1. Sets and Probability Chapter 7

2. Ch. 7 Sets and Probabilities 4-2 Basic Concepts of Probability 4-3 Addition Rule 4-4 Multiplication Rule 4-5 Multiplication Rule: Complements and Conditional Probability 4-6 Counting 4-7 Bayes’ Theorem B. Potter - ECO 34012

3. 4-2 Basic Concepts of Probability Probability is a numerical measure of the likelihood that an event will occur. Probability values are always assigned on a scale from 0 to 1. A probability near 0 indicates an event is very unlikely to occur. A very small probability (0.001 for example) corresponds to an event that is unusual. A probability near 1 indicates an event is almost certain to occur. A probability of 0.5 indicates the occurrence of the event is just as likely as it is unlikely. B. Potter - ECO 34013

4. A procedure is any process that generates well- defined outcomes. ProcedureOutcome Toss a coinHeads, Tails Roll a die1, 2, 3, 4, 5, 6 Take a testA, B, C, D, F B. Potter - ECO 3401 Basics of Probability 4

5. A procedure is any process that generates well- defined outcomes. An event is any collection of results or outcomes of a procedure. B. Potter - ECO 3401 Basics of Probability ProcedureOutcome Toss a coinHeads, Tails Roll a die1, 2, 3, 4, 5, 6 Take a testA, B, C, D, F Event: odd numbers 5

6. A procedure is any process that generates well- defined outcomes. An event is any collection of results or outcomes of a procedure. A simple event is an outcome, or an event that cannot be further broken down into simpler components. B. Potter - ECO 3401 Basics of Probability ProcedureOutcome Toss a coinHeads, Tails Roll a die1, 2, 3, 4, 5, 6 Take a testA, B, C, D, F Simple event: tails 6

7. A procedure is any process that generates well- defined outcomes. An event is any collection of results or outcomes of a procedure. A simple event is an outcome, or an event that cannot be further broken down into simpler components. The sample space for a procedure consists of all possible simple events. B. Potter - ECO 3401 Basics of Probability ProcedureOutcome Take a testA, B, C, D, F Sample space 7

8. Basics of Probability Notation for Probabilities P denotes a probability A, B, and C denote specific events P ( A ) denotes the probability of event A occurring. B. Potter - ECO 34018

9. Assigning Probabilities Classical Method Assigning probabilities based on the assumption of equally likely outcomes. If event A can occur in s ways, then B. Potter - ECO 34019

10. Assigning Probabilities Classical Method Assigning probabilities based on the assumption of equally likely outcomes. Relative Frequency Method Conduct a procedure or observe historical data, and count the number of times event A occurred. B. Potter - ECO 340110

11. Assigning Probabilities Classical Method Assigning probabilities based on the assumption of equally likely outcomes. Relative Frequency Method Conduct a procedure or observe historical data, and count the number of times event A occurred. Law of Large Numbers – Relative frequency approximations tend to improve with more observations. B. Potter - ECO 340111

12. Assigning Probabilities Classical Method Assigning probabilities based on the assumption of equally likely outcomes. Relative Frequency Method Conduct a procedure or observe historical data, and count the number of times event A occurred. Subjective Method Assigning probabilities based on the assignor’s judgment. P ( A ) is estimated by using knowledge of the relevant circumstances. B. Potter - ECO 340112

13. Classical Method If an experiment has n possible outcomes, and event A can occur s different ways, this method would assign a probability of s / n to each outcome. Example Experiment: Rolling a die Sample Space: S = {1, 2, 3, 4, 5, 6}, n = 6 B. Potter - ECO 340113

14. Relative Frequency Method Example: Lucas Tool Rental Lucas would like to assign probabilities to the number of floor polishers it rents per day. Office records show the following frequencies of daily rentals for the last 40 days. Number of Number Polishers Rentedof Days 0 4 1 6 2 18 3 10 4 2 B. Potter - ECO 340114

15. Example: Lucas Tool Rental The probability assignments are given by dividing the number-of-days frequencies by the total number of observations (total number of days). Number of Number Polishers Rentedof Days Probability 0 4.10 = 4/40 1 6.15 = 6/40 2 18.45 etc. 3 10.25 4 2.05 401.00 Relative Frequency Method B. Potter - ECO 340115

16. Now You Try USA Today reported on a survey of office workers who were asked how much time the spend on personal phone calls at work per day. Among the responses, 1065 reported times between 1 and 10 minutes, 240 reported times between 11 and 30 minutes, 14 reported times between 31 and 60 minutes, and 66 said they do not make personal calls at work. If a worker is randomly selected, what is the probability the worker does not make personal calls at work? B. Potter - ECO 340116

17. Now You Try Time (minutes)# of WorkersProbability 0 66 1 – 101065 11 – 30 240 31 – 60 14 B. Potter - ECO 340117

18. Subjective Method When economic conditions and circumstances change rapidly it might be inappropriate to assign probabilities based solely on historical data. Past results are no guarantee of future performance! B. Potter - ECO 340118

19. Subjective Method When economic conditions and circumstances change rapidly it might be inappropriate to assign probabilities based solely on historical data. We can use any data available as well as our experience and intuition, but ultimately a probability value should express our degree of belief that the experimental outcome will occur. B. Potter - ECO 340119

20. Subjective Method When economic conditions and circumstances change rapidly it might be inappropriate to assign probabilities based solely on historical data. We can use any data available as well as our experience and intuition, but ultimately a probability value should express our degree of belief that the experimental outcome will occur. The best probability estimates often are obtained by combining the estimates from the classical or relative frequency approach with the subjective estimates. B. Potter - ECO 340120

21. Complementary Events The complement of event A, denoted consists of all outcomes in which event A does not occur. B. Potter - ECO 3401 Event A Sample Space S Venn Diagram 21

22. The union of events A and B is the event containing all sample points that are in A or B or both. The union is denoted by A  B  The union of A and B is illustrated below. P( A  B) = The probability of the occurrence of Event A or Event B or both. Sample Space S Union of Two Events Event A Event B B. Potter - ECO 340122

23. Intersection of Two Events The intersection of events A and B is the set of all sample points that are in both A and B. The intersection is denoted by A  The intersection of A and B is the area of overlap in the illustration below. P( A  ) = The probability of the occurrence of Event A and Event B. Sample Space S Event A Event B Intersection 23

24. Addition Rule The addition rule provides a way to compute the probability of the union of event A and B, P ( A or B ) The law is written as: P ( A  B ) = P ( A ) + P ( B ) – P ( A  B ) Event A B. Potter - ECO 340124

25. Addition Rule The addition rule provides a way to compute the probability of the union of event A and B, P ( A or B ) The law is written as: P ( A  B ) = P ( A ) + P ( B ) – P ( A  B ) Event B B. Potter - ECO 340125

26. Addition Rule The addition rule provides a way to compute the probability of the union of event A and B, P ( A or B ) The law is written as: P ( A  B ) = P ( A ) + P ( B ) – P ( A  B ) Event A Event B B. Potter - ECO 340126

27. Addition Rule The following table contains data from a class of 312 students. Find the probability of randomly selecting a student who has red hair or green eyes. B. Potter - ECO 3401 Hair/EyesBrownBlueGreenHazel Totals Blonde 3015182689 Brunette 46182435123 Black 34351658 Red 1869942 Totals 128425686312 27

28. Addition Rule Find the probability of randomly selecting a student who has red hair or green eyes. B. Potter - ECO 3401 Hair/EyesBrownBlueGreenHazel Totals Blonde 3015182689 Brunette 46182435123 Black 34351658 Red 1869942 Totals 128425686312 Let R h = the event that a student has red hair, and let G e = the event that a student has green eyes. Then, P ( R h  G e ) = P ( R h ) + P ( G e ) – P ( R h  G e ) 28

29. Addition Rule Find the probability of randomly selecting a student who has red hair or green eyes. B. Potter - ECO 3401 Hair/EyesBrownBlueGreenHazel Totals Blonde 3015182689 Brunette 46182435123 Black 34351658 Red 1869942 Totals 128425686312 P ( R h  G e ) = P ( R h ) + P ( G e ) – P ( R h  G e ) 29

30. Addition Rule Find the probability of randomly selecting a student who has red hair or green eyes. B. Potter - ECO 3401 Hair/EyesBrownBlueGreenHazel Totals Blonde 3015182689 Brunette 46182435123 Black 34351658 Red 1869942 Totals 128425686312 P ( R h  G e ) = 0.135 + 0.18 – P ( R h  G e ) 30

31. Addition Rule Find the probability of randomly selecting a student who has red hair or green eyes. B. Potter - ECO 3401 Hair/EyesBrownBlueGreenHazel Totals Blonde 3015182689 Brunette 46182435123 Black 34351658 Red 1869942 Totals 128425686312 P ( R h  G e ) = 0.135 + 0.18 – P ( R h  G e ) 31

32. Addition Rule Find the probability of randomly selecting a student who has red hair or green eyes. B. Potter - ECO 3401 Hair/EyesBrownBlueGreenHazel Totals Blonde 3015182689 Brunette 46182435123 Black 34351658 Red 1869942 Totals 128425686312 P ( R h  G e ) = 0.135 + 0.18 – 0.029 = 0.286 32

33. Addition Law for Mutually Exclusive (Disjoint) Events Two events are said to be mutually exclusive if the events have no sample points in common. That is, two events are mutually exclusive if, when one event occurs, the other cannot occur. Addition Law for Mutually Exclusive Events: P( A  B ) = P( A ) + P( B ) Sample Space S Event A Event B B. Potter - ECO 340133

34. Roll the Dice If you roll 2 dice, what’s the probability of rolling a 7 or 11?

35. 123456 1234567 2345678 3456789 45678910 56789 11 6789101112 Die 1 Die 2 B. Potter - ECO 340135

36. 123456 1234567 2345678 3456789 45678910 56789 11 6789101112 Die 1 Die 2 P (7) = 6/36 =.167 P (11) = 2/36 =.056 B. Potter - ECO 340136

37. Roll the Dice If you roll 2 dice, what’s the probability of rolling a 7 or 11?

38. B. Potter - ECO 3401 Addition Law for Complementary Events Event A Total Area = 1 38

39. Conditional Probability The probability of an event given that another event has occurred is called a conditional probability. The conditional probability of A given B is denoted by P( A | B ). B. Potter - ECO 340139

40. ManWomanTotals Promoted 28836324 Not Promoted 672204876 Totals 9602401200 Conditional Probability B. Potter - ECO 340140

41. ManWomanTotals Promoted P = 288/1200 36324 Not Promoted 672204876 Totals 9602401200 Conditional Probability B. Potter - ECO 340141

42. ManWomanTotals Promoted P = 0.2436324 Not Promoted 672204876 Totals 9602401200 Conditional Probability B. Potter - ECO 340142

43. Joint Probability Table Man ( M )Woman ( W ) Totals Promoted ( A ).24.03.27 Not Promoted.56.17.73 Totals.80.201.00 P( A  ) P(  ) B. Potter - ECO 340143

44. Conditional Probability The probability of an event given that another event has occurred is called a conditional probability. The conditional probability of A given B is denoted by P( A | B ). A conditional probability is computed as follows: If P(A|B) = 0, then event A and event B are mutually exclusive. B. Potter - ECO 340144

45. Conditional Probability Officer promoted given the officer is a man: Officer promoted given the officer is a woman: B. Potter - ECO 340145

46. Multiplication Rule The multiplication law provides a way to compute the probability of an intersection of two events. The probability that event A and event B will occur. The law is written as: Event A Event B B. Potter - ECO 340146

47. Multiplication Rule: Example B. Potter - ECO 3401 Suppose 26% of the population are 66 or over, 35% of those 66 or over have a loan. Find the probability that a randomly selected person is 66 or older and has a loan. Let L represent the event that a person has a loan, and O represent the event that a person is 66 or older. Then 47

48. Multiplication Law for Independent Events Events A and B are independent if P( A | B ) = P( A ). Multiplication Law for Independent Events: P( A  B ) = P( A )P( B ) The multiplication law also can be used as a test to see if two events are independent. B. Potter - ECO 340148

49. You are given the following information on Events A, B, C, and D (A, B, C, and D are not the entire sample space). P(A) =.4P(A  D) =.6 P(B) =.2P(A  B) =.3 P(C) =.1P(A  C) =.04 P(A  D) =.03 h. Are A and C independent? d. Compute the probability of the complement of C. g. Are A and C mutually exclusive? c. Compute P(A f. Are A and B independent? e. Are A and B mutually exclusive? a. Compute P(D). h. Are A and C independent? d. Compute the probability of the complement of C.complement of C. g. Are A and C mutually exclusive? c. Compute P(A  C)Compute P(A  C) f. Are A and B independent? e. Are A and B mutually exclusive? a. Compute P(D).Compute P(D). b. Compute P(A  B)Compute P(A  B) Hint: Use the addition rule B. Potter - ECO 340149

50. Bayes’ Theorem The probability of an event A | B is generally different from the probability of B | A. The probability of an event A | B is generally different from the probability of B | A. However, there is a definite relationship between the two. However, there is a definite relationship between the two. Bayes' theorem is the statement of that relationship. Bayes' theorem is the statement of that relationship. Medical researchers know that the probability of getting lung cancer if a person smokes is.34. Medical researchers know that the probability of getting lung cancer if a person smokes is.34. The probability that a nonsmoker will get lung cancer is.03. The probability that a nonsmoker will get lung cancer is.03. With Bayes’ theorem, we can calculate the probability that a person with lung cancer is (or was) a smoker. 50

51. n Prior Probabilities Let: Bayes’ Theorem S = Person is a smoker S = Person is a smoker N = Person is a non-smoker N = Person is a non-smoker S = Person is a smoker S = Person is a smoker N = Person is a non-smoker N = Person is a non-smoker P( S ) =.22, P( N ) =.78 According to the Center for Disease Control and Prevention, approximately 22% of the population 18 years or older smokes tobacco products regularly. B. Potter - ECO 340151

52. nConditional Probabilities Let: Let: Bayes’ Theorem P ( C | S ) =.34 P ( C | N ) =.03 P (H| S ) =.66 P (H| N ) =.97 Hence: C = Person has (or will have) lung cancer H = Person will not have lung cancer C = Person has (or will have) lung cancer H = Person will not have lung cancer Based upon medical research: B. Potter - ECO 340152

53. Bayes’ Theorem We can illustrate the different possible outcomes with a tree diagram (2-step experiment). H S N C H C ( S, C ) ( N, C ) ( N, H ) ( S, H ) Step 1 Smoker or non-smoker Step 2 Health ExperimentalOutcomes B. Potter - ECO 340153

54. P( H | S ) =.66 P( S ) =.22 P( N ) =.78 P( C | N ) =.03 P( H | N ) =.97 P( C | S ) =.34  P( S  C ) = P( S )P( C | S ) =.07  P( N  C ) =.02  P( N  H ) =.76  P( S  H ) =.15 Step 1 Smoker or non-smoker Step 2 Health ExperimentalOutcomes Bayes’ Theorem Now we can fill in the probabilities: B. Potter - ECO 340154

55. Bayes’ Theorem Now suppose we want to determine the probability that a person who has been diagnosed with lung cancer is a smoker. In other words, From the law of conditional probabilities, we know that From the probability tree, we know that  Event C can occur in only two ways: ( S  C ) and ( N  C ) Numerator, Equation 1 Denominator, Equation 2 Posterior probability B. Potter - ECO 340155

56. Bayes’ Theorem (2 events) B. Potter - ECO 340156

57. Bayes’ Theorem To find the posterior probability that event A i will To find the posterior probability that event A i will occur given that event B has occurred, we apply occur given that event B has occurred, we apply Bayes’ theorem. Bayes’ theorem. Bayes’ theorem is applicable when the events for Bayes’ theorem is applicable when the events for which we want to compute posterior probabilities which we want to compute posterior probabilities are mutually exclusive and their union is the entire are mutually exclusive and their union is the entire sample space. sample space. B. Potter - ECO 340157

58. Bayes’ Theorem, example A local bank reviewed its credit card policy with the intention of recalling some of its credit cards. In the past, approximately 5% of cardholders defaulted, leaving the bank unable to collect the outstanding balance. Hence, management established a prior probability of.05 that any particular cardholder will default. The bank also found that the probability of missing a monthly payment is.20 for customers who do not default. Of course, the probability of missing a monthly payment for those who default is 1. Given that a customer missed a monthly payment, compute the posterior probability that the customer will default. B. Potter - ECO 340158

59. Bayes’ Theorem M = missed payment D 1 = customer defaults D 2 = customer does not default P(D 1 ) =.05 P(D 2 ) =.95 P(M|D 2 ) =.2 P(M|D 1 ) = 1 B. Potter - ECO 340159

60. Now You Try In 2001, 26.2% of Americans were college graduates. The probability that a college graduate used direct deposit with his or her financial institution was.78. For non- college graduates, the probability was.62. Suppose a person used direct deposit. What is the probability that that person was a college graduate? B. Potter - ECO 340160

61. End of Chapter 7 B. Potter - ECO 340161