probability of drawing a red ball is twice the other colors due to its size. The number of green balls are 3 more than twice the number of blue balls, and blue are 5 less  than the twice the red. What is the probability that 1 st  two balls drawn from the box randomly will be red? 

a. 10/102 b. 11/102 c. 1/29 d. 1/30

8. Sarah has two children and we know that she has a daughter. What is the probability that the other child is a girl as well?

a. 1/4 b. 1/3 c. 1/2 d. 1

Basic Probability and Statistics Answer Key

1. A There are 8 balls in the bag in total. It is important that two balls are taken out of the bag one by one. We can first take the blue then the white, or first white, then the blue. So, we will have two possibilities to be summed up. Since the balls are taken consecutively, we should be careful with the total number of balls for each case:

First blue, then white ball: There are 3 blue balls; so, having a blue ball is 3/8 possible. Then, we have 7 balls left in the bag. The possibility to have a white ball is 1/7.

P = (3/8) * (1/7) = 3/56

First white, then blue ball: There is only 1 white ball; so, having a white ball is 1/8 possible. Then, we have 7 balls left in the bag. The possibility to have a blue ball is 3/7.

P = (1/8) * (3/7) = 3/56

Overall probability is: 3/56 + 3/56 = 3/28

2. A Probability that the 1st ball is red: 4/11 Probability the 2nd ball is green: 5/10 Combined probability is 4/11 * 5/10 = 20/110 = 2/11

3. D Assume that the first book chosen is red. Since we need to choose the second book in green or blue, there are 10 possible books to be chosen out of 15 – 1(that is the red book chosen first) = 14 books. There are equal number of books in each color, so the results will be the same if we think that blue or green book is the first book.

So, the probability will be 10/14 = 5/7.

4. B Ignoring the order means this is a combination problem, not permutation. The reader will choose 3 books out of 4. So, C(4, 3) = 4! / (3! * (4 – 3)!) = 4! / (3! * 1!) = 4

There are 4 different ways. Ignoring the order means this is a combination problem, not permutation. The reader will choose 3 books out of 4. So,

C(4, 3) = 4! / (3! * (4 – 3)!) = 4! / (3! * 1!) = 4

There are 4 different ways.

5. A If the die is rolled for once, it can be 4, 5 or 6 since we are searching for 4 successive heads. We need to think each case separately. There are two possibilities for a coin; heads (H) or tails (T), each possibility of 1/2; we are searching for H. The possibility for a number to appear on the top of the die is 1/6. Die and coin cases are disjoint events. Also, each flip of coin is independent from the other:

Die: 4 coin: HHHH : 1 permutation P = (1/6) * (1/2) * (1/2) * (1/2) * (1/2) = (1/6) * (1/16)

Die: 5 coin: HHHHT, THHHH, HHHHH : 3 permutations P = (1/6) * 3 * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = (1/6) * (3/32)

Die: 6 coin: HHHHTT, TTHHHH, THHHHT, HHHHHT, THHHHH, HTHHHH, HHHHTH, HHHHHH : 8 permutations P = (1/6) * 8 * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = (1/6) * (8/64)

The overall probability is: Pall = (1/6) * (1/16) + (1/6) * (3/32) + (1/6) * (8/64) = (1/6) * (1/16 + 3/32 + 8/64) = (1/6) * (4 + 6 + 8) / 64 = (1/6) * (18/64) = 3/64

6. B There are 52 cards in total. If we closely observe Smith has  16 cards in which he can win. So his winning probability in a single game will be 16/52 on the other hand Simon has 20 cards of wining so his probability on win in single draw is 20/52.

7. A Let the number of red balls be x Then number of blue balls = 2x – 5 Then number of green balls= 2(2x – 5) + 3 = 4x – 10 + 3

= 4x – 7

As there are total 30 balls so the equation becomes x + 2x – 5 + 4x – 7 = 30 x = 6 Red balls are 6, blue are 7 and green are 17. As the probability of drawing a red ball is twice than the others, let’s take them as 12. So the total number of balls will be 36.

Probability of drawing the 1st red: 12/36 Probability of drawing the 2nd red: 10/34 Combined probability = 12/36 X 10/34 = 10/102

8. B At first glance; we can think that a child can be either a girl or a boy, so the probability for the other child to be a girl is 1/2. However, we need to think deeper. The combinations of two children can be as follows:

boy + girl boy + boy girl + boy girl + girl So, the sample space is S = {BG, BB, GB, GG} where the sequence is important. Sarah has a girl; this is the fact. So, calling this as event A, here are the possibilities:

boy + girl girl + boy girl + girl

We eliminate boy + boy, since one child is a girl. A = {BG, GB, GG} The event that Sarah has two girls: B = {GG} We need to compute: P(B|A. = P(B ∩ A. / P(A. = 1/3

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Hi there – I will be back to read much more

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The answer to number 2 should be 4/11, which isn’t an option. The wording implies that the order in which the balls are picked out doesn’t matter. The balls can be picked out as red then green, or green then red.

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The answer is correct – Combined probability is 4/11 * 5/10 = 20/110 = 10/55 = 2/11

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I don’t get it, in #7, why can you take the number of red balls as 12?

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4th is confusing. i need more explanation pls.

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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.

  • Probability Practice Questions

Probability Practice Questions section is here and we have collected all the different question types for you. The Probability Practice Questions has questions that will check your grasp on the concepts of probability and also provide you with the crucial practice opportunity. Let us see more.

Browse more Topics Under Probability

  • Probability of Random Event
  • Mutually Exclusive Events
  • Equally Likely Events
  • Independent Events
  • Compound Events
  • Total Probability

Q1: In a throw of a coin, the probability of getting a head is?

A) 1            B) 1/2                 C) 1/4                    D) 2

Q2: Two unbiased coins are tossed. What is the probability of getting at most one head?

A) 2/3       B) 1/2             C) 3/4             D) 4/3

Q3: An unbiased die is tossed. Find the probability of getting a multiple of 3.

A) 1/4            B) 1/3               C) 1/2           D) 1

Q4: In a simultaneous throw of a pair of dice, find the probability of getting a total more than 7.

A) 3/2           B) 4/7            C) 5/12           D) 6/13

Q5: A bag contains 6 white and 4 black balls. two balls are drawn at random. Find the probability that they are of the same color.

A) 3/4         B) 5/3           C) 7/15             D) 8/17

Q6: Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is divisible by 4 or 6?

A) 13/14         B) 5/3           C) 7/16             D) 7/18

Q7: Two cards are drawn at random from a pack of 52 cards. What is the probability that either both are black or both are queens?

A) 5/21       B) 55/221              C) 555/2221          D) 5555/22221

Q8: A box contains 5 green, 4 yellow and 3 white marbles. Three marbles are drawn at random. What is the probability that they are not of the same color?           [Bank P.O. 2000]

A) 3/44          B) 3/55           C) 52/55           D) 41/44

Q9: A bag contains 4 white, 5 red, and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red is:      [M.B.A. 2002]

A) 1/22           B) 3/22        C) 2/91             D) 2/77

Find Your Answers Here

Q1: B), Q2: C), Q3: B), Q4: C), Q5: C), Q6: D), Q7: B), Q8: D), Q9: C)

Q1: A bag contains 2 red, 3 green, and 2 blue balls. two balls are drawn at random. What is the probability that none of the balls drawn is blue?             [Bank PO 2003]

A) 10/21          B) 11/21         C) 2/7           D) 5/7

Q2: A bag contains 6 white and 4 red balls. Three balls are drawn at random. What is the probability that one ball is red and the other two are white?

A) 1/2            B) 1/12          C) 3/10            D) 7/12

Q3: In a box, there are 8 red, 7 blue, and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?            [Bank P.O. 2002]

A) 2/3          B) 3/4             C) 7/19           D) 8/21          E) 9/21

Q4: A box contains 10 black and 10 white balls. The probability of drawing two balls of the same color is:

A) 9/19         B) 9/38           C) 10/19          D) 5/19

Q5: A box contains 4 red balls, 5 green balls, and 6 white balls. A ball is drawn at random from the box. What is the probability that the ball is drawn is either red or green?

A) 2/5       B0 3/5         C) 1/5          D) 7/15

Q6: In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

A) 21/46               B) 25/117            C) 1/50             D) 3/25

Q7: Four persons are chosen at random from a group of 3 men, 2 women, and 4 children. The chance that exactly 2 of them are children , is:

A) 1/9               B) 1/5            C) 1/12              D) 10/21

Probability Practice Questions

Q1: A), Q2: A), Q3: D), Q4: A), Q5: B), Q6: A), Q7: D)

Q1: A box contains 20 electric bulbs , out of which 4 are defective. Two bulbs are chosen at random from this box. the probability that at least one of these is defective is:

A) 4/19          B) 7/19         C) 12/19        D) 21/95

Q2: In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both. If a student is selected at random, what is the probability that he has offered English or Hindi?

A) 2/5            B) 3/4                  C) 3/5                  D) 3/10

Q3: Two dice are tossed. the probability that a total score is a prime number is:

A) 1/6          B) 5/12            C) 1/2              D) 7/9

Q4: A speaks the truth in 75% of cases and B in 80% of the cases. In what percentage of cases are they likely to contradict each other, narrating the same incident?    [ Bank PO 2000]

A) 5%            B) 15%           C) 35%           D) 45%

Q5: A man and his wife appear in an interview for two vacancies in the same post. The probability of husband’s selection is (1/7) and the probability of wife’s selection is (1/5). What is the probability that only one of them is selected?

A) 4/5          B) 2/7               C) 8/15            D) 4/7

Q6: From a pack of 52 cards, one card is drawn at random. What is the probability that the card drawn is a ten or a spade?

A) 4/13            B) 1/4         C) 1/13         D) 1/26

Q7: The probability that a card drawn from a pack of 52 cards will be a diamond or a king is:

A) 2/13          B) 4/13      C) 1/13            D) 1/52

Q8: From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?        [M.B.A. 2002, Railways, 2002]

A) 1/15           B) 25/57            C) 35/256            D) 1/221

Q9: Two cards are drawn together from a pack of 52 cards. The probability that one is spade and one is a heart is:            [MBA 2000]

A) 3/20          B) 29/34            C) 47/100       D) 13/102

Q1: B), Q2: A), Q3: B), Q4: C), Q5: B), Q6: A), Q7: B), Q8: D), Q9: D)

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15 Probability Questions And Practice Problems (KS3 & KS4): Harder GCSE 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 KS3 and KS4 including several GCSE exam style questions.

What are some real life examples of probability?

How to calculate probabilities, year 7 probability questions, year 8 probability questions, year 9 probability questions, year 10 probability questions, gcse foundation probability questions, gcse higher probability questions, looking for more probability questions and resources, looking for more ks3 and ks4 maths questions.

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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.

The probability of something happening is given by:

We can also use the following formulae 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)

Download this 15 Probability Questions And Practice Problems (KS3 & KS4) Worksheet

Help your students prepare for their Maths GSCE with this free Probability worksheet of 15 multiple choice questions and answers.

KS3 probability questions

In KS3 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?

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 labelled 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 blue 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 blue ball is 4x. Work out how many blue 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 maths 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.

KS4 probability questions

In KS4 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

4 + 3 = 7

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 colour 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 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}

13. 200 people were asked about which athletics 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:

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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.

practice problems for probability

Probability Important Questions

practice 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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Statistics Problems

One of the best ways to learn statistics is to solve practice problems. These problems test your understanding of statistics terminology and your ability to solve common statistics problems. Each problem includes a step-by-step explanation of the solution.

  • Use the dropdown boxes to describe the type of problem you want to work on.
  • click the Submit button to see problems and solutions.

Main topic:

Problem description:

In one state, 52% of the voters are Republicans, and 48% are Democrats. In a second state, 47% of the voters are Republicans, and 53% are Democrats. Suppose a simple random sample of 100 voters are surveyed from each state.

What is the probability that the survey will show a greater percentage of Republican voters in the second state than in the first state?

The correct answer is C. For this analysis, let P 1 = the proportion of Republican voters in the first state, P 2 = the proportion of Republican voters in the second state, p 1 = the proportion of Republican voters in the sample from the first state, and p 2 = the proportion of Republican voters in the sample from the second state. The number of voters sampled from the first state (n 1 ) = 100, and the number of voters sampled from the second state (n 2 ) = 100.

The solution involves four steps.

  • Make sure the sample size is big enough to model differences with a normal population. Because n 1 P 1 = 100 * 0.52 = 52, n 1 (1 - P 1 ) = 100 * 0.48 = 48, n 2 P 2 = 100 * 0.47 = 47, and n 2 (1 - P 2 ) = 100 * 0.53 = 53 are each greater than 10, the sample size is large enough.
  • Find the mean of the difference in sample proportions: E(p 1 - p 2 ) = P 1 - P 2 = 0.52 - 0.47 = 0.05.

σ d = sqrt{ [ P1( 1 - P 1 ) / n 1 ] + [ P 2 (1 - P 2 ) / n 2 ] }

σ d = sqrt{ [ (0.52)(0.48) / 100 ] + [ (0.47)(0.53) / 100 ] }

σ d = sqrt (0.002496 + 0.002491) = sqrt(0.004987) = 0.0706

z p 1 - p 2 = (x - μ p 1 - p 2 ) / σ d = (0 - 0.05)/0.0706 = -0.7082

Using Stat Trek's Normal Distribution Calculator , we find that the probability of a z-score being -0.7082 or less is 0.24.

Therefore, the probability that the survey will show a greater percentage of Republican voters in the second state than in the first state is 0.24.

See also: Difference Between Proportions

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  • Probability Practice Questions: Level 01

practice problems for probability

  • Determine the probability that a digit chosen at random from digits 1, 2, 3, ….. 13 will be even. 1. 1/2 2. 1/9 3. 5/9 4. 6/13 Answer & Explanation Sol: Option 4 Explanation: Here out of six digits (2, 4, 6, 8,10, 12), any digits chosen will be considered favorable. Probability = 6 / 13
  • A coin is tossed four times, if H = head and T = tail, what is the probability of the tosses coming up in the order HTHH? 1. 3/16 2. 1/16 3. 5/16 4. 7/16 Answer & Explanation Sol: Option 2 Explanation: Coin is tossed four times. The total possible outcomes = 16. Favorable outcomes = 1.Because HTHH can come only in one way, Probability = 1 / 16
  • A dice is thrown, what is the probability that the number obtained is a prime number. 1. 1/6 2. 1/8 3. 1/2 4. 1/3 Answer & Explanation Sol: Option 3 Explanation: Dice is thrown, the total possible outcomes = 6. Favorable outcomes = 3 i.e. (2,3,5). Probability = 3 / 6 = 1 / 2
  • Find the probability of throwing a total of 8 in a single throw with two dice. 1. 1/36 2. 5/36 3. 25/36 4. 12/36 Answer & Explanation Sol: Option 2 Explanation: Two Dice are thrown, the total possible outcomes = 36. Favorable outcomes = 5 i.e. (2, 6), (6, 2), (3, 5), (5, 3), (4, 4). Therefore, Probability = 5 / 36
  • Probability: Concepts & Tricks
  • Probability: Solved Examples
  • Probability Practice Questions: Level 02
  • A card is drawn from a pack of 52 cards. What is the probability that the card is a Queen? 1. 1/52 2. 1/4 3. 1/16 4. None of these Answer & Explanation Sol: Option 4 Explanation: Total possible outcomes = 52. Favorable outcomes = 4. Probability = 4 / 52 = 1 / 13
  • Two cards are drawn in succession from a pack of 52 cards, without replacement. What is the probability, that the first is a Queen and the second is a Jack of a different suit? 1. 1/52 2. 1/13 3. 4/13 4. 1/221 Answer & Explanation Suggested Action: Kick start Your Preparations with FREE access to 25+ Mocks, 75+ Videos & 100+ Chapterwise Tests. Sign Up Now Sol: Option 4 Explanation: The probability of first Queen = 4 / 52 The probability of Second Jack of different suit = 3 / 51 Reqd. Probability = (4/52) x (3/51) = (1/13) x (1/17) = (1/221)
  • Two dice are thrown, what is the probability that both the dices are not having the same number. 1. 1/4 2. 5/6 3. 1/9 4. 1/12 Answer & Explanation Sol: Option 2 Explanation: Total possible outcomes = 36. Favorable outcomes of having same number = 6 [(1, 1) (2, 2) (3, 3) (4, 4) (5, 5) (6, 6)]. Probability of both the dices having same number = 6 / 36 = 1 / 6 Required probability = 1 - 1 / 6 = 5 / 6

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Mathematics LibreTexts

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

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  • 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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Unit 7: Probability

About this unit, estimating probabilities using simulation.

  • Intro to theoretical probability (Opens a modal)
  • 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)
  • Interpret results of simulations Get 3 of 4 questions to level up!

Mutually exclusive events and unions of events

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

Conditional probability

  • Conditional probability and independence (Opens a modal)
  • Conditional probability with Bayes' Theorem (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)
  • Calculate conditional probability Get 3 of 4 questions to level up!

Independent versus dependent events and the multiplication rule

  • Compound probability of independent events (Opens a modal)
  • Independent events example: test taking (Opens a modal)
  • General multiplication rule example: independent events (Opens a modal)
  • Dependent probability introduction (Opens a modal)
  • General multiplication rule example: dependent events (Opens a modal)
  • The general multiplication rule (Opens a modal)
  • "At least one" probability with coin flipping (Opens a modal)
  • Probabilities involving "at least one" success (Opens a modal)
  • Probability with general multiplication rule Get 3 of 4 questions to level up!
  • Probability of "at least one" success Get 3 of 4 questions to level up!

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  1. Probability Practice Problems

    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.

  2. 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.

  3. 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.

  4. Statistics and Probability

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  5. Probability

    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; the probability of the coin landing T is ½ . Throwing Dice

  6. Simple probability (practice)

    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) ? If necessary, round your answer to 2 decimal places.

  7. Exams with Solutions

    18.05 Introduction to Probability and Statistics (S22), Practice Final Exam Solutions. 18.05 Introduction to Probability and Statistics (S22), Practice Post Exam 2 Solutions. MIT OpenCourseWare is a web based publication of virtually all MIT course content. OCW is open and available to the world and is a permanent MIT activity.

  8. Probability: the basics (article)

    Tips The probability of an event can only be between 0 and 1 and can also be written as a percentage. The probability of event A is often written as P ( A) . If P ( A) > P ( B) , then event A has a higher chance of occurring than event B . If P ( A) = P ( B)

  9. PDF Twenty problems in probability

    necessarily in order) A,B,C so that A ≤ B ≤ C. Let an be the probability that A = B = C and let bn be the probability that B = A+1 and C = B +1. Show that for every n ≥ 1, either 4an ≤ bn or 4an+1 ≤ bn+1. 16. [Putnam Exam] Four points are chosen on the unit sphere. What is the probability that the

  10. Basic Probability and Statistics Quick Review and Practice Questions

    Correct Answer: A First determine the possible number of outcomes, the sample space of this event will be: S = { (H,1), (H,2), (H,3), (H,4), (H,5), (H,6) (T,1), (T,2), (T,3), (T,4), (T,5), (T,6) } So there are a total of 12 outcomes and 8 winning outcomes. The probability of a win in a single event is P (W) = 8/12 = 2/3.

  11. Probabilities: Problems with Solutions

    Probabilities Easy Normal Probabilities: Problems with Solutions Problem 1 Throw a dice 3 times. What's the probability that we have three 6? \displaystyle P (A) = \frac {1} {6^3} P (A) = 631 \displaystyle P (A) = \frac {1} {6^2} P (A) = 621 \displaystyle P (A) = \frac {1} {6} P (A) = 61 \displaystyle P (A) = \frac {1} {3} P (A) = 31 Problem 2

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    Part A Q1: In a throw of a coin, the probability of getting a head is? A) 1 B) 1/2 C) 1/4 D) 2 Q2: Two unbiased coins are tossed. What is the probability of getting at most one head? A) 2/3 B) 1/2 C) 3/4 D) 4/3 Q3: An unbiased die is tossed. Find the probability of getting a multiple of 3. A) 1/4 B) 1/3 C) 1/2 D) 1

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  18. 15 Probability Questions And Practice Problems (KS3, KS4, GCSE)

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  20. Statistics Problems

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  21. Probability Practice Questions with Answers

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  22. 5.3: Probability Rules- "And" and "Or"

    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.

  23. Probability

    500 possible mastery points Mastered Proficient Familiar Attempted Not started Quiz Unit test About this unit If you're curious about the mathematical ins and outs of probability, you've come to the right unit! Here, we'll take a deep dive into the many ways we can calculate the likelihood of different outcomes.

  24. Probability Practice Problems Flashcards

    If the probability of getting turkey is .3 and the probability of getting ham is .4, what is the probability of getting turkey or ham if the probability of getting both is .12. We use the equation [P (AUB)= P (A) + P (B) - P (AΩB)] to find that the probability of getting either turkey or ham is .3 + .4 - .12 which equals .58. Study with ...