1. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Chapter Probability 5 5.1 5.2 5.3 5.4 5.5 5.6 5.7

2. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section Probability Rules 5.1

3. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1.Apply the rules of probabilities 2.Compute and interpret probabilities using the empirical method 3.Compute and interpret probabilities using the classical method 4.Use simulation to obtain data based on probabilities 5.Recognize and interpret subjective probabilities 5-3

4. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 5-4 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. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 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. 5-5

6. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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, e i. In general, events are denoted using capital letters such as E. 5-6

7. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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”. EXAMPLEIdentifying Events and the Sample Space of a Probability Experiment e 1 = boy, boy, e 2 = boy, girl, e 3 = girl, boy, e 4 = girl, girl 5-7 {(boy, boy), (boy, girl), (girl, boy), (girl, girl)} {(boy, girl), (girl, boy)}

8. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 Apply Rules of Probabilities 5-8 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. 0 ≤ P(E) ≤ 1 P(e 1 ) + P(e 2 ) + … + P(e n ) = 1

9. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. ColorProbability Brown0.12 Yellow0.15 Red0.12 Blue0.23 Orange0.23 Green0.15 All probabilities are between 0 and 1, inclusive. Because 0.12 + 0.15 + 0.12 + 0.23 + 0.23 + 0.15 = 1, (the sum of all probabilities must equal 1) is satisfied. 5-9

10. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. If an event is a certainty, the probability of the event is 1. If an event is impossible, the probability of the event is 0. An unusual event is an event that has a low probability of occurring. 5-10

11. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Compute and Interpret Probabilities Using the Empirical Method 5-11 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 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

12. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Pass the Pigs TM 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 EXAMPLE Building a Probability Model OutcomeFreq Side with no dot1344 Side with dot1294 Razorback767 Trotter365 Snouter137 Leaning Jowler32 5-12 (a)Use the results of the experiment to build a probability model for the way the pig lands. (b)Estimate the probability that a thrown pig lands on the “side with dot”. (c)Would it be unusual to throw a “Leaning Jowler”? Prob 0.341 0.329 0.195 0.093 0.035 0.008 0.329 yes

13. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 3 Compute and Interpret Probabilities Using the Classical Method 5-13 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.

14. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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? 5-14 P(yellow) = 6/30 = 1/5 P(blue) = 2/30 = 1/15

15. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 4 Use Simulation to Obtain Data Based on Probabilities 5-15 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. EXAMPLEUsing Simulation

16. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 5 Recognize and Interpret Subjective Probabilities 5-16 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.

17. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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? EXAMPLEEmpirical, 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. 5-17

18. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 5.1 Summary 1.Apply the rules of probabilities 2.Compute and interpret probabilities using the empirical method (based on past data values) 3.Compute and interpret probabilities using the classical method 4.Use simulation to obtain data based on probabilities 5.Recognize and interpret subjective probabilities 5-18

19. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section The Addition Rule and Complements 5.2

20. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1.Use the Addition Rule for Disjoint Events 2.Use the General Addition Rule 3.Compute the probability of an event using the Complement Rule 5-20

21. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 Use the Addition Rule for Disjoint Events 5-21 Two events are disjoint if they have no outcomes in common. Another name for disjoint events is mutually exclusive events. Addition Rule for Disjoint Events If E and F are disjoint (or mutually exclusive) events, then Addition Rule for Disjoint Events If E and F are disjoint (or mutually exclusive) events, then

22. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. We often draw pictures of events using Venn diagrams. 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.” Let F represent the event “choose a number greater than or equal to 8.” These events are disjoint as shown in the figure. 5-22 P(E) = P(F) = P(E or F) = 3 / 10 2 / 10 = 1 / 5 P(E) + P(F) = 5 / 10 = 1 / 2

23. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 5-23

24. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. The probability model to the right shows the distribution of the number of rooms in housing units in the United States. Number of Rooms in Housing Unit Probability One0.010 Two0.032 Three0.093 Four0.176 Five0.219 Six0.189 Seven0.122 Eight0.079 Nine or more0.080 EXAMPLE The Addition Rule for Disjoint Events (a) Verify that this is a probability model. All probabilities are between 0 and 1, inclusive. 0.010 + 0.032 + … + 0.080 = 1 5-24

25. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Number of Rooms in Housing Unit Probability One0.010 Two0.032 Three0.093 Four0.176 Five0.219 Six0.189 Seven0.122 Eight0.079 Nine or more0.080 EXAMPLE The Addition Rule for Disjoint Events 5-25 (b) What is the probability a randomly selected housing unit has two or three rooms? P(two or three) = P(two) + P(three) = 0.032 + 0.093 = 0.125

26. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Number of Rooms in Housing Unit Probability One0.010 Two0.032 Three0.093 Four0.176 Five0.219 Six0.189 Seven0.122 Eight0.079 Nine or more0.080 EXAMPLE The Addition Rule for Disjoint Events 5-26 (c) What is the probability a randomly selected housing unit has one or two or three rooms? P(one or two or three) = P(one) + P(two) + P(three) = 0.010 + 0.032 + 0.093 = 0.135

27. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Use the General Addition Rule 5-27 The General Addition Rule For any two events E and F, The General Addition Rule For any two events E and F,

28. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Suppose that a pair of dice are thrown. Let E = “the first die is a two” Let F = “the sum <5” Find P(E or F) EXAMPLEIllustrating the General Addition Rule 5-28

29. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 3 Compute the Probability of an Event Using the Complement Rule 5-29 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 E C, is all outcomes in the sample space S that are not outcomes in the event E.

30. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Complement Rule If E represents any event and E C represents the complement of E, then P(E C ) = 1 – P(E) 5-30 Entire region The area outside the circle represents E c

31. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 EXAMPLE Illustrating the Complement Rule 5-31

32. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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? EXAMPLE Computing Probabilities Using Complements Source: United States Census Bureau 5-32 Total = 393,186 (b) Compute the probability that a randomly selected resident will have a commute time less than 90 minutes. P( 90 minutes) = 1 – 0.012 = 0.988

33. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 5.2 Summary 1.Use the Addition Rule for Disjoint Events 2.Use the General Addition Rule 3.Compute the probability of an event using the Complement Rule 5-33

34. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section Independence and the Multiplication Rule 5.3

35. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1.Identify independent events 2.Use the Multiplication Rule for Independent Events 3.Compute at-least probabilities 5-35

36. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 1.Identify Independent Events 5-36 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.

37. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. EXAMPLE Independent or Not? (a)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. (c)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. 5-37

38. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Use the Multiplication Rule for Independent Events 5-38 Multiplication Rule for Independent Events If E and F are independent events, then Multiplication Rule for Independent Events If E and F are independent events, then

39. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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? EXAMPLE Computing Probabilities of Independent Events The survival of the first female is independent of the survival of the second female. We also have that P(survive) = 0.99186. 5-39

40. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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? EXAMPLE Computing Probabilities of Independent Events We assume that the defectiveness of the equipment is independent of the use of the equipment. So, 5-40

41. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Multiplication Rule for n Independent Events If E 1, E 2, E 3, … and E n are independent events, then Multiplication Rule for n Independent Events If E 1, E 2, E 3, … and E n are independent events, then 5-41

42. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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? EXAMPLE Illustrating the Multiplication Principle for Independent Events 5-42 P(all 4 survive) = P(1 st survives and 2 nd survives and 3 rd survives and 4 th survives) = P(1 st survives). P(2 nd survives). P(3 rd survives). P(4 th survives) = (0.99186) (0.99186) (0.99186) (0.99186) = 0.9678

43. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 3 Compute At-Least One Probabilities 5-43 P(at least one) = 1 – P(none)

44. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 EXAMPLE Computing “at least” Probabilities 5-44

45. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Summary: Rules of Probability 5-45 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). 0 ≤ P(E) ≤ 1 P(e 1 ) + P(e 2 ) + … + P(e n ) = 1 P( E C ) = 1 – P(E). If E and F are independent events, then

46. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section Conditional Probability and the General Multiplication Rule 5.4

47. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1.Compute conditional probabilities 2.Compute probabilities using the General Multiplication Rule 5-47

48. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 5-48 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.

49. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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} 5-49

50. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 5-50

51. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. EXAMPLE Conditional Probabilities on Belief about God and Region of the Country 5-51 Believe in God Believe in universal spirit Don’t believe in either East2043615 Midwest2122913 South219269 West1527626 (a)What is the probability that a randomly selected adult American who lives in the East believes in God?

52. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. EXAMPLE Conditional Probabilities on Belief about God and Region of the Country 5-52 Believe in God Believe in universal spirit Don’t believe in either East2043615 Midwest2122913 South219269 West1527626 (b) What is the probability that a randomly selected adult American who believes in God lives in the East?

53. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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? EXAMPLE Murder Victims 5-53

54. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Compute Probabilities Using the General Multiplication Rule 5-54

55. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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? EXAMPLE Murder Victims 5-55

56. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 5.4 Summary 1.Compute conditional probabilities 2.Compute probabilities using the General Multiplication Rule 5-56

57. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section Counting Techniques 5.5

58. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1.Solve counting problems using the Multiplication rule 2.Solve counting problems using permutations 3.Solve counting problems using combinations 4.Solve counting problems involving permutations with nondistinct items 5.Compute probabilities involving permutations and combinations 5-58

59. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 5-59 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. 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.

60. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. You have a choice of 2 appetizers, 4 entrées, and 2 desserts. How many different meals with 1 appetizer, 1 entrée, and 1 dessert may be ordered? 2 4 2 = 16 EXAMPLE Counting the Number of Possible Meals 5-60

61. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 5-61 A permutation is an ordered arrangement in which r objects are chosen from n distinct (different) objects and repetitions are not allowed. The symbol n P r represents the number of permutations of r objects selected from n objects.

62. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 1. the n objects are distinct, 2. repetition of objects is not allowed, and 3. order is important, is given by the formula 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 1. the n objects are distinct, 2. repetition of objects is not allowed, and 3. order is important, is given by the formula 5-62

63. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 5-63

64. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. In how many ways can horses in a 10-horse race finish first, second, and third? EXAMPLE Betting on the Trifecta 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 5-64

65. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 3 5-65 A combination is a collection, without regard to order, of n distinct objects without repetition. The symbol n C r represents the number of combinations of n distinct objects taken r at a time.

66. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 1. the n objects are distinct, 2. repetition of objects is not allowed, and 3. order is not important, is given by the formula 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 1. the n objects are distinct, 2. repetition of objects is not allowed, and 3. order is not important, is given by the formula 5-66

67. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. How many different simple random samples of size 4 can be obtained from a population whose size is 20? EXAMPLE Simple Random Samples 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. 5-67

68. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 4 Solve Counting Problems Involving Permutations with Nondistinct Items 5-68

69. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Permutations with Nondistinct Items The number of permutations of n objects of which n 1 are of one kind, n 2 are of a second kind,..., and n k are of a k th kind is given by where n = n 1 + n 2 + … + n k. Permutations with Nondistinct Items The number of permutations of n objects of which n 1 are of one kind, n 2 are of a second kind,..., and n k are of a k th kind is given by where n = n 1 + n 2 + … + n k. 5-69

70. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. How many different vertical arrangements are there of 10 flags if 5 are white, 3 are blue, and 2 are red? EXAMPLE Arranging Flags 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 5-70

71. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 5 Compute Probabilities Involving Permutations and Combinations 5-71

72. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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? EXAMPLE Winning the Lottery 5-72 The probability of winning is given by the number of ways a ticket could win divided by the size of the sample space. P(win) = 2 / (52C6) There is about a 1 in 10,000,000 chance of winning the Illinois Lottery!

73. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 5.5 Summary 1.Solve counting problems using the Multiplication rule 2.Solve counting problems using permutations 3.Solve counting problems using combinations 4.Solve counting problems involving permutations with nondistinct items 5.Compute probabilities involving permutations and combinations 5-73

74. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section Putting It Together: Which Method Do I Use? 5.6

75. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1.Determine the appropriate probability rule to use 2.Determine the appropriate counting technique to use 5-75

76. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 Determine the Appropriate Probability Rule to Use 5-76

77. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 5-77

78. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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: Find the probability of randomly drawing a suitcase worth at least $100,000. EXAMPLE Probability: Which Rule Do I Use? 5-78 Answer: 7/26

79. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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? EXAMPLE Probability: Which Rule Do I Use? 5-79

80. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Determine the appropriate counting technique to use 5-80

81. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 5-81

82. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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) EXAMPLE Counting: Which Technique Do I Use? 5-82 = 5C3 4C2 = 60

83. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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? EXAMPLE Counting: Which Technique Do I Use? 5-83 We can use either the multiplication rule or Permutations 43424140 or 43P4 = 2,961,840

84. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 5.6 summary 1.Determine the appropriate probability rule to use 2.Determine the appropriate counting technique to use 5-84

85. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Section Bayes’s Rule (on CD) 5.7

86. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1.Use the Rule of Total Probabilities 2.Use Bayes’s Rule to compute probabilities 5-86 The material in this section is available on the CD that accompanies the text.

87. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 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 5-87

88. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. General Multiplication Rule The probability that two events E and F both occur is General Multiplication Rule The probability that two events E and F both occur is 5-88

89. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 Use the Rule of Total Probabilities 5-89

90. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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? 5-90

91. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 5-91

92. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. ● 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 EXAMPLE Introduction to the Rule of Total Probability 5-92

93. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Another way to analyze this problem is to use a tree diagram Female Male 0.55 0.45 EXAMPLE Introduction to the Rule of Total Probability 5-93

94. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. EXAMPLE Introduction to the Rule of Total Probability Female Male 0.55 0.45 Business Not Business Business Not Business 0.550.15 0.550.85 0.450.20 0.450.80 5-94

95. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Multiply out, and add the two business branches EXAMPLE Introduction to the Rule of Total Probability Business Not Business Business Not Business Female Male 0.55 0.45 0.550.15 0.550.85 0.450.20 0.450.80 0.0825 0.0900 0.4675 0.3600 0.0825 0.0900 Total = 0.1725 0.4675 0.3600 5-95

96. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 5-96

97. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. ● A partition of the sample space S are two non-empty sets A 1 and A 2 that divide up S ● In other words  A 1 ≠ Ø  A 2 ≠ Ø  A 1 ∩ A 2 = Ø (there is no overlap)  A 1 U A 2 = S (they cover all of S) 5-97

98. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. ● Let E be any event in the sample space S ● Because A 1 and A 2 are disjoint, E ∩ A 1 and E ∩ A 2 are also disjoint ● Because A 1 and A 2 cover all of S, E ∩ A 1 and E ∩ A 2 cover all of E ● This means that we have divided E into two disjoint pieces E = (E ∩ A 1 ) U (E ∩ A 2 ) 5-98

99. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. ● Because E ∩ A 1 and E ∩ A 2 are disjoint, we can use the Addition Rule P(E) = P(E ∩ A 1 ) + P(E ∩ A 2 ) ● We now use the General Multiplication Rule on each of the P(E ∩ A 1 ) and P(E ∩ A 2 ) terms P(E) = P(A 1 ) P(E | A 1 ) + P(A 2 ) P(E | A 2 ) 5-99

100. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. P(E) = P(A 1 ) P(E | A 1 ) + P(A 2 ) P(E | A 2 ) ● This is the Rule of Total Probability (for a partition into two sets A 1 and A 2 ) ● It is useful when we want to compute a probability (P(E)) but we know only pieces of it (such as P(E | A 1 )) ● The Rule of Total Probability tells us how to put the probabilities together 5-100

101. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 5-101

102. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. ● The general Rule of Total Probability assumes that we have a partition (the general definition) of S into n different subsets A 1, A 2, …, A n  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 5-102

103. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Rule of Total Probability Let E be an event that is a subset of a sample space S. Let A 1, A 2, A 3, …, A n be partitions of a sample space S. Then, Rule of Total Probability Let E be an event that is a subset of a sample space S. Let A 1, A 2, A 3, …, A n be partitions of a sample space S. Then, 5-103

104. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. ● 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) EXAMPLE The Rule of Total Probability 5-104

105. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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) 5-105

106. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. ● 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 5-106

107. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Use Bayes’s Rule to Compute Probabilities 5-107

108. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 5-108

109. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 5-109

110. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. ● For this example  We first choose a random business student (event E)  What is the probability that this student is female? (partition element A 1 ) ● This question is asking for the value of P(A 1 | E) ● Before, we were working with P(E | A 1 ) instead  The probability (15%) that a female student is a business major 5-110

111. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. The Rule of Total Probability  Know P(A i ) and P(E | A i )  Solve for P(E) Bayes’s Rule  Know P(E) and P(E | A i )  Solve for P(A i | E) 5-111

112. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Bayes’s Rule, for a partition into two sets U 1 and U 2, is This rule is very useful when P(U 1 |B) is difficult to compute, but P(B|U 1 ) is easier 5-112

113. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Bayes’s Rule Let A 1, A 2, A 3, …, A n be a partition of a sample space S. Then, for any event E for which P(E) > 0, the probability of event A i for i = 1, 2, …, n given the event E, is Bayes’s Rule Let A 1, A 2, A 3, …, A n be a partition of a sample space S. Then, for any event E for which P(E) > 0, the probability of event A i for i = 1, 2, …, n given the event E, is 5-113

114. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. The business majors example from before Business Not Business Business Not Business Female Male 0.55 0.45 0.550.15 0.550.85 0.450.20 0.450.80 0.0825 0.0900 Total = 0.1725 0.4675 0.3600 EXAMPLE Bayes’s Rule 5-114

115. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. ● If we chose a random business major, what is the probability that this student is female?  A 1 = Female student  A 2 = Male student  E = business major ● We want to find P(A 1 | E), the probability that the student is female (A 1 ) given that this is a business major (E) EXAMPLE Bayes’s Rule 5-115

116. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. ● 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% EXAMPLE Bayes’s Rule (continued) 5-116

117. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. ● Now do it using Bayes’s Rule – it’s the same calculation ● Bayes’s Rule for a partition into two sets (n = 2)  P(A 1 ) =.55, P(A 2 ) =.45  P(E | A 1 ) =.15, P(E | A 2 ) =.20  We know all of the numbers we need EXAMPLE Bayes’s Rule (continued) 5-117

118. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. EXAMPLE Bayes’s Rule (continued) 5-118