Chapter 5 Probability

Section 5.1 Probability Rules

Objectives Apply the rules of probabilities Compute and interpret probabilities using the empirical method Compute and interpret probabilities using the classical method Use simulation to obtain data based on probabilities Recognize and interpret subjective probabilities 3

Probability is a measure of the likelihood of a random phenomenon or chance behavior. Probability describes the long-term proportion with which a certain outcome will occur in situations with short-term uncertainty. Use the probability applet to simulate flipping a coin 100 times. Plot the proportion of heads against the number of flips. Repeat the simulation. 4

Probability deals with experiments that yield random short-term results or outcomes, yet reveal long-term predictability. The long-term proportion in which a certain outcome is observed is the probability of that outcome. 5

The Law of Large Numbers As the number of repetitions of a probability experiment increases, the proportion with which a certain outcome is observed gets closer to the probability of the outcome. 6

In probability, an experiment is any process that can be repeated in which the results are uncertain. The sample space, S, of a probability experiment is the collection of all possible outcomes. An event is any collection of outcomes from a probability experiment. An event may consist of one outcome or more than one outcome. We will denote events with one outcome, sometimes called simple events, ei. In general, events are denoted using capital letters such as E. 7

Consider the probability experiment of having two children. EXAMPLE Identifying Events and the Sample Space of a Probability Experiment Consider the probability experiment of having two children. (a) Identify the outcomes of the probability experiment. (b) Determine the sample space. (c) Define the event E = “have one boy”. e1 = boy, boy, e2 = boy, girl, e3 = girl, boy, e4 = girl, girl {(boy, boy), (boy, girl), (girl, boy), (girl, girl)} {(boy, girl), (girl, boy)} 8

Objective 1 Apply Rules of Probabilities 9

Rules of probabilities 1. The probability of any event E, P(E), must be greater than or equal to 0 and less than or equal to 1. That is, 0 ≤ P(E) ≤ 1. 2. The sum of the probabilities of all outcomes must equal 1. That is, if the sample space S = {e1, e2, …, en}, then P(e1) + P(e2) + … + P(en) = 1 10

A probability model lists the possible outcomes of a probability experiment and each outcome’s probability. A probability model must satisfy rules 1 and 2 of the rules of probabilities. 11

All probabilities are between 0 and 1, inclusive. EXAMPLE A Probability Model In a bag of peanut M&M milk chocolate candies, the colors of the candies can be brown, yellow, red, blue, orange, or green. Suppose that a candy is randomly selected from a bag. The table shows each color and the probability of drawing that color. Verify this is a probability model. Color Probability Brown 0.12 Yellow 0.15 Red Blue 0.23 Orange Green All probabilities are between 0 and 1, inclusive. Because 0.12 + 0.15 + 0.12 + 0.23 + 0.23 + 0.15 = 1, rule 2 (the sum of all probabilities must equal 1) is satisfied. 12

If an event is impossible, the probability of the event is 0. If an event is a certainty, the probability of the event is 1. An unusual event is an event that has a low probability of occurring. 13

Objective 2 Compute and Interpret Probabilities Using the Empirical Method 14

Approximating Probabilities Using the Empirical Approach The probability of an event E is approximately the number of times event E is observed divided by the number of repetitions of the experiment. P(E) ≈ relative frequency of E 15

EXAMPLE Building a Probability Model Outcome Frequency Side with no dot 1344 Side with dot 1294 Razorback 767 Trotter 365 Snouter 137 Leaning Jowler 32 Pass the PigsTM is a Milton-Bradley game in which pigs are used as dice. Points are earned based on the way the pig lands. There are six possible outcomes when one pig is tossed. A class of 52 students rolled pigs 3,939 times. The number of times each outcome occurred is recorded in the table at right 16

(c) Would it be unusual to throw a “Leaning Jowler”? EXAMPLE Building a Probability Model Use the results of the experiment to build a probability model for the way the pig lands. Estimate the probability that a thrown pig lands on the “side with dot”. Outcome Frequency Side with no dot 1344 Side with dot 1294 Razorback 767 Trotter 365 Snouter 137 Leaning Jowler 32 (c) Would it be unusual to throw a “Leaning Jowler”? 17

(a) Outcome Probability Side with no dot Side with dot 0.329 Razorback 0.195 Trotter 0.093 Snouter 0.035 Leaning Jowler 0.008 18

(a) Outcome Probability Side with no dot 0.341 Side with dot 0.329 Razorback 0.195 Trotter 0.093 Snouter 0.035 Leaning Jowler 0.008 (b) The probability a throw results in a “side with dot” is 0.329. In 1000 throws of the pig, we would expect about 329 to land on a “side with dot”. 19

(a) Outcome Probability Side with no dot 0.341 Side with dot 0.329 Razorback 0.195 Trotter 0.093 Snouter 0.035 Leaning Jowler 0.008 (c) A “Leaning Jowler” would be unusual. We would expect in 1000 throws of the pig to obtain “Leaning Jowler” about 8 times. 20

Objective 3 Compute and Interpret Probabilities Using the Classical Method 21

The classical method of computing probabilities requires equally likely outcomes. An experiment is said to have equally likely outcomes when each simple event has the same probability of occurring. 22

Computing Probability Using the Classical Method If an experiment has n equally likely outcomes and if the number of ways that an event E can occur is m, then the probability of E, P(E) is 23

Computing Probability Using the Classical Method So, if S is the sample space of this experiment, where N(E) is the number of outcomes in E, and N(S) is the number of outcomes in the sample space. 24

(a) What is the probability that it is yellow? EXAMPLE Computing Probabilities Using the Classical Method Suppose a “fun size” bag of M&Ms contains 9 brown candies, 6 yellow candies, 7 red candies, 4 orange candies, 2 blue candies, and 2 green candies. Suppose that a candy is randomly selected. (a) What is the probability that it is yellow? (b) What is the probability that it is blue? (c) Comment on the likelihood of the candy being yellow versus blue. 25

There are a total of 9 + 6 + 7 + 4 + 2 + 2 = 30 candies, so N(S) = 30. EXAMPLE Computing Probabilities Using the Classical Method There are a total of 9 + 6 + 7 + 4 + 2 + 2 = 30 candies, so N(S) = 30. (b) P(blue) = 2/30 = 0.067. (c) Since P(yellow) = 6/30 and P(blue) = 2/30, selecting a yellow is three times as likely as selecting a blue. 26

Objective 4 Use Simulation to Obtain Data Based on Probabilities 27

EXAMPLE Using Simulation Use the probability applet to simulate throwing a 6-sided die 100 times. Approximate the probability of rolling a 4. How does this compare to the classical probability? Repeat the exercise for 1000 throws of the die. 28

Objective 5 Recognize and Interpret Subjective Probabilities 29

The subjective probability of an outcome is a probability obtained on the basis of personal judgment. For example, an economist predicting there is a 20% chance of recession next year would be a subjective probability. 30

EXAMPLE Empirical, Classical, or Subjective Probability In his fall 1998 article in Chance Magazine, (“A Statistician Reads the Sports Pages,” pp. 17-21,) Hal Stern investigated the probabilities that a particular horse will win a race. He reports that these probabilities are based on the amount of money bet on each horse. When a probability is given that a particular horse will win a race, is this empirical, classical, or subjective probability? Subjective because it is based upon people’s feelings about which horse will win the race. The probability is not based on a probability experiment or counting equally likely outcomes. 31

The Addition Rule and Complements Section 5.2 The Addition Rule and Complements

Objectives Use the Addition Rule for Disjoint Events Use the General Addition Rule Compute the probability of an event using the Complement Rule 33

Objective 1 Use the Addition Rule for Disjoint Events 34

Two events are disjoint if they have no outcomes in common Two events are disjoint if they have no outcomes in common. Another name for disjoint events is mutually exclusive events. 35

We often draw pictures of events using Venn diagrams We often draw pictures of events using Venn diagrams. These pictures represent events as circles enclosed in a rectangle. The rectangle represents the sample space, and each circle represents an event. For example, suppose we randomly select a chip from a bag where each chip in the bag is labeled 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Let E represent the event “choose a number less than or equal to 2,” and let F represent the event “choose a number greater than or equal to 8.” These events are disjoint as shown in the figure. 36

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Addition Rule for Disjoint Events If E and F are disjoint (or mutually exclusive) events, then 38

The Addition Rule for Disjoint Events can be extended to more than two disjoint events. In general, if E, F, G, . . . each have no outcomes in common (they are pairwise disjoint), then 39

(a) Verify that this is a probability model. EXAMPLE The Addition Rule for Disjoint Events Number of Rooms in Housing Unit Probability One 0.010 Two 0.032 Three 0.093 Four 0.176 Five 0.219 Six 0.189 Seven 0.122 Eight 0.079 Nine or more 0.080 The probability model to the right shows the distribution of the number of rooms in housing units in the United States. (a) Verify that this is a probability model. All probabilities are between 0 and 1, inclusive. 0.010 + 0.032 + … + 0.080 = 1 40

EXAMPLE The Addition Rule for Disjoint Events (b) What is the probability a randomly selected housing unit has two or three rooms? Number of Rooms in Housing Unit Probability One 0.010 Two 0.032 Three 0.093 Four 0.176 Five 0.219 Six 0.189 Seven 0.122 Eight 0.079 Nine or more 0.080 P(two or three) = P(two) + P(three) = 0.032 + 0.093 = 0.125 41

= P(one) + P(two) + P(three) = 0.010 + 0.032 + 0.093 = 0.135 EXAMPLE The Addition Rule for Disjoint Events (c) What is the probability a randomly selected housing unit has one or two or three rooms? Number of Rooms in Housing Unit Probability One 0.010 Two 0.032 Three 0.093 Four 0.176 Five 0.219 Six 0.189 Seven 0.122 Eight 0.079 Nine or more 0.080 P(one or two or three) = P(one) + P(two) + P(three) = 0.010 + 0.032 + 0.093 = 0.135 42

Objective 2 Use the General Addition Rule 43

The General Addition Rule For any two events E and F, 44

EXAMPLE Illustrating the General Addition Rule Suppose that a pair of dice are thrown. Let E = “the first die is a two” and let F = “the sum of the dice is less than or equal to 5”. Find P(E or F) using the General Addition Rule. 45

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Objective 3 Compute the Probability of an Event Using the Complement Rule 47

Complement of an Event Let S denote the sample space of a probability experiment and let E denote an event. The complement of E, denoted EC, is all outcomes in the sample space S that are not outcomes in the event E. 48

If E represents any event and EC represents the complement of E, then Complement Rule If E represents any event and EC represents the complement of E, then P(EC) = 1 – P(E) Entire region The area outside the circle represents Ec 49

P(do not own a dog) = 1 – P(own a dog) = 1 – 0.316 = 0.684 EXAMPLE Illustrating the Complement Rule According to the American Veterinary Medical Association, 31.6% of American households own a dog. What is the probability that a randomly selected household does not own a dog? P(do not own a dog) = 1 – P(own a dog) = 1 – 0.316 = 0.684 50

EXAMPLE Computing Probabilities Using Complements The data to the right represent the travel time to work for residents of Hartford County, CT. (a) What is the probability a randomly selected resident has a travel time of 90 or more minutes? Source: United States Census Bureau 51

residents in Hartford County EXAMPLE Computing Probabilities Using Complements There are a total of 24,358 + 39,112 + … + 4,895 = 393,186 residents in Hartford County The probability a randomly selected resident will have a commute time of “90 or more minutes” is Source: United States Census Bureau 52

P(less than 90 minutes) = 1 – P(90 minutes or more) = 1 – 0.012 (b) Compute the probability that a randomly selected resident of Hartford County, CT will have a commute time less than 90 minutes. P(less than 90 minutes) = 1 – P(90 minutes or more) = 1 – 0.012 = 0.988 53

Independence and the Multiplication Rule Section 5.3 Independence and the Multiplication Rule

Objectives Identify independent events Use the Multiplication Rule for Independent Events Compute at-least probabilities 55

Objective 1 Identify Independent Events 56

Two events E and F are independent if the occurrence of event E in a probability experiment does not affect the probability of event F. Two events are dependent if the occurrence of event E in a probability experiment affects the probability of event F. 57

EXAMPLE Independent or Not? Suppose you draw a card from a standard 52-card deck of cards and then roll a die. The events “draw a heart” and “roll an even number” are independent because the results of choosing a card do not impact the results of the die toss. (b) Suppose two 40-year old women who live in the United States are randomly selected. The events “woman 1 survives the year” and “woman 2 survives the year” are independent. Suppose two 40-year old women live in the same apartment complex. The events “woman 1 survives the year” and “woman 2 survives the year” are dependent. 58

Objective 2 Use the Multiplication Rule for Independent Events 59

Multiplication Rule for Independent Events If E and F are independent events, then 60

EXAMPLE Computing Probabilities of Independent Events The probability that a randomly selected female aged 60 years old will survive the year is 99.186% according to the National Vital Statistics Report, Vol. 47, No. 28. What is the probability that two randomly selected 60 year old females will survive the year? The survival of the first female is independent of the survival of the second female. We also have that P(survive) = 0.99186. 61

EXAMPLE Computing Probabilities of Independent Events A manufacturer of exercise equipment knows that 10% of their products are defective. They also know that only 30% of their customers will actually use the equipment in the first year after it is purchased. If there is a one-year warranty on the equipment, what proportion of the customers will actually make a valid warranty claim? We assume that the defectiveness of the equipment is independent of the use of the equipment. So, 62

Multiplication Rule for n Independent Events If E1, E2, E3, … and En are independent events, then 63

EXAMPLE Illustrating the Multiplication Principle for Independent Events The probability that a randomly selected female aged 60 years old will survive the year is 99.186% according to the National Vital Statistics Report, Vol. 47, No. 28. What is the probability that four randomly selected 60 year old females will survive the year? 64

= P(1st survives and 2nd survives and 3rd survives and 4th survives) EXAMPLE Illustrating the Multiplication Principle for Independent Events P(all 4 survive) = P(1st survives and 2nd survives and 3rd survives and 4th survives) = P(1st survives) . P(2nd survives) . P(3rd survives) . P(4th survives) = (0.99186) (0.99186) (0.99186) (0.99186) = 0.9678 65

Objective 3 Compute At-Least Probabilities 66

P(at least one dies) = 1 – P(none die) = 1 – P(all survive) EXAMPLE Computing “at least” Probabilities The probability that a randomly selected female aged 60 years old will survive the year is 99.186% according to the National Vital Statistics Report, Vol. 47, No. 28. What is the probability that at least one of 500 randomly selected 60 year old females will die during the course of the year? P(at least one dies) = 1 – P(none die) = 1 – P(all survive) = 1 – (0.99186)500 = 0.9832 67

Summary: Rules of Probability The probability of any event must be between 0 and 1. If we let E denote any event, then 0 ≤ P(E) ≤ 1. 2.The sum of the probabilities of all outcomes in the sample space must equal 1. That is, if the sample space S = {e1, e2, …, en}, then P(e1) + P(e2) + … + P(en) = 1 68

Summary: Rules of Probability 3. If E and F are disjoint events, then P(E or F) = P(E) + P(F). If E and F are not disjoint events, then P(E or F) = P(E) + P(F) – P(E and F). 69

Summary: Rules of Probability 4. If E represents any event and EC represents the complement of E, then P(EC) = 1 – P(E). 70

Summary: Rules of Probability 5. If E and F are independent events, then 71

Conditional Probability and the General Multiplication Rule Section 5.4 Conditional Probability and the General Multiplication Rule

Objectives Compute conditional probabilities Compute probabilities using the General Multiplication Rule 73

Objective 1 Compute Conditional Probabilities 74

Conditional Probability The notation P(F|E) is read “the probability of event F given event E”. It is the probability that the event F occurs given that event E has occurred. 75

EXAMPLE An Introduction to Conditional Probability Suppose that a single six-sided die is rolled. What is the probability that the die comes up 4? Now suppose that the die is rolled a second time, but we are told the outcome will be an even number. What is the probability that the die comes up 4? First roll: S = {1, 2, 3, 4, 5, 6} Second roll: S = {2, 4, 6} 76

Conditional Probability Rule If E and F are any two events, then The probability of event F occurring, given the occurrence of event E, is found by dividing the probability of E and F by the probability of E, or by dividing the number of outcomes in E and F by the number of outcomes in E. 77

EXAMPLE Conditional Probabilities on Belief about God and Region of the Country A survey was conducted by the Gallup Organization conducted May 8 – 11, 2008 in which 1,017 adult Americans were asked, “Which of the following statements comes closest to your belief about God – you believe in God, you don’t believe in God, but you do believe in a universal spirit or higher power, or you don’t believe in either?” The results of the survey, by region of the country, are given in the table on the next slide. 78

Believe in universal spirit Don’t believe in either EXAMPLE Conditional Probabilities on Belief about God and Region of the Country Believe in God Believe in universal spirit Don’t believe in either East 204 36 15 Midwest 212 29 13 South 219 26 9 West 152 76 What is the probability that a randomly selected adult American who lives in the East believes in God? What is the probability that a randomly selected adult American who believes in God lives in the East? 79

Believe in universal spirit Don’t believe in either EXAMPLE Conditional Probabilities on Belief about God and Region of the Country Believe in God Believe in universal spirit Don’t believe in either East 204 36 15 Midwest 212 29 13 South 219 26 9 West 152 76 80

Believe in universal spirit Don’t believe in either EXAMPLE Conditional Probabilities on Belief about God and Region of the Country Believe in God Believe in universal spirit Don’t believe in either East 204 36 15 Midwest 212 29 13 South 219 26 9 West 152 76 81

EXAMPLE Murder Victims In 2005, 19.1% of all murder victims were between the ages of 20 and 24 years old. Also in 2005, 16.6% of all murder victims were 20 – 24 year old males. What is the probability that a randomly selected murder victim in 2005 was male given that the victim is 20 – 24 years old? 82

Objective 2 Compute Probabilities Using the General Multiplication Rule 83

General Multiplication Rule The probability that two events E and F both occur is In words, the probability of E and F is the probability of event E occurring times the probability of event F occurring, given the occurrence of event E. 84

EXAMPLE Murder Victims In 2005, 19.1% of all murder victims were between the ages of 20 and 24 years old. Also in 2005, 86.9% of murder victims were male given that the victim was 20 – 24 years old. What is the probability that a randomly selected murder victim in 2005 was a 20 – 24 year old male? 85

Section 5.5 Counting Techniques

Objectives Solve counting problems using the Multiplication rule Solve counting problems using permutations Solve counting problems using combinations Solve counting problems involving permutations with nondistinct items Compute probabilities involving permutations and combinations 87

Objective 1 Solve Counting Problems Using the Multiplication Rule 88

Multiplication Rule of Counting If a task consists of a sequence of choices in which there are p selections for the first choice, q selections for the second choice, r selections for the third choice, and so on, then the task of making these selections can be done in different ways. 89

different meals can be ordered. EXAMPLE Counting the Number of Possible Meals For each choice of appetizer, we have 4 choices of entrée, and for each of these 2 • 4 = 8 parings, there are 2 choices for dessert. A total of 2 • 4 • 2 = 16 different meals can be ordered. 90

If n ≥ 0 is an integer, the factorial symbol, n If n ≥ 0 is an integer, the factorial symbol, n!, is defined as follows: n! = n(n – 1) • • • • • 3 • 2 • 1 0! = 1 1! = 1 91

Objective 2 Solve Counting Problems Using Permutations 92

A permutation is an ordered arrangement in which r objects are chosen from n distinct (different) objects and repetition is not allowed. The symbol nPr represents the number of permutations of r objects selected from n objects. 93

Number of permutations of n Distinct Objects Taken r at a Time The number of arrangements of r objects chosen from n objects, in which the n objects are distinct, repetition of objects is not allowed, and order is important, is given by the formula 94

The top three horses can finish a 10-horse race in EXAMPLE Betting on the Trifecta In how many ways can horses in a 10-horse race finish first, second, and third? The 10 horses are distinct. Once a horse crosses the finish line, that horse will not cross the finish line again, and, in a race, order is important. We have a permutation of 10 objects taken 3 at a time. The top three horses can finish a 10-horse race in 95

Objective 3 Solve Counting Problems Using Combinations 96

A combination is a collection, without regard to order, of n distinct objects without repetition. The symbol nCr represents the number of combinations of n distinct objects taken r at a time. 97

Number of Combinations of n Distinct Objects Taken r at a Time The number of different arrangements of r objects chosen from n objects, in which the n objects are distinct, repetition of objects is not allowed, and order is not important, is given by the formula 98

Use Formula (2) with n = 20 and r = 4: EXAMPLE Simple Random Samples How many different simple random samples of size 4 can be obtained from a population whose size is 20? The 20 individuals in the population are distinct. In addition, the order in which individuals are selected is unimportant. Thus, the number of simple random samples of size 4 from a population of size 20 is a combination of 20 objects taken 4 at a time. Use Formula (2) with n = 20 and r = 4: There are 4,845 different simple random samples of size 4 from a population whose size is 20. 99

Objective 4 Solve Counting Problems Involving Permutations with Nondistinct Items 100

Permutations with Nondistinct Items The number of permutations of n objects of which n1 are of one kind, n2 are of a second kind, . . . , and nk are of a kth kind is given by where n = n1 + n2 + … + nk. 101

Using Formula (3), we find that there are EXAMPLE Arranging Flags How many different vertical arrangements are there of 10 flags if 5 are white, 3 are blue, and 2 are red? We seek the number of permutations of 10 objects, of which 5 are of one kind (white), 3 are of a second kind (blue), and 2 are of a third kind (red). Using Formula (3), we find that there are different vertical arrangements 102

Objective 5 Compute Probabilities Involving Permutations and Combinations 103

EXAMPLE Winning the Lottery In the Illinois Lottery, an urn contains balls numbered 1 to 52. From this urn, six balls are randomly chosen without replacement. For a $1 bet, a player chooses two sets of six numbers. To win, all six numbers must match those chosen from the urn. The order in which the balls are selected does not matter. What is the probability of winning the lottery? 104

EXAMPLE Winning the Lottery The probability of winning is given by the number of ways a ticket could win divided by the size of the sample space. Each ticket has two sets of six numbers, so there are two chances of winning for each ticket. The sample space S is the number of ways that 6 objects can be selected from 52 objects without replacement and without regard to order, so N(S) = 52C6. 105

The size of the sample space is EXAMPLE Winning the Lottery The size of the sample space is Each ticket has two sets of 6 numbers, so a player has two chances of winning for each $1. If E is the event “winning ticket,” then There is about a 1 in 10,000,000 chance of winning the Illinois Lottery! 106

Putting It Together: Which Method Do I Use? Section 5.6 Putting It Together: Which Method Do I Use?

Objectives Determine the appropriate probability rule to use Determine the appropriate counting technique to use 108

Objective 1 Determine the Appropriate Probability Rule to Use 109

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EXAMPLE Probability: Which Rule Do I Use? In the game show Deal or No Deal?, a contestant is presented with 26 suitcases that contain amounts ranging from $0.01 to $1,000,000. The contestant must pick an initial case that is set aside as the game progresses. The amounts are randomly distributed among the suitcases prior to the game. Consider the following breakdown: 113

EXAMPLE Probability: Which Rule Do I Use? The probability of this event is not compound. Decide among the empirical, classical, or subjective approaches. Each prize amount is randomly assigned to one of the 26 suitcases, so the outcomes are equally likely. From the table we see that 7 of the cases contain at least $100,000. Letting E = “worth at least $100,000,” we compute P(E) using the classical approach. 114

EXAMPLE Probability: Which Rule Do I Use? The chance the contestant selects a suitcase worth at least $100,000 is 26.9%. In 100 different games, we would expect about 27 games to result in a contestant choosing a suitcase worth at least $100,000. 115

EXAMPLE Probability: Which Rule Do I Use? According to a Harris poll in January 2008, 14% of adult Americans have one or more tattoos, 50% have pierced ears, and 65% of those with one or more tattoos also have pierced ears. What is the probability that a randomly selected adult American has one or more tattoos and pierced ears? 116

EXAMPLE Probability: Which Rule Do I Use? The probability of a compound event involving ‘AND’. Letting E = “one or more tattoos” and F = “ears pierced,” we are asked to find P(E and F). The problem statement tells us that P(F) = 0.50 and P(F|E) = 0.65. Because P(F) ≠ P(F|E), the two events are not independent. We can find P(E and F) using the General Multiplication Rule. So, the chance of selecting an adult American at random who has one or more tattoos and pierced ears is 9.1%. 117

Objective 2 Determine the appropriate counting technique to use 118

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N(subcommittees) = N(ways to pick 3 men) • N(ways to pick 2 women) EXAMPLE Counting: Which Technique Do I Use? The Hazelwood city council consists of 5 men and 4 women. How many different subcommittees can be formed that consist of 3 men and 2 women? Sequence of events to consider: select the men, then select the women. Since the number of choices at each stage is independent of previous choices, we use the Multiplication Rule of Counting to obtain N(subcommittees) = N(ways to pick 3 men) • N(ways to pick 2 women) 120

EXAMPLE Counting: Which Technique Do I Use? To select the men, we must consider the number of arrangements of 5 men taken 3 at a time. Since the order of selection does not matter, we use the combination formula. To select the women, we must consider the number of arrangements of 3 women taken 2 at a time. Since the order of selection does not matter, we use the combination formula again. N(subcommittees) = 10 • 3 = 30. There are 30 possible subcommittees that contain 3 men and 2 women. 121

EXAMPLE Counting: Which Technique Do I Use? On February 17, 2008, the Daytona International Speedway hosted the 50th running of the Daytona 500. Touted by many to be the most anticipated event in racing history, the race carried a record purse of almost $18.7 million. With 43 drivers in the race, in how many different ways could the top four finishers (1st, 2nd, 3rd, and 4th place) occur? 122

EXAMPLE Counting: Which Technique Do I Use? The number of choices at each stage is independent of previous choices, so we can use the Multiplication Rule of Counting. The number of ways the top four finishers can occur is N(top four) = 43 • 42 • 41 • 40 = 2,961,840 We could also approach this problem as an arrangement of units. Since each race position is distinguishable, order matters in the arrangements. We are arranging the 43 drivers taken 4 at a time, so we are only considering a subset of r = 4 distinct drivers in each arrangement. Using our permutation formula, we get 123

Section 5.7 Bayes’s Rule (on CD)

Objectives Use the Rule of Total Probabilities Use Bayes’s Rule to compute probabilities The material in this section is available on the CD that accompanies the text. 125

Addition Rule for Disjoint Events If E and F are disjoint (mutually exclusive) events, then In general, if E, F, G, … are disjoint (mutually exclusive) events, then 126

General Multiplication Rule The probability that two events E and F both occur is 127

Objective 1 Use the Rule of Total Probabilities 128

55% of the students are female and 45% are male EXAMPLE Introduction to the Rule of Total Probability At a university 55% of the students are female and 45% are male 15% of the female students are business majors 20% of the male students are business majors What percent of students, overall, are business majors? 129

55% of the students are female EXAMPLE Introduction to the Rule of Total Probability The percent of the business majors in the university contributed by females 55% of the students are female 15% of those students are business majors Thus 15% of 55%, or 0.55 • 0.15 = 0.0825 or 8.25% of the total student body are female business majors Contributed by males In the same way, 20% of 45%, or 0.45 • 0.20 = .09 or 9% are male business majors 130

8.25% of the total student body are female business majors EXAMPLE Introduction to the Rule of Total Probability Altogether 8.25% of the total student body are female business majors 9% of the total student body are male business majors So … 17.25% of the total student body are business majors 131

Another way to analyze this problem is to use a tree diagram EXAMPLE Introduction to the Rule of Total Probability Another way to analyze this problem is to use a tree diagram Female Male 0.55 0.45 132

EXAMPLE Introduction to the Rule of Total Probability Business Not Business 0.55•0.15 0.55•0.85 0.45•0.20 0.45•0.80 Female 0.55 0.45 Male 133

Multiply out, and add the two business branches EXAMPLE Introduction to the Rule of Total Probability Multiply out, and add the two business branches Female Male 0.55 0.45 0.55•0.15 0.55•0.85 0.45•0.20 0.45•0.80 Business Not Business 0.0825 0.0900 0.4675 0.3600 0.0825 0.0900 Total = 0.1725 0.4675 0.3600 134

This is an example of the Rule of Total Probability P(Bus) = 55% • 15% + 45% • 20% = P(Female) • P(Bus | Female) + P(Male) • P(Bus | Male) This rule is useful when the sample space can be divided into two (or more) disjoint parts 135

A1 ∩ A2 = Ø (there is no overlap) A1 U A2 = S (they cover all of S) A partition of the sample space S are two non-empty sets A1 and A2 that divide up S In other words A1 ≠ Ø A2 ≠ Ø A1 ∩ A2 = Ø (there is no overlap) A1 U A2 = S (they cover all of S) 136

Let E be any event in the sample space S Because A1 and A2 are disjoint, E ∩ A1 and E ∩ A2 are also disjoint Because A1 and A2 cover all of S, E ∩ A1 and E ∩ A2 cover all of E This means that we have divided E into two disjoint pieces E = (E ∩ A1) U (E ∩ A2) 137

Because E ∩ A1 and E ∩ A2 are disjoint, we can use the Addition Rule P(E) = P(E ∩ A1) + P(E ∩ A2) We now use the General Multiplication Rule on each of the P(E ∩ A1) and P(E ∩ A2) terms P(E) = P(A1) • P(E | A1) + P(A2) • P(E | A2) 138

P(E) = P(A1) • P(E | A1) + P(A2) • P(E | A2) This is the Rule of Total Probability (for a partition into two sets A1 and A2) It is useful when we want to compute a probability (P(E)) but we know only pieces of it (such as P(E | A1)) The Rule of Total Probability tells us how to put the probabilities together 139

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Each subset is non-empty None of the subsets overlap The general Rule of Total Probability assumes that we have a partition (the general definition) of S into n different subsets A1, A2, …, An Each subset is non-empty None of the subsets overlap S is covered completely by the union of the subsets This is like the partition before, just that S is broken up into many pieces, instead of just two pieces 141

Rule of Total Probability Let E be an event that is a subset of a sample space S. Let A1, A2, A3, …, An be partitions of a sample space S. Then, 142

30% of the voters are Republican 30% of the voters are Democrats EXAMPLE The Rule of Total Probability In a particular town 30% of the voters are Republican 30% of the voters are Democrats 40% of the voters are independents This is a partition of the voters into three sets There are no voters that are in two sets (disjoint) All voters are in one of the sets (covers all of S) 143

90% of the Republicans favor it 60% of the Democrats favor it EXAMPLE The Rule of Total Probability For a particular issue 90% of the Republicans favor it 60% of the Democrats favor it 70% of the independents favor it These are the conditional probabilities E = {favor the issue} The above probabilities are P(E | political party) 144

The total proportion of votes who favor the issue 0.3 • 0.9 + 0.3 • 0.6 + 0.4 • 0.7 = 0.73 So 73% of the voters favor this issue 145

Objective 2 Use Bayes’s Rule to Compute Probabilities 146

“What percent of students are business majors?” In our male / female and business / non-business majors examples before, we used the rule of total probability to answer the question: “What percent of students are business majors?” We solved this problem by analyzing male students and female students separately 147

“What percent of business majors are female?” We could turn this problem around We were told the percent of female students who are business majors We could also ask: “What percent of business majors are female?” This is the situation for Bayes’s Rule 148

We first choose a random business student (event E) For this example We first choose a random business student (event E) What is the probability that this student is female? (partition element A1) This question is asking for the value of P(A1 | E) Before, we were working with P(E | A1) instead The probability (15%) that a female student is a business major 149

The Rule of Total Probability Know P(Ai) and P(E | Ai) Solve for P(E) Bayes’s Rule Know P(E) and P(E | Ai) Solve for P(Ai | E) 150

Bayes’s Rule, for a partition into two sets U1 and U2, is This rule is very useful when P(U1|B) is difficult to compute, but P(B|U1) is easier 151

Bayes’s Rule Let A1, A2, A3, …, An be a partition of a sample space S. Then, for any event E for which P(E) > 0, the probability of event Ai for i = 1, 2, …, n given the event E, is 152

The business majors example from before EXAMPLE Bayes’s Rule The business majors example from before Business Not Business Female Male 0.55 0.45 0.55•0.15 0.55•0.85 0.45•0.20 0.45•0.80 0.0825 0.4675 0.0900 0.3600 Total = 0.1725 153

EXAMPLE Bayes’s Rule If we chose a random business major, what is the probability that this student is female? A1 = Female student A2 = Male student E = business major We want to find P(A1 | E), the probability that the student is female (A1) given that this is a business major (E) 154

Do it in a straight way first EXAMPLE Bayes’s Rule (continued) Do it in a straight way first We know that 8.25% of the students are female business majors We know that 9% of the students are male business majors Choosing a business major at random is choosing one of the 17.25% The probability that this student is female is 8.25% / 17.25% = 47.83% 155

Now do it using Bayes’s Rule – it’s the same calculation EXAMPLE Bayes’s Rule (continued) Now do it using Bayes’s Rule – it’s the same calculation Bayes’s Rule for a partition into two sets (n = 2) P(A1) = .55, P(A2) = .45 P(E | A1) = .15, P(E | A2) = .20 We know all of the numbers we need 156

EXAMPLE Bayes’s Rule (continued) 157