1. Copyright © 2014, 2013, 2010 and 2007 Pearson Education, Inc. Chapter Probability 5

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. Use the probability applet to simulate flipping a coin 100 times. Plot the proportion of heads against the number of flips. Repeat the simulation. 5-4

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

6. 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. 5-6

7. 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. 5-7

8. 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 (a) e 1 = boy, boy, e 2 = boy, girl, e 3 = girl, boy, e 4 = girl, girl (b) {(boy, boy), (boy, girl), (girl, boy), (girl, girl)} (c) {(boy, girl), (girl, boy)} 5-8

9. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 Apply Rules of Probabilities 5-9

10. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 = {e 1, e 2, …, e n }, then P(e 1 ) + P(e 2 ) + … + P(e n ) = 1 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 = {e 1, e 2, …, e n }, then P(e 1 ) + P(e 2 ) + … + P(e n ) = 1 5-10

11. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 5-11

12. 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, rule 2 (the sum of all probabilities must equal 1) is satisfied. 5-12

13. 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. 5-13

14. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Compute and Interpret Probabilities Using the Empirical Method 5-14

15. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 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 5-15

16. 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 OutcomeFrequency Side with no dot1344 Side with dot1294 Razorback767 Trotter365 Snouter137 Leaning Jowler32 5-16

17. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. EXAMPLE Building a Probability Model (c)Would it be unusual to throw a “Leaning Jowler”? 5-17 OutcomeFrequency Side with no dot1344 Side with dot1294 Razorback767 Trotter365 Snouter137 Leaning Jowler32 (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”.

18. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. (a) OutcomeProbability Side with no dot Side with dot0.329 Razorback0.195 Trotter0.093 Snouter0.035 Leaning Jowler0.008 5-18

19. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. (a) OutcomeProbability Side with no dot0.341 Side with dot0.329 Razorback0.195 Trotter0.093 Snouter0.035 Leaning Jowler0.008 (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”. 5-19

20. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. (a) OutcomeProbability Side with no dot0.341 Side with dot0.329 Razorback0.195 Trotter0.093 Snouter0.035 Leaning Jowler0.008 (c)A “Leaning Jowler” would be unusual. We would expect in 1000 throws of the pig to obtain “Leaning Jowler” about 8 times. 5-20

21. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 3 Compute and Interpret Probabilities Using the Classical Method 5-21

22. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 5-22

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

24. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 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. 5-24

25. 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? (c) Comment on the likelihood of the candy being yellow versus blue. 5-25

26. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. EXAMPLE Computing Probabilities Using the Classical Method (a)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. 5-26

27. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 5 Recognize and Interpret Subjective Probabilities 5-27

28. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 5-28

29. 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. 5-29

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

31. 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 5-31

32. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 Use the Addition Rule for Disjoint Events 5-32

33. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Two events are disjoint if they have no outcomes in common. Another name for disjoint events is mutually exclusive events. 5-33

34. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 5-34

35. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 5-35

36. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 5-36

37. 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 5-37

38. 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 5-38

39. 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 5-39 (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

40. 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 5-40 (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

41. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Use the General Addition Rule 5-41

42. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. The General Addition Rule For any two events E and F, The General Addition Rule For any two events E and F, 5-42

43. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. EXAMPLEIllustrating the General Addition Rule 5-43

44. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 5-44

45. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 3 Compute the Probability of an Event Using the Complement Rule 5-45

46. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 5-46

47. 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) 5-47 Entire region The area outside the circle represents E c

48. 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 5-48

49. 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 5-49

50. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. EXAMPLE Computing Probabilities Using Complements Source: United States Census Bureau 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 5-50

51. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. (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 5-51

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

53. 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 5-53

54. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 1.Identify Independent Events 5-54

55. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 5-55

56. 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. 5-56

57. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Use the Multiplication Rule for Independent Events 5-57

58. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 5-58

59. 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. 5-59

60. 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, 5-60

61. 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 5-61

62. 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 5-62

63. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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 EXAMPLE Illustrating the Multiplication Principle for Independent Events 5-63

64. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 3 Compute At-Least Probabilities 5-64

65. 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 5-65

66. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Summary: Rules of Probability 1.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 = {e 1, e 2, …, e n }, then P(e 1 ) + P(e 2 ) + … + P(e n ) = 1 Summary: Rules of Probability 1.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 = {e 1, e 2, …, e n }, then P(e 1 ) + P(e 2 ) + … + P(e n ) = 1 5-66

67. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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). 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). 5-67

68. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Summary: Rules of Probability 4.If E represents any event and E C represents the complement of E, then P( E C ) = 1 – P(E). Summary: Rules of Probability 4.If E represents any event and E C represents the complement of E, then P( E C ) = 1 – P(E). 5-68

69. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Summary: Rules of Probability 5.If E and F are independent events, then Summary: Rules of Probability 5.If E and F are independent events, then 5-69

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

71. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objectives 1.Compute conditional probabilities 2.Compute probabilities using the General Multiplication Rule 5-71

72. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 Compute Conditional Probabilities 5-72

73. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 5-73

74. 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} 5-74

75. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 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. 5-75

76. 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. 5-76

77. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. EXAMPLE Conditional Probabilities on Belief about God and Region of the Country Believe in God Believe in universal spirit Don’t believe in either East2043615 Midwest2122913 South219269 West1527626 5-77 (a)What is the probability that a randomly selected adult American who lives in the East believes in God? (b)What is the probability that a randomly selected adult American who believes in God lives in the East?

78. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. EXAMPLE Conditional Probabilities on Belief about God and Region of the Country Believe in God Believe in universal spirit Don’t believe in either East2043615 Midwest2122913 South219269 West1527626 5-78

79. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. EXAMPLE Conditional Probabilities on Belief about God and Region of the Country Believe in God Believe in universal spirit Don’t believe in either East2043615 Midwest2122913 South219269 West1527626 5-79

80. 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 5-80

81. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Compute Probabilities Using the General Multiplication Rule 5-81

82. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 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. 5-82

83. 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 5-83

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

85. 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 5-85

86. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 1 Solve Counting Problems Using the Multiplication Rule 5-86

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

88. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. EXAMPLE Counting the Number of Possible Meals 5-88

89. 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 5-89

90. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 2 Solve Counting Problems Using Permutations 5-90

91. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. A permutation is an ordered arrangement in which r objects are chosen from n distinct (different) objects and repetition is not allowed. The symbol n P r represents the number of permutations of r objects selected from n objects. 5-91

92. 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 5-92

93. 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 5-93

94. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 3 Solve Counting Problems Using Combinations 5-94

95. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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. 5-95

96. 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 5-96

97. 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. 5-97

98. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 4 Solve Counting Problems Involving Permutations with Nondistinct Items 5-98

99. 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. 5-99

100. 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 5-100

101. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. Objective 5 Compute Probabilities Involving Permutations and Combinations 5-101

102. 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 5-102

103. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. 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) = 52 C 6. 5-103

104. Copyright © 2013, 2010 and 2007 Pearson Education, Inc. The size of the sample space is EXAMPLE Winning the Lottery 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! 5-104