1. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 1 of 103 Chapter 6 Probability

2. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 2 of 103 Outline 6.1 Introduction 6.2 Experiments, Outcomes, and Events 6.3 Assignment of Probabilities 6.4 Calculating Probabilities of Events 6.5 Conditional Probability and Independence 6.6 Tree Diagrams 6.7 Bayes’ Theorem 6.8 Simulation

3. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 3 of 103 6.1 Introduction 1.Mathematical Probability 2.Probability of an Event 3.An Example Problem

4. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 4 of 103 Mathematical Probability Many events in the world exhibit a random character. Yet, by repeated observations of such events we can often discern long-term patterns that persist despite random, short-term fluctuations. Probability is the branch of mathematics devoted to the study of such events.

5. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 5 of 103 Probability of an Event The probability of an event is a number between 0 and 1 that expresses the long-run likelihood that the event will occur. An event having probability.1 is rather unlikely to occur. An event with probability.9 is very likely to occur. An event with probability.5 is just as likely to occur as not.

6. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 6 of 103 An Example Problem Medical DiagnosisA clinic tests for active pulmonary tuberculosis. If a person has tuberculosis, the probability of a positive test result is.98. If a person does not have tuberculosis, the probability of a negative test result is.99. The incidence of tuberculosis in a certain city is 2 cases per 10,000 population. What is the probability that an individual who tests positive actually has pulmonary tuberculosis?

7. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 7 of 103 6.2 Experiments, Outcomes, and Events 1.Experiment, Trial and Outcome 2.Sample Space 3.Event 4.Special Events 5.Events As Sets 6.Mutually Exclusive Events

8. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 8 of 103 Experiment, Trial and Outcome An experiment is an activity with an observable outcome. Each repetition of the experiment is called a trial. In each trial we observe the outcome of the experiment.

9. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 9 of 103 Example Experiment, Trial and Outcome Experiment 1: Flip a coin Trial: One coin flipOutcome: Heads Experiment 2: Allow a conditioned rat to run a maze containing three possible paths Trial: One runOutcome: Path 1 Experiment 3: Tabulate the amount of rainfall in New York, NY in a year Trial: One yearOutcome: 37.23 in.

10. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 10 of 103 Sample Space The set of all possible outcomes of an experiment is called the sample space of the experiment. So each outcome is an element of the sample space.

11. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 11 of 103 Example Sample Space An experiment consists of throwing two dice, one red and one green, and observing the numbers on the uppermost face on each. What is the sample space S of this experiment? Each outcome of the experiment can be regarded as an ordered pair of numbers, the first representing the number on the red die and the second the number on the green die.

12. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 12 of 103 Example Sample Space (2) S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6) (2,1), (2,2), (2,3), (2,4), (2,5), (2,6) (3,1), (3,2), (3,3), (3,4), (3,5), (3,6) (4,1), (4,2), (4,3), (4,4), (4,5), (4,6) (5,1), (5,2), (5,3), (5,4), (5,5), (5,6) (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}

13. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 13 of 103 Event An event E is a subset of the sample space. We say that the event occurs when the outcome of the experiment is an element of E.

14. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 14 of 103 Example Event For the experiment of rolling two dice, describe the events E 1 = {The sum of the numbers is greater than 9}; E 2 = {The sum of the numbers is 7 or 11}. E 1 = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)} E 2 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (5,6), (6,5)}

15. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 15 of 103 Special Events Let S be the sample space of an experiment. The event corresponding to the empty set, , is called the impossible event, since it can never occur. The event corresponding to the sample space itself, S, is called the certain event because the outcome must be in S.

16. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 16 of 103 Events As Sets Let E and F be two events of the sample space S. The event where either E or F or both occurs is designated by E  F. The event where both E and F occurs is designated by E  F. The event where E does not occur is designated by E '.

17. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 17 of 103 Example Events As Sets For the experiment of rolling two dice, let E 1 = “The sum of the numbers is greater than 9” and E 3 = “The numbers on the two dice are equal”. Determine the sets E 1  E 3, E 1  E 3, and (E 1  E 3 )'.

18. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 18 of 103 Example Events As Sets (2) E 1 = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)} E 3 = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} E 1  E 3 = {(1,1), (2,2), (3,3), (4,4), (4,6), (5,5), (5,6), (6,4), (6,5), (6,6)} E 1  E 3 = {(5,5), (6,6)} (E 1  E 3 )' = {(1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,5), (5,1), (5,2), (5,3), (5,4), (6,1), (6,2), (6,3)}

19. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 19 of 103 Mutually Exclusive Events Let E and F be events in a sample space S. Then E and F are mutually exclusive (or disjoint) if E  F = . If E and F are mutually exclusive, then E and F cannot simultaneously occur; if E occurs, then F does not; and if F occurs, then E does not.

20. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 20 of 103 Example Mutually Exclusive Events For the experiment of rolling two dice, which of the following events are mutually exclusive? E 1 = “The sum of the dots is greater than 9” E 2 = “The sum of the dots is 7 or 11” E 3 = “The dots on the two dice are equal”

21. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 21 of 103 Example Mutually Exclusive Events (2) E 1 = {(4,6), (5,5), (5,6), (6,4), (6,5), (6,6)} E 2 = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (5,6), (6,5)} E 3 = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} E 1  E 2 = {(5,6), (6,5)} - Not mutually exclusive E 1  E 3 = {(5,5), (6,6)} - Not mutually exclusive E 2  E 3 =  - Mutually exclusive

22. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 22 of 103  The sample space of an experiment is the set of all possible outcomes of the experiment. Each subset of the sample space is called an event. We say that an event occurs when the outcome is an element of the event.  The event E  F occurs when either E or F or both occurs. The event E  F occurs when both E and F occur. The event E' occurs when E does not occur. Summary Section 6.2 - Part 1

23. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 23 of 103  Two events are mutually exclusive if they cannot both occur at the same time. Summary Section 6.2 - Part 2

24. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 24 of 103 6.3 Assignment of Probabilities 1.Probability of an Outcome 2.Experimental Probability 3.Fundamental Properties of Probabilities 4.Addition Principle 5.Inclusion-Exclusion Principle 6.Odds

25. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 25 of 103 Probability of an Outcome Let a sample space S consist of a finite number of outcomes s 1, s 2, …,s N. To each outcome we associate a number, called the probability of the outcome, which represents the relative likelihood that the outcome will occur. A chart showing the outcomes and the assigned probability is called the probability distribution for the experiment.

26. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 26 of 103 Example Probability of an Outcome Toss an unbiased coin and observe the side that faces upward. Determine the probability distribution for this experiment. Since the coin is unbiased, each outcome is equally likely to occur. OutcomeProbability Heads½ Tails½

27. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 27 of 103 Experimental Probability Let a sample space S consist of a finite number of outcomes s 1, s 2, …,s N. The relative frequency, or experimental probability, of each outcome is calculated after many trials. The experimental probability could be different for a different set of trials and different from the probability of the events.

28. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 28 of 103 Example Experimental Probability Traffic engineers measure the volume of traffic on a major highway during the rush hour. Generate a probability distribution using the data generated over 300 consecutive weekdays.

29. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 29 of 103 Example Experimental Probability We will use the experimental probability for the distribution.

30. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 30 of 103 Fundamental Properties of Probabilities Let an experiment have outcomes s 1, s 2, …, s N with probabilities p 1, p 2, …, p N. Then the numbers p 1, p 2, …, p N must satisfy: Fundamental Property 1 Each of the numbers p 1, p 2, …, p N is between 0 and 1; Fundamental Property 2 p 1 + p 2 + … + p N = 1.

31. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 31 of 103 Example Fundamental Properties Verify the fundamental properties for the following distribution. All probabilities are between 0 and 1 Total: 1.00

32. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 32 of 103 Addition Principle Addition PrincipleSuppose that an event E consists of the finite number of outcomes s, t, u, …,z. That is E = {s, t, u, …,z }. Then Pr(E) = Pr(s) + Pr(t) + Pr(u) + … + Pr(z), where Pr(A) is the probability of event A.

33. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 33 of 103 Example Addition Principle Suppose that we toss a red die and a green die and observe the numbers on the sides that face upward. a) Calculate the probabilities of the elementary events. b) Calculate the probability that the two dice show the same number.

34. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 34 of 103 a)As shown previously, the sample space consists of 36 pairs of numbers S = {(1,1), (1,2), …, (6,5), (6,6)}. Each of these pairs is equally likely to occur. The probability of each pair is b) E = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)} Example Addition Principle (2)

35. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 35 of 103 Inclusion-Exclusion Principle Let E and F be any events. Then If E and F are mutually exclusive, then

36. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 36 of 103 Example Inclusion-Exclusion Principle A factory needs two raw materials. The probability of not having an adequate supply of material A is.05 and the probability of not having an adequate supply of material B is.03. A study determines that the probability of a shortage of both materials is.01. What proportion of the time will the factory not be able to operate from lack of materials?

37. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 37 of 103 Example Inclusion-Exclusion Principle (2) The factory will not operate 7% of the time.

38. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 38 of 103 Odds If the odds in favor of an event E are a to b, then On average, for every a + b trials, E will occur a times and E will not occur b times.

39. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 39 of 103 Example Odds In the two dice problem, what are the odds of rolling a pair with the same number on the faces? The probability of obtaining a pair with the same number on the faces is 1/6. The probability of not obtaining a pair with the same number on the faces is 5/6. The odds are

40. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 40 of 103 Summary Section 6.3 - Part 1  A probability distribution for a finite sample space associates a probability with each outcome of the sample space. Each probability is a number between 0 and 1, and the sum of the probabilities is 1. The probability of an event is the sum of the probabilities of outcomes in the event.

41. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 41 of 103 Summary Section 6.3 - Part 2  The inclusion-exclusion principle states that the probability of the union of two events is the sum of the probabilities of the events minus the probability of their intersection. If the two events are mutually exclusive, the probability of the union is just the sum of the probabilities of the events.

42. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 42 of 103 Summary Section 6.3 - Part 3  We say that the odds in favor of an event are a to b if the probability of the event is a/(a + b). Intuitively, the event is expected to occur a times for every b times it does not occur.

43. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 43 of 103 6.4 Calculating Probabilities of Events 1.Probability of Equally Likely Outcomes 2.Complement Rule

44. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 44 of 103 Probability of Equally Likely Outcomes Let S be a sample space consisting of N equally likely outcomes. Let E be any event. Then

45. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 45 of 103 Example Equally Likely Outcomes Suppose that a cruise ship returns to the US from the Far East. Unknown to anyone, 4 of its 600 passengers have contracted a rare disease. Suppose that the Public Health Service screens 20 passengers, selected at random, to see whether the disease is present aboard ship. What is the probability that the presence of the disease will escape detection?

46. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 46 of 103 Example Equally Likely Outcomes (2) The sample space consists of samples of 20 drawn from among the 600 passengers. N = C(600, 20) There are 600 - 4 = 596 non infected passengers. The disease is not detected if the 20 passengers are chosen from this group. Let E be the event that the disease is not detected.

47. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 47 of 103 Example Equally Likely Outcomes (3) The number of outcomes in E is n (E ) = C(596, 20). Thus, Pr (E) = n (E) / N =.87

48. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 48 of 103 Complement Rule Complement Rule Let E be any event, E ' its complement. Then Pr(E) = 1 - Pr(E ').

49. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 49 of 103 Example Complement Rule A group of 5 people is to be selected at random. What is the probability that 2 or more of them have the same birthday? For simplicity we will ignore leap years and assume that each of the 365 days of the year are equally likely. Choosing a person at random is equivalent to choosing a birthday at random.

50. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 50 of 103 Example Complement Rule (2) There are 365 5 ways of choosing 5 birthdays. It will also be easier to first find the probability that the 5 birthdays are different. Let this be E '. The number of outcomes where the 5 birthdays are different is 365  364  363  362  361.

51. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 51 of 103 Example Complement Rule (3) Pr(E) = 1- Pr(E ') = 1 -.973 =.027 This example could be worked for a general group of r people. In this case Probability that, in a randomly selected group of r people, at least two will have the same birthday r5101520222325304050 Pr(E).027.117.253.411.476.507.569.706.891.970

52. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 52 of 103 Summary Section 6.4  For a sample space with a finite number of equally likely outcomes, the probability of an event is the number of elements in the event divided by the number of elements in the sample space.  The probability of the complement of an event is 1 minus the probability of the event.

53. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 53 of 103 6.5 Conditional Probability and Independence 1.Conditional Probability 2.Conditional Probability of Equally Likely Outcomes 3.Product Rule 4.Independence 5.Independence of a Set of Events

54. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 54 of 103 Conditional Probability Let E and F be events is a sample space S. The conditional probability Pr(E | F) is the probability of event E occurring given the condition that event F has occurred. In calculating this probability, the sample space is restricted to F. provided that Pr(F) ≠ 0.

55. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 55 of 103 Example Conditional Probability Twenty percent of the employees of Acme Steel Company are college graduates. Of all its employees, 25% earn more than $50,000 per year, and 15% are college graduates earning more than $50,000. What is the probability that an employee selected at random earns more than $50,000 per year, given that he or she is a college graduate?

56. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 56 of 103 Example Conditional Probability (2) Let H = "earns more than $50,000 per year" and C = "college graduate." From the problem, Pr(H) =.25Pr(C) =.20, Pr (H  C) =.15. Therefore,

57. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 57 of 103 Conditional Probability - Equally Likely Outcomes Conditional Probability in Case of Equally Likely Outcomes provided that [number of outcomes in F] ≠ 0.

58. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 58 of 103 Example Conditional Probability A sample of two balls are selected from an urn containing 8 white balls and 2 green balls. What is the probability that the second ball selected is white given that the first ball selected was white?

59. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 59 of 103 Example Conditional Probability The number of outcomes in "the first ball is white" is 8  9 = 72. That is, the first ball must be among the 8 white balls and the second ball can be any of the 9 balls left. The number of outcomes in "the first ball is white and the second ball is white" is 8  7 = 56.

60. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 60 of 103 Product Rule Product RuleIf Pr(F) ≠ 0, Pr (E  F) = Pr(F)  Pr(E | F). The product rule can be extended to three events. Pr(E 1  E 2  E 3 ) = Pr(E 1 )  Pr(E 2 | E 1 )  Pr(E 3 | E 1  E 2 )

61. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 61 of 103 Example Product Rule A sequence of two playing cards is drawn at random (without replacement) from a standard deck of 52 cards. What is the probability that the first card is red and the second is black? Let F = "the first card is red," and E = "the second card is black."

62. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 62 of 103 Example Product Rule (2) Pr(F) = ½ since half the deck is red cards. If we know the first card is red, then the probability the second is black is Therefore,

63. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 63 of 103 Independence Let E and F be events. We say that E and F are independent provided that Pr (E  F) = Pr(E)  Pr (F). Equivalently, they are independent provided that Pr(E | F) = Pr(E) and Pr(F | E) = Pr (F).

64. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 64 of 103 Example Independence Let an experiment consist of observing the results of drawing two consecutive cards from a 52-card deck. Let E = "second card is black" and F = "first card is red". Are these two events independent? From the previous example, Pr(E | F) = 26/51. Note that Pr(E) = 1/2. Since they are not equal, E and F are not independent.

65. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 65 of 103 Independence of a Set of Events A set of events is said to be independent if, for each collection of events chosen from them, say E 1, E 2, …, E n, we have Pr(E 1  E 2  …  E n ) = Pr(E 1 )  Pr(E 2 )  …  Pr (E n ).

66. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 66 of 103 Example Independence of a Set A company manufactures stereo components. Experience shows that defects in manufacture are independent of one another. Quality control studies reveal that 2% of CD players are defective, 3% of amplifiers are defective, and 7% of speakers are defective. A system consists of a CD player, an amplifier, and 2 speakers. What is the probability that the system is not defective?

67. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 67 of 103 Example Independence of a Set (2) Let C, A, S 1, and S 2 be events corresponding to defective CD player, amplifier, speaker 1, and speaker 2, respectively. Then Pr(C) =.02, Pr(A) =.03, Pr(S 1 ) = Pr(S 2 ) =.07 Pr(C') =.98, Pr(A') =.97, Pr(S 1 ') = Pr(S 2 ') =.93 Pr(C'  A'  S 1 '  S 2 ') =.98 .97 .93 2 =.822

68. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 68 of 103 Summary Section 6.5 - Part 1  Pr(E|F), the conditional probability that E occurs given that F occurs, is computed as Pr (E  F)/Pr (F). For a sample space with a finite number of equally likely outcomes, it can be computed as n (E  F) /n (F).  The product rule states that if Pr(F) ≠ 0, then Pr (E  F) = Pr(F)  Pr(E|F).

69. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 69 of 103 Summary Section 6.5 - Part 2  E and F are independent events if Pr (E  F) = Pr(F)  Pr(E). Equivalently, E and F [with Pr(F) ≠ 0] are independent events if Pr(E|F) = Pr(E).  A collection of events is said to be independent if for each collection of events chosen from them, the probability that all the events occur equals the product of the probabilities that each occurs.

70. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 70 of 103 6.6 Tree Diagrams 1.Tree Diagram 2.Medical Example 3.Quality Control Example

71. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 71 of 103 Tree Diagram A tree diagram helps us represent the various events and their associated probabilities. The various outcomes of each experiment are represented as branches emanating from a point. Each branch is labeled with the probability of the associated outcome. For example:

72. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 72 of 103 Tree Diagram (2) We represent experiments performed one after another by stringing together diagrams of the sort given in the previous slide. The probabilities for the second set of branches are conditional probabilities given the outcome from which the branches are emanating.

73. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 73 of 103 Tree Diagram (3) For example:

74. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 74 of 103 Example Medical The reliability of a skin test for active pulmonary tuberculosis (TB) is as follows: Of people with TB, 98% have a positive reaction and 2% have a negative reaction; of people free of TB, 99% have a negative reaction and 1% have a positive reaction. From a large population of which 2 per 10,000 persons have TB, a person is selected at random and given a skin test, which turns out positive. What is the probability that the person has active TB?

75. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 75 of 103 Example Medical (2) Make a tree diagram using: Pr(TB) = 2/10000 =.0002, Pr(not TB) = 1 -.0002 =.9998, Pr(POS|TB) =.98, Pr(NEG|TB) =.02, PR(NEG|not TB) =.99 and PR(POS|not TB) =.01.

76. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 76 of 103 Example Medical (3)

77. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 77 of 103 Example Quality Control A box contains 5 good light bulbs and 2 defective ones. Bulbs are selected one at a time (without replacement) until a good bulb is found. Find the probability that the number of bulbs selected is (i) 1, (ii) 2, (iii) 3.

78. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 78 of 103 Example Quality Control (2) For the first draw, a bulb selected at random is good (G) with probability 5/7 For the first draw, a bulb selected at random is defective (D) with probability 2/7

79. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 79 of 103 Example Quality Control (3) If the first bulb was good, the activity stops. If the first bulb was defective, then the second selection is good with probability 5/6 and defective with probability 1/6. If the second selection was good, the activity stops. If it was defective, then the third selection is good with probability 1.

80. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 80 of 103 Example Quality Control (4) The tree diagram is:

81. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 81 of 103 Summary Section 6.6  Tree diagrams provide a useful device for determining probabilities of combined outcomes in a sequence of experiments.

82. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 82 of 103 6.7 Bayes' Theorem 1.Bayes' Theorem (n = 2) 2.Bayes' Theorem (n = 3) 3.Bayes' Theorem

83. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 83 of 103 If B 1 and B 2 are mutually exclusive events, and B 1  B 2 = S, then for any event A in S, Bayes' Theorem (n = 2)

84. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 84 of 103 Example Bayes' Theorem (n = 2) Rework the medical problem using Bayes' Theorem. A = POS, B 1 = TB and B 2 = not TB Pr(TB) =.0002, Pr(not TB) =.9998, Pr(POS|TB) =.98, Pr(POS|not TB) =.01,

85. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 85 of 103 Example Bayes' Theorem (n = 2) (2)

86. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 86 of 103 If B 1, B 2 and B 3 are mutually exclusive events, and B 1  B 2  B 3 = S, then for any event A in S, Bayes' Theorem (n = 3)

87. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 87 of 103 If B 1, B 2, …, B n are mutually exclusive events, and B 1  B 2  …  B n = S, then for any event A in S, Bayes' Theorem

88. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 88 of 103 Example Bayes' Theorem A printer has seven book-binding machines with the proportion of production and probability of defective binding listed in the following table.

89. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 89 of 103 Example Bayes' Theorem (2) If a book is selected at random and found to have a defective binding, what is the probability that it was bound by machine 1? A = "book has a defective binding" B i = "book was bound by machine i" The second column in the table gives Pr(B i ). The last column in the table gives Pr(A|B i ).

90. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 90 of 103 Example Bayes' Theorem (3)

91. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 91 of 103 Summary Section 6.7  Bayes' theorem states that if B 1, B 2, …, B n are mutually exclusive events whose union is the entire sample space, then for any event A,

92. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 92 of 103 6.8 Simulation 1.Simulation 2.Simulation - Dice 3.Simulation - Balls From Urn 4.Simulation - Free Throw

93. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 93 of 103 Simulation Simulation is a method of imitating an experiment by using an artificial device to substitute for the real thing.

94. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 94 of 103 1. Simulate the toss of a single die. randInt(1,6)TI-83 rand(6)TI-89 2. Simulate the toss of a pair of dice. randInt(1,6) + randInt(1,6)TI-83 rand(6) + rand(6)TI-89 3. Simulate ten tosses of a pair of dice. randInt(1,6,10) + randInt(1,6,10)TI-83 Seq(rand(6) + rand(6),x,1,10)TI-89 Example Simulation - Dice

95. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 95 of 103 1. Simulate the selection of a ball from an urn containing 7 white balls and 3 red balls. 1-7 = white balls and 8-10 = red balls randInt(1,10)TI-83 rand(10)TI-89 Example Simulation - Balls From Urn

96. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 96 of 103 1. Simulate the outcome of a free throw by Michael Jordan, who had an 83% free-throw average. 1-83 = succeeded and 84-100 = missed randInt(1,100)TI-83 rand(100)TI-89 Example Simulation - Free Throw

97. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 97 of 103 Customers steadily arrive at a bank during the hour from 9 A.M. to 10 A.M. so that the line of customers is never empty. There are three tellers, and each customer requires a varying amount of time with a teller. Surveys have shown that 40% of the customers need 3 minutes, 50% need 5 minutes, and 10% need 8 minutes. Each customer enters the queue at the end and goes to the first available teller when reaching the front. Example Simulation - Queue

98. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 98 of 103 Simulate the service process. (a) Show how many of the first 20 customers each of the tellers is able to service on a random day and in that hour. (b) If all 20 customers were at the bank when it opened at 9 A.M., what was the average time spent at the bank once it opened? Example Simulation - Queue (2)

99. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 99 of 103 Example Simulation - Queue (a) Generate 20 random whole numbers from 1 - 10. The numbers 1-4 represent a customer requiring 3 min., 5-9 represent a customer requiring 5 min., and 10 represent a customer requiring 8 min.

100. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 100 of 103 Example Simulation - Queue (2a) To determine the schedule, let the tellers be A, B, and C. Then the first customer goes to teller A. Since her random numbers is 10, she requires 8 min. occupying teller A until 9:08. The second customer, with random number 7, goes to teller B and stays till 9:05. The third person goes to teller C and stays till 9:03. The fourth customer goes to teller C at 9 :03 and stays till 9:06. Continue in this fashion until the last customer leaves.

101. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 101 of 103 Example Simulation - Queue (3a)

102. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 102 of 103 Example Simulation - Queue (b) Total the number of minutes each customer spent in the bank (using End time) and divide by 20.

103. Goldstein/Schnieder/Lay: Finite Math & Its Applications, 9e 103 of 103 Summary Section 6.8  The ability to generate random numbers allows us to simulate the outcomes of experiments.