# Nwheatleyschaller s The Next Step...Conditional Probability

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1 CK-12 FOUNDATION Nwheatleyschaller s The Next Step...Conditional Probability Say Thanks to the Authors Click (No sign in required) Meery

2 To access a customizable version of this book, as well as other interactive content, visit CK-12 Foundation is a non-profit organization with a mission to reduce the cost of textbook materials for the K-12 market both in the U.S. and worldwide. Using an open-content, web-based collaborative model termed the FlexBook, CK-12 intends to pioneer the generation and distribution of high-quality educational content that will serve both as core text as well as provide an adaptive environment for learning, powered through the FlexBook Platform. Copyright 2011 CK-12 Foundation, The names CK-12 and CK12 and associated logos and the terms FlexBook, and FlexBook Platform, (collectively CK-12 Marks ) are trademarks and service marks of CK-12 Foundation and are protected by federal, state and international laws. Any form of reproduction of this book in any format or medium, in whole or in sections must include the referral attribution link (placed in a visible location) in addition to the following terms. Except as otherwise noted, all CK-12 Content (including CK-12 Curriculum Material) is made available to Users in accordance with the Creative Commons Attribution/Non-Commercial/Share Alike 3.0 Unported (CC-by-NC-SA) License ( as amended and updated by Creative Commons from time to time (the CC License ), which is incorporated herein by this reference. Complete terms can be found at Printed: November 13, 2011

3 Author Brenda Meery i

4 Contents 1 The Next Step...Conditional Probability What are Tree Diagrams? Order and Probability Conditional Probability Review Questions ii

5 Chapter 1 The Next Step...Conditional Probability Introduction This chapter builds on the concepts learned in the previous chapter on probability. Starting with tree diagrams as a means of displaying outcomes for various trials, we will learn how to read the diagrams and ﬁnd probabilities. We will also ﬁnd that order does not matter unless working with permutations. Permutations, such as the combination of your lock at the gym, require their own special formula. When outcomes for permutations have repetitions, these repetitions must also be included in the calculations in order to account for the multiple entries. Combinations, like permutations, also have their own special formula. Combinations, such as the number of teams of 4 that can be arranged in a class of 15 students, are diﬀerent from permutations, because the order in combinations is insigniﬁcant. In this chapter, we will also learn about conditional probability. Conditional probability comes into play when the probability of the second event occurring is dependent on the probability of the ﬁrst event. 1.1 What are Tree Diagrams? Learning Objectives Know the deﬁnition of conditional probability. Use conditional probability to solve for probabilities in ﬁnite sample spaces. 1

6 In the last chapter, we studied independent and dependent events, as well as mutually inclusive and mutually exclusive events. We used the Addition Rule for dependent events, as well as mutually inclusive and mutually exclusive events. The Addition Rule, or Addition Principle, is used to ﬁnd P(A or B), while the Multiplication Rule is used for independent events. Addition Rule For 2 events, A and B, the probability of selecting one event or another is given by: P(A or B) = P(A) + P(B) P(A and B). Multiplication Rule For 2 independent events, A and B, where the outcome of A does not change the probability of B, the probability of A and B is given by: P(A and B) = P(A) P(B). Tree diagrams are another way to show the outcomes of simple probability events. In a tree diagram, each outcome is represented as a branch on a tree. Let s say you were going to toss a coin 2 times and wanted to ﬁnd the probability of getting 2 heads. This is an example of independent events, because the outcome of one event does not aﬀect the outcome of the second event. What does this mean? Well, when you ﬂip the coin once, you have an equal chance of getting a head (H) or a tail (T). On the second ﬂip, you also have an equal chance of getting a a head or a tail. In other words, whether the ﬁrst ﬂip was heads or tails, the second ﬂip could just as likely be heads as tails. You can represent the outcomes of these events on a tree diagram. 2

7 From the tree diagram, you can see that the probability of getting a head on the first flip is 1 2. Starting with heads, the probability of getting a second head will again be 1 2. But how do we calculate the probability of getting 2 heads? These are independent events, since the outcome of tossing the first coin in no way affects the outcome of tossing the second coin. Therefore, we can calculate the probability as follows: P(A and B) = P(A and B) = 1 4 Therefore, we can conclude that the probability of getting 2 heads when tossing a coin twice is 1 4, or 25%. Let s try an example that is a little more challenging. Example 1 Irvin opens up his sock drawer to get a pair socks to wear to school. He looks in the sock drawer and sees 4 red socks, 8 white socks, and 6 brown socks. Irvin reaches in the drawer and pulls out a red sock. He is wearing blue shorts, so he replaces it. He then draws out a white sock. What is the probability that Irvin pulls out a red sock, replaces it, and then pulls out a white sock? First let s draw a tree diagram. 3

8 There are 18 socks in Irvin s sock drawer. The probability of getting a red sock when he pulls out the first sock is: P(red) = 4 18 P(red) = 2 9 Irvin puts the sock back in the drawer and pulls out the second sock. The probability of getting a white sock on the second draw is: P(white) = 6 18 P(white) = 1 3 Therefore, the probability of getting a red sock and then a white sock when the first sock is replaced is: P(red and white) = P(red and white) = 2 27 One important part of these types of problems is that order is not important. Let s say Irvin picked out a white sock, replaced it, and then picked out a red sock. probability. Calculate this P(white and red) = P(white and red) =

9 So regardless of the order in which he takes the socks out, the probability is the same. In other words, P(red and white) = P(white and red). Example 2 In Example 1, what happens if the first sock is not replaced? The probability that the first sock is red is: P(red) = 4 18 P(red) = 2 9 The probability of picking a white sock on the second pick is now: So now, the probability of selecting a red sock and then a white sock, without replacement, is: P(red and white) = P(red and white) = P(red and white) = 4 51 If the first sock is white, will P(red and white) = P(white and red) as we found in Example 1? Let s find out. P(white) = 6 18 P(white) = 1 3 The probability of picking a red sock on the second pick is now: As with the last example, P(red and white) = P(white and red). So when does order really matter? 5

10 1.2 Order and Probability Permutations and combinations are the next step in the learning of probability. It is by using permutations and combinations that we can find the probabilities of various events occurring at the same time, such as choosing 3 boys and 3 girls from a class of grade 12 math students. In mathematics, we use more precise language: If the order doesn t matter, it is a combination. If the order does matter, it is a permutation. Say, for example, you are making a salad. You throw in some lettuce, carrots, cucumbers, and green peppers. The order in which you throw in these vegetables doesn t really matter. Here we are talking about a combination. For combinations, you are merely selecting. Say, though, that Jack went to the ATM to get out some money and that he has to put in his PIN number. Here the order of the digits in the PIN number is quite important. In this case, we are talking about a permutation. For permutations, you are ordering objects in a specific manner. Permutations The Fundamental Counting Principle states that if an event can be chosen in p different ways and another independent event can be chosen in q different ways, the number of different ways the 2 events can occur is p q. In other words, the Fundamental Counting Principle tells you how many ways you can arrange items. Permutations are the number of possible arrangements in an ordered set of objects. Example 3 How many ways can you arrange the letters in the word MATH? You have 4 letters in the word, and you are going to choose 1 letter at a time. When you choose the first letter, you have 4 possibilities ( M, A, T, or H ). Your second choice will have 3 possibilities, your third choice will have 2 possibilities, and your last choice will have only 1 possibility. Therefore, the number of arrangements is: = 24 possible arrangements. The notation for a permutation is: n P r, where: n is the total number of objects. r is the number of objects chosen. For simplifying calculations, when n = r, then n P r = n!. The factorial function (!) requires us to multiply a series of descending natural numbers. Examples: Note: It is a general rule that 0! = 1. 5! = = 120 4! = = 24 1! = 1 With TI calculators, you can find the factorial function using: MATH (PRB) ( 4 ) Example 4 6

11 Solve for 4 P 4. 4P 4 = = 24 This represents the number of ways to arrange 4 objects that are chosen from a set of 4 different objects. Example 5 Solve for 6 P 3. Starting with 6, multiply the first 3 numbers of the factorial: 6P 3 = = 120 This represents the number of ways to arrange 3 objects that are chosen from a set of 6 different objects. The formula to solve permutations like these is: np r = n! (n r)! Look at Example 5 above. In this example, the total number of objects (n) is 6, while the number of objects chosen (r) is 3. We can use these 2 numbers to calculate the number of possible permutations (or the number of arrangements) of 6 objects chosen 3 at a time. np r = n! (n r)! 6! 6P 3 = (6 3)! 6P 3 = 6! 3! = P 3 = P 3 = 120 Example 6 What is the total number of possible 4-letter arrangements of the letters s, n, o, and w if each letter is used only once in each arrangement? In this problem, there are 4 letters to choose from, so n = 4. We want 4-letter arrangements; therefore, we are choosing 4 objects at a time. In this example, r =

12 Example 7 A committee is to be formed with a president, a vice president, and a treasurer. If there are 10 people to select from, how many committees are possible? In this problem, there are 10 committee members to choose from, so n = 10. We want to choose 3 members to be president, vice-president, and treasurer; therefore, we are choosing 3 objects at a time. In this example, r = 3. np r = n! (n r)! 10! 10P 3 = (10 3)! 10P 3 = 10! 7! 10P 3 = P 3 = 720 = Permutations with Repetition There is a subset of permutations that takes into account that there are double objects or repetitions in a permutation problem. In general, repetitions are taken care of by dividing the permutation by the factorial of the number of objects that are identical. If you look at the word TOOTH, there are 2 O s in the word. Both O s are identical, and it does not matter in which order we write these 2 O s, since they are the same. In other words, if we exchange O for O, we still spell TOOTH. The same is true for the T s, since there are 2 T s in the word TOOTH as well. If we were to ask the question, In how many ways can we arrange the letters in the word TOOTH? we must account for the fact that these 2 O s are identical and that the 2 T s are identical. We do this using the formula: np r x 1!x 2!, where x is the number of times a letter is repeated. 8

13 np r x 1!x 2! = 5 P 5 2!2! 5P 5 2!2! = P 5 2!2! = P 5 2!2! = 30 Tech Tip: Calculating Permutations on the Calculator Permutations (npr) Enter the n value. Press MATH. You should see menus across the top of the screen. You want the fourth menu: PRB (arrow right 3 times). You will see several options. npr is the second. Press 2. Enter the r value. Press ENTER. Example 8 Compute 9 P 5. 9 MATH (PRB) 2 (npr) 5 ENTER 9P 5 = 15, 120 Example 9 How many different 5-letter arrangements can be formed from the word APPLE? There are 5 letters in the word APPLE, so n = 5. We want 5-letter arrangements; therefore, we are choosing 5 objects at a time. In this example, r = 5, and we are using a word with letters that repeat. In the word APPLE, there are 2 P s, so x 1 = 2. Example 10 How many different 6-digit numerals can be written using the following 7 digits? Assume the repeated digits are all used. 3, 3, 4, 4, 4, 5, 6 9

14 There are 7 digits, so n = 7. We want 6-digit arrangements; therefore, we are choosing 6 objects at a time. In this example, r = 6, and we are using a group of digits with numbers that repeat. In the group of 7 digits (3, 3, 4, 4, 4, 5, 6), there are two 3 s and three 4 s, so x 1 = 2 and x 2 = 3. When there are no repetitions, remember that we use the standard permutation formula: np r = n! (n r)! Example 11 In how many ways can first and second place be awarded to 10 people? There are 10 people (n = 10), and there are 2 prize winners (r = 2). Example

15 In how many ways can 3 favorite desserts be listed in order from a menu of 10 (i.e., permutations without repetition)? There are 10 menu items (n = 10), and you are choosing 3 favorite desserts (r = 3) in order. Combinations If you think about the lottery, you choose a group of lucky numbers in hopes of winning millions of dollars. When the numbers are drawn, the order in which they are drawn does not have to be the same order as on your lottery ticket. The numbers drawn simply have to be on your lottery ticket in order for you to win. You can imagine how many possible combinations of numbers exist, which is why your odds of winning are so small! Combinations are arrangements of objects without regard to order and without repetition, selected from a distinct number of objects. A combination of n objects taken r at a time ( n C r ) can be calculated using the formula: nc r = n! r!(n r)! Example 13 Evaluate: 7 C

16 Example 14 7! 7C 2 = 2!(7 2)! 7C 2 = 7! 2!(5)! 7C 2 = (2 1)( ) 5, 040 7C 2 = (2)(120) 5, 040 7C 2 = 240 7C 2 = 21 In how many ways can 3 desserts be chosen in any order from a menu of 10? There are 10 menu items (n = 10), and you are choosing 3 desserts in any order (r = 3). 10! 10C 3 = 3!(10 3)! 10C 3 = 10! 3!(7)! 10C 3 = C 3 = 120 Example 15 There are 12 boys and 14 girls in Mrs. Cameron s math class. Find the number of ways Mrs. Cameron can select a team of 3 students from the class to work on a group project. The team must consist of 2 girls and 1 boy. There are groups of both boys and girls to consider. From the 14 girls (n = 14) in the class, we are choosing 2 (r = 2). Girls: 14! 14C 2 = 2!(14 2)! 14C 2 = 14! 2!(12)! 87, 178, 291, C 2 = 2(479, 001, 600) 87, 178, 291, C 2 = 958, 003, C 2 = 91 From the 12 boys (n = 12) in the class, we are choosing 1(r = 1). 12

17 Boys: 12! 12C 1 = 1!(12 1)! 12C 1 = 12! 1!(11)! 479, 001, C 1 = 1(39, 916, 800) 479, 001, C 1 = 39, 916, C 1 = 12 Therefore, the number of ways Mrs. Cameron can select a team of 3 students (2 girls and 1 boy) from the class of 26 students to work on a group project is: Total combinations = 14 C 2 12 C 1 = = 1, 092 Example 16 If there are 20 rock songs and 20 rap songs to choose from, in how many different ways can you select 12 rock songs and 7 rap songs for a mix CD? As in Example 15, we have multiple groups from which we are required to select, so we have to calculate the possible combinations for each group (rock songs and rap songs in this example) separately and then multiply together. Using TI technology: for n C r, type the n value (the total number of items), and then press MATH (PRB) (to number 3) ENTER. Then type the r value (the number of items your want to choose), and finally, press ENTER. Rock: 20! 20C 12 = 12!(20 12)! 20C 12 = 20! 12!(8)! 20C 12 = 125, 970 Rap: 20! 20C 7 = 7!(20 7)! 20C 7 = 20! 7!(13)! 20C 7 = 77, 520 Therefore, the total number of possible combinations is: Rock Rap 20C C 7 = 125, , 520 = possible combinations 13

18 Tech Tip: Calculating Combinations on the Calculator Combinations (ncr) Enter the n value. Press MATH. You should see modes across the top of the screen. You want the fourth menu: PRB (arrow right 3 times). You will see several options. ncr is the third. Press 3. Enter the r value. Press ENTER. Example 17 Compute 10 C 6. 10C MATH (PRB) 3 (ncr) 6 ENTER 10C 6 = Conditional Probability What if the probability of a second event is affected by the probability of the first event? This type of probability calculation is known as conditional probability. When working with events that are conditionally probable, you are working with 2 events, where the probability of the second event is conditional on the first event occurring. Say, for example, that you want to know the probability of drawing 2 kings from a deck of cards. As we have previously learned, here is how you would calculate this: P(first king) = 1 13 P(second king) = 3 51 P(2 kings) = P(2 kings) = P(2 kings) = Now let s assume you are playing a game where you need to draw 2 kings to win. You draw the first card and get a king. What is the probability of getting a king on the second card? The probability of getting a king on the second card can be thought of as a conditional probability. The formula for calculating conditional probability is given as: P(B A) = P(A B) P(A) P(A B) = P(A) P(B A) Another way to look at the conditional probability formula is as follows. Assuming the first event has occurred, the probability of the second event occurring is: P(second event first event) = P(first event and second event) P(first event) 14

19 Let s work through a few problems using the formula for conditional probability. Example 18 You are playing a game of cards where the winner is determined when a player gets 2 cards of the same suit. You draw a card and get a club ( ). What is the probability that the second card will be a club? Step 1: List what you know. First event = drawing the first club Second event = drawing the second club P(first club) = P(second club) = P(club and club) = P(club and club) = 156 2, 652 P(club and club) = 1 17 Step 2: Calculate the probability of choosing a club as the second card when a club is chosen as the first card. Step 3: Write your conclusion. P(club and club) Probability of drawing the second club = P(first club) P(club club) = P(club club) = P(club club) = P(club club) = 4 17 Therefore, the probability of selecting a club as the second card when a club is chosen as the first card is 24%. Example

20 In the next round of the game, the first person to be dealt a black ace wins the game. You get your first card, and it is a queen. What is the probability of obtaining a black ace? Step 1: List what you know. First event = being dealt the queen Second event = being dealt the black ace P(queen) = 4 52 P(black ace) = 2 51 P(black ace and queen) = P(black ace and queen) = 8 2, 652 P(black ace and queen) = Step 2: Calculate the probability of choosing black ace as a second card when a queen is chosen as a first card. P(black ace queen) = P(black ace queen) = P(black ace and queen) P(queen) P(black ace queen) = P(black ace queen) = 104 2, 652 P(black ace queen) = 2 51 Step 3: Write your conclusion. Therefore, the probability of selecting a black ace as the second card when a queen is chosen as the first card is 3.9%. Example 20 At Bluenose High School, 90% of the students take physics and 35% of the students take both physics and 16

21 statistics. What is the probability that a student from Bluenose High School who is taking physics is also taking statistics? Step 1: List what you know. P(physics) = 0.90 P(physics and statistics) = 0.35 Step 2: Calculate the probability of choosing statistics as a second course when physics is chosen as a first course. P(physics and statistics) P(statistics physics) = P(physics) P(statistics physics) = P(statistics physics) = P(statistics physics) = 39% Step 3: Write your conclusion. Therefore, the probability that a student from Bluenose High School who is taking physics is also taking statistics is 39%. Example 21 Sandra went out for her daily run. She goes on a path that has alternate routes to give her a variety of choices to make her run more enjoyable. The path has 3 turns where she can go left or right at each turn. The probability of turning right the first time is 1 2. Based on past runs, the probability of turning right the second time is 2 3. Draw a tree diagram to represent the path. What is the probability that she will turn left the second time after turning right the first time? 17

22 Step 1: List what you know. P(right the first time) = 1 2 P(right the second time) = 2 3 P(left the second time) = = 1 3 P(right the first time and left the second time) = P(right the first time and left the second time) = 1 6 Step 2: Calculate the probability of choosing left as the second turn when right is chosen as the first turn. P(left the second time right the first time) = P(left the second time right the first time) = P(left the second time right the first time) = P(right the first time and left the second time) P(right the first time) P(left the second time right the first time) = 2 6 P(left the second time right the first time) = 1 3 P(left the second time right the first time) = P(left the second time right the first time) = 33% Step 3: Write your conclusion. Therefore, the probability of choosing left as the second turn when right was chosen as the first turn is 33%. Points to Consider How does a permutation differ from a combination? How are tree diagrams helpful for determining probabilities? 18

23 Vocabulary Combinations The number of possible arrangements ( n C r ) of objects (r) without regard to order and without repetition selected from a distinct number of objects (n). Conditional probability The probability of a particular dependent event, given the outcome of the event on which it depends. Factorial function (!) To multiply a series of consecutive descending natural numbers. Fundamental Counting Principle If an event can be chosen in p different ways and another independent event can be chosen in q different ways, the probability of the 2 events occurring is p q. Permutations The number of possible arrangements ( n P r ) in an ordered set of objects, where n = the number of objects and r = the number of objects selected. Tree diagrams A way to show the outcomes of simple probability events, where each outcome is represented as a branch on a tree. 1.4 Review Questions Answer the following questions and show all work (including diagrams) to create a complete answer. 1. A bag contains 3 red balls and 4 blue balls. Thomas reaches in the bag and picks a ball out at random from the bag. He places it back into the bag. Thomas then reaches in the bag and picks another ball at random. (a) Draw a tree diagram to represent this problem. (b) What is the probability that Thomas picks: i. 2 red balls ii. a red ball in his second draw 2. A teacher has a prize box on her front desk for when students do exceptional work in math class. Inside the box there are 20 math pencils and 10 very cool erasers. Janet completed a challenge problem for Ms. Cameron, and Ms. Cameron rewarded Janet s innovative problem-solving approach with a trip to the prize box. Janet reaches into the box and picks out a prize and then drops it back in. Then she reaches in again and picks out a prize a second time. (a) Draw a tree diagram to represent this problem. (b) What is the probability that Janet reaches into the box and picks out an eraser on the second pick? 19

25 19. 2 fair dice are rolled. What is the probability that the sum is even given that the first die that is rolled is a 2? fair dice are rolled. What is the probability that the sum is even given that the first die rolled is a 5? fair dice are rolled. What is the probability that the sum is odd given that the first die rolled is a 5? 22. Steve and Scott are playing a game of cards with a standard deck of playing cards. Steve deals Scott a black king. What is the probability that Scott s second card will be a red card? 23. Sandra and Karen are playing a game of cards with a standard deck of playing cards. Sandra deals Karen a red seven. What is the probability that Karen s second card will be a black card? 24. Donna discusses with her parents the idea that she should get an allowance. She says that in her class, 55% of her classmates receive an allowance for doing chores, and 25% get an allowance for doing chores and are good to their parents. Her mom asks Donna what the probability is that a classmate will be good to his or her parents given that he or she receives an allowance for doing chores. What should Donna s answer be? 25. At a local high school, the probability that a student speaks English and French is 15%. The probability that a student speaks French is 45%. What is the probability that a student speaks English, given that the student speaks French? 26. At a local high school, the probability that a student takes statistics and art is 10%. The probability that a student takes art is 65%. What is the probability that a student takes statistics, given that the student takes art? 21

26 22

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