1. Ka-fu Wong © 2003 Chap 5- 1 Dr. Ka-fu Wong ECON1003 Analysis of Economic Data

2. Ka-fu Wong © 2003 Chap 5- 2 Chapter Five A Survey of Probability Concepts l GOALS 1.Define probability. 2.Describe the classical, empirical, and subjective approaches to probability. 3.Understand the terms: experiment, event, outcome, permutations, and combinations. 4.Define the terms: conditional probability and joint probability. 5.Calculate probabilities applying the rules of addition and the rules of multiplication. 6.Use a tree diagram to organize and compute probabilities. 7.Calculate a probability using Bayes’ theorem.

3. Ka-fu Wong © 2003 Chap 5- 3 Definitions A probability is a measure of the likelihood that an event in the future will happen. It can only assume a value between 0 and 1. A value near zero means the event is not likely to happen. A value near one means it is likely. There are three definitions of probability: classical, empirical, and subjective.

4. Ka-fu Wong © 2003 Chap 5- 4 Definitions continued The classical definition applies when there are n equally likely outcomes. The empirical definition applies when the number of times the event happens is divided by the number of observations, based on data. Subjective probability is based on whatever information is available, based on subjective feelings.

5. Ka-fu Wong © 2003 Chap 5- 5 A fair die is rolled once. Peter is concerned with whether the resulted number is even, i.e., 2, 4, 6. Paul is concerned with whether the resulted number is less than or equal to 3, i.e., 1, 2, 3. Mary is concerned with whether the resulted number is 6. Sonia is concerned with whether the resulted number is odd, i.e., 1, 3, 5. A fair die is rolled twice. John is concerned with whether the resulted number of first roll is even, i.e., 2, 4, 6. Sarah is concerned with whether the resulted number of second roll is even, i.e., 2, 4, 6. Example 1 (to be used to illustrate the definitions)

6. Ka-fu Wong © 2003 Chap 5- 6 Definitions continued An experiment is the observation of some activity or the act of taking some measurement. The experiment is rolling the die. An outcome is the particular result of an experiment. The possible outcomes are the numbers 1, 2, 3, 4, 5, and 6. An event is the collection of one or more outcomes of an experiment. For Peter: the occurrence of an even number, i.e., 2, 4, 6. For Paul: the occurrence of a number less than or equal to 3, i.e., 1, 2, 3. For Mary: the occurrence of a number 6. For Sonia: the occurrence of an odd number, i.e., 1, 3, 5.

7. Ka-fu Wong © 2003 Chap 5- 7 Mutually Exclusive, Independent and Exhaustive events Events are mutually exclusive if the occurrence of any one event means that none of the others can occur at the same time. Peter’s event and Paul’s event are not mutually exclusive – both contains 2. Peter’s event and Mary’s event are not mutually exclusive – both contains 6. Paul’s event and Mary’s event are mutually exclusive – no common numbers. Peter’s event and Sonia’s event are mutually exclusive – no common numbers.

8. Ka-fu Wong © 2003 Chap 5- 8 Mutually Exclusive, Independent and Exhaustive events Events are independent if the occurrence of one event does not affect the occurrence of another. P(A&B) = P(A)*P(B) Not independent: Peter’s event and Paul’s event. Peter’s event and Mary’s event. Paul’s event and Mary’s event. Peter’s event and Sonia’s event. Independent: John’s event and Sarah’s event are independent. P(John & Sarah) = P(John)*P(Sarah)

9. Ka-fu Wong © 2003 Chap 5- 9 Mutually Exclusive, Independent and Exhaustive events Events are collectively exhaustive if at least one of the events must occur when an experiment is conducted. Peter ’ s event (even numbers) and Sonia ’ s event (odd numbers) are collectively exhaustive. Peter ’ s event (even numbers) and Mary ’ s event (number 6) are not collectively exhaustive.

10. Ka-fu Wong © 2003 Chap 5- 10 Example 2 Throughout her teaching career Professor Jones has awarded 186 A’s out of 1,200 students. What is the probability that a student in her section this semester will receive an A? This is an example of the empirical definition of probability. To find the probability a selected student earned an A: This number may be interpreted as “unconditional probability”. In most cases, we are interested in the probability of earning an A for a selected student who study 10 hours or more per week. We call this “conditional probability”. P(A | study 10 or more hours per week)

11. Ka-fu Wong © 2003 Chap 5- 11 Subjective Probability Examples of subjective probability are: Estimating the probability mortgage rates for home loans will top 8 percent this year. Estimating the probability that HK’s economic growth will be 3% this year. Estimating the probability that HK will solve its deficit problem in 2006.

12. Ka-fu Wong © 2003 Chap 5- 12 Learning exercise 1: University Demographics Current enrollments by college and by sex appear in the following table. CollegeAg-ForArts-SciBus-EconEducEngrLawUndeclTotals Female500150040010002001008004500 Male9001200500 13002009005500 Totals140027009001500 300170010000 If I select a student at random, answer the following: Find P(Female or Male) Find P(not-Ag-For) Find P(Female |BusEcon) Find P(Male and Arts-Sci) Are “Female” and “Educ” Statistical independent? Why or Why not?

13. Ka-fu Wong © 2003 Chap 5- 13 Learning exercise 1: University Demographics CollegeAg-ForArts-SciBus-EconEducEngrLawUndeclTotals Female500150040010002001008004500 Male9001200500 13002009005500 Totals140027009001500 300170010000 P(Female or Male)=(4500 + 5500)/10000 = 1 P(not-Ag-For)=(10000 – 1400) /10000 = 0.86 P(Female | Bus Econ)= 400 /900 = 0.44 P(Male and Arts-Sci)=1200 /10000 = 0.12 Are Female and Educ Statistical independent? P(female and Educ)=1000 /10000 = 0.1 P(female)=4500 /10000 = 0.45 P(Educ)=1500 /10000 = 0.15 P(female and Educ)> P(female)*P(Educ) = 0.0675 NO!

14. Ka-fu Wong © 2003 Chap 5- 14 Learning exercise 2: Predicting Sex of Babies Many couples take advantage of ultrasound exams to determine the sex of their baby before it is born. Some couples prefer not to know beforehand. In any case, ultrasound examination is not always accurate. About 1 in 5 predictions are wrong. In one medical group, the proportion of girls correctly identified is 9 out of 10 and the number of boys correctly identified is 3 out of 4. The proportion of girls born is 48 out of 100. What is the probability that a baby predicted to be a girl actually turns out to be a girl? Formally, find P(girl | test says girl).

15. Ka-fu Wong © 2003 Chap 5- 15 Learning exercise 2: Predicting Sex of Babies P(girl | test says girl) In one medical group, the proportion of girls correctly identified is 9 out of 10 and the number of boys correctly identified is 3 out of 4. The proportion of girls born is 48 out of 100. Think about the next 1000 births handled by this medical group. 480 = 1000*0.48 are girls 520 = 1000*0.52 are boys Of the girls, 432 (=480*0.9) tests indicate that they are girls. Of the boys, 130 (=520*0.25) tests indicate that they are girls. In total, 562 (=432+130) tests indicate girls. Out of these 462 babies, 432 are girls. Thus P(girl | test says girl ) = 432/562 = 0.769

16. Ka-fu Wong © 2003 Chap 5- 16 Learning exercise 2: Predicting Sex of Babies 480 = 1000*0.48 are girls 520 = 1000*0.52 are boys Of the girls, 432 (=480*0.9) tests indicate that they are girls. Of the boys, 130 (=520*0.25) tests indicate that they are girls. In total, 562 tests indicate girls. Out of these 562 babies, 432 are girls. Thus P(girls | test syas girls ) = 432/562 = 0.769 1000*P(girls) 1000*P(boys) 1000*[P(girls)*P(test says girls|girls) + P(boys)*P(test says girls|boys)] 1000*P(boys)*P(test says girls | boys) 1000*P(girls)*P(test says girls|girls) 1000*[P(girls)*P(test says girls|girls) + P(boys)*P(test says girls|boys)]

17. Ka-fu Wong © 2003 Chap 5- 17 Learning exercise 3: Putting in Extra Trunk Lines Between [insert local names for Town A and Town B] Given recent flooding (or other condition more appropriate to your area) between Town A and Town B, the local telephone company is assessing the value of adding an independent trunk line between the two towns. The second line will fail independently of the first because it will depend on different equipment and routing (we assume a regional disaster is highly unlikely). Under current conditions, the present line works 98 out of 100 times someone wishes to make a call. If the second line performs as well, what is the chance that a caller will be able to get through? Formally, P( Line 1 works ) = 98/100 P( Line 2 works ) = 98/100 Find P( Line 1 or Line 2 works ).

18. Ka-fu Wong © 2003 Chap 5- 18 Learning exercise 3: Putting in Extra Trunk Lines Between [insert local names for Town A and Town B] P( Line 1 works ) = 98/100 P( Line 2 works ) = 98/100 Find P( Line 1 or Line 2 works ). P( Line 1 or Line 2 works ) = 1 – P(Both line1 and Line 2 fail) = 1 – P(Line 1 fails)*P(line 2 fails) = 1 – 0.02*0.02 = 0.9996. Line 2 worksLine 2 fails Line 1 works0.98*0.980.98*0.02 Line 1 fails0.02*0.980.02*0.02

19. Ka-fu Wong © 2003 Chap 5- 19 Learning exercise 4: Part-time Work on Campus A student has been offered part-time work in a laboratory. The professor says that the work will vary from week to week. The number of hours will be between 10 and 20 with a uniform probability density function, represented as follows: How tall is the rectangle? What is the probability of getting less than 15 hours in a week? Given that the student gets at least 15 hours in a week, what is the probability that more than 17.5 hours will be available?

20. Ka-fu Wong © 2003 Chap 5- 20 Learning exercise 4: Part-time Work on Campus How tall is the rectangle? (20-10)*h = 1 h=0.1 What is the probability of getting less than 15 hours in a week? 0.1*(15-10) = 0.5 Given that the student gets at least 15 hours in a week, what is the probability that more than 17.5 hours will be available? 0.1*(20-17.5) = 0.25 0.25/0.5 = 0.5 P(hour>17.5)/P(hour>15)

21. Ka-fu Wong © 2003 Chap 5- 21 Learning exercise 5: Customer Complaints You are the manager of the complaint department for a large mail order company. Your data and experience indicate that the time it takes to handle a single call has the following probability density function, Show that the area under the triangle is 1. Find the probability that a call will take longer than 10 minutes. That is, find P( Time > 10 ). Given that the call takes at least 5 minutes, what is the probability that it will take longer than 10 minutes? That is, find P( Time > 10 | Time > 5 ). Find P( Time < 10 ). h=2/15

22. Ka-fu Wong © 2003 Chap 5- 22 Learning exercise 5: Customer Complaints Show that the area under the triangle is 1. (2/15)*(15-0)/2 = 1 h=2/15

23. Ka-fu Wong © 2003 Chap 5- 23 Learning exercise 5: Customer Complaints Find the probability that a call will take longer than 10 minutes. That is, find P( Time > 10 ). (2/15)/x = 10/5 X=2/30 = 0.067 P(time>10)=(2/30)*5/2=1/6 h=2/15 x

24. Ka-fu Wong © 2003 Chap 5- 24 Learning exercise 5: Customer Complaints Find P( Time > 10 | Time > 5 ). P(time>10)=(2/30)*5/2=1/6 P(time>5) = 2/15*10/2=2/3 P( Time > 10 | Time > 5 ) = (1/6)/(2/3) = 1/4. Find P( Time < 10 ). P( Time 10) = 1-1/6 = 5/6. h=2/15 x

25. Ka-fu Wong © 2003 Chap 5- 25 Learning exercise 6: Clutch Sizes in Boreal Owl Nests The number of eggs in Boreal owl nests has a probability mass function with P(0) = 0.2, P(1) = 0.1, P(2) = 0.1, P(3) = 0.3, and P(4) = 0.3. Draw the probability mass function (pmf) correctly labeling the axes. What is the probability that a randomly chosen nest will have no eggs? If you examine two nests and they are independent, what is the probability that neither nest will have eggs?

26. Ka-fu Wong © 2003 Chap 5- 26 Learning exercise 6: Clutch Sizes in Boreal Owl Nests What is the probability that a randomly chosen nest will have no eggs? P(0) = 0.2 P(0) = 0.2, P(1) = 0.1, P(2) = 0.1, P(3) = 0.3, and P(4) = 0.3. Draw the probability mass function (pmf) correctly labeling the axes. If you examine two nests and they are independent, what is the probability that neither nest will have eggs? P(0 & 0) = P(0)*P(0) = 0.04.

27. Ka-fu Wong © 2003 Chap 5- 27 Basic Rules of Probability If two events A and B are mutually exclusive, the special rule of addition states that the probability of A or B occurring equals the sum of their respective probabilities: P(A or B) = P(A) + P(B)

28. Ka-fu Wong © 2003 Chap 5- 28 EXAMPLE 3 New England Commuter Airways recently supplied the following information on their commuter flights from Boston to New York:

29. Ka-fu Wong © 2003 Chap 5- 29 EXAMPLE 3 continued If A is the event that a flight arrives early, then P(A) = 100/1000 =.10. If B is the event that a flight arrives late, then P(B) = 75/1000 =.075. The probability that a flight is either early or late is: P(A or B) = P(A) + P(B) =.10 +.075 =.175.

30. Ka-fu Wong © 2003 Chap 5- 30 The Complement Rule The complement rule is used to determine the probability of an event occurring by subtracting the probability of the event not occurring from 1. If P(A) is the probability of event A and P(~A) is the complement of A, P(A) + P(~A) = 1 or P(A) = 1 - P(~A).

31. Ka-fu Wong © 2003 Chap 5- 31 The Complement Rule continued A Venn diagram illustrating the complement rule would appear as: A ~A

32. Ka-fu Wong © 2003 Chap 5- 32 EXAMPLE 4 Recall EXAMPLE 3. Use the complement rule to find the probability of an early (A) or a late (B) flight P(A or B) = 1 - P(C or D) If C is the event that a flight arrives on time, then P(C) = 800/1000 =.8. If D is the event that a flight is canceled, then P(D) = 25/1000 =.025. P(A or B) = 1 - P(C or D) = 1 - [.8 +.025] =.175

33. Ka-fu Wong © 2003 Chap 5- 33 EXAMPLE 4 continued P(A or B) = 1 - P(C or D) = 1 - [.8 +.025] =.175 C.8 D.025 ~(C or D) = (A or B).175

34. Ka-fu Wong © 2003 Chap 5- 34 The General Rule of Addition If A and B are two events that are not mutually exclusive, then P(A or B) is given by the following formula: P(A or B) = P(A) + P(B) - P(A and B)

35. Ka-fu Wong © 2003 Chap 5- 35 The General Rule of Addition The Venn Diagram illustrates this rule: A and B A B

36. Ka-fu Wong © 2003 Chap 5- 36 EXAMPLE 5 In a sample of 500 students, 320 said they had a stereo, 175 said they had a TV, and 100 said they had both: Stereo 320 Both 100 TV 175

37. Ka-fu Wong © 2003 Chap 5- 37 EXAMPLE 5 continued If a student is selected at random, what is the probability that the student has only a stereo, only a TV, and both a stereo and TV? P(S) = 320/500 =.64. P(T) = 175/500 =.35. P(S and T) = 100/500 =.20. In a sample of 500 students, 320 said they had a stereo, 175 said they had a TV, and 100 said they had both.

38. Ka-fu Wong © 2003 Chap 5- 38 EXAMPLE 5 continued If a student is selected at random, what is the probability that the student has either a stereo or a TV in his or her room? P(S or T) = P(S) + P(T) - P(S and T) =.64 +.35 -.20 =.79. In a sample of 500 students, 320 said they had a stereo, 175 said they had a TV, and 100 said they had both. P(S) = 320/500 =.64. P(T) = 175/500 =.35. P(S and T) = 100/500 =.20.

39. Ka-fu Wong © 2003 Chap 5- 39 Joint Probability A joint probability measures the likelihood that two or more events will happen concurrently. An example would be the event that a student has both a stereo and TV in his or her dorm room. P(dice=5, and dice=6) =0 because the two outcomes are mutually exclusive.

40. Ka-fu Wong © 2003 Chap 5- 40 Special Rule of Multiplication The special rule of multiplication requires that two events A and B are independent. Two events A and B are independent if the occurrence of one has no effect on the probability of the occurrence of the other. This rule is written: P(A and B) = P(A)P(B)

41. Ka-fu Wong © 2003 Chap 5- 41 EXAMPLE 6 Chris owns two stocks, IBM and General Electric (GE). The probability that IBM stock will increase in value next year is.5 and the probability that GE stock will increase in value next year is.7. Assume the two stocks are independent. What is the probability that both stocks will increase in value next year? P(IBM and GE) = (.5)(.7) =.35.

42. Ka-fu Wong © 2003 Chap 5- 42 EXAMPLE 6 continued What is the probability that at least one of these stocks increase in value during the next year? (This means that either one can increase or both.) Approach 1: P(at least one) = (.5)(.3) + (.5)(.7) +(.7)(.5) =.85. Approach 2: P(at least one) = 0.5 + 0.7 – 0.35 =.85. The probability that IBM stock will increase in value next year is.5 and the probability that GE stock will increase in value next year is.7.

43. Ka-fu Wong © 2003 Chap 5- 43 Conditional Probability A conditional probability is the probability of a particular event occurring, given that another event has occurred. The probability of the event A given that the event B has occurred is written P(A|B). P(female | BusEcon) P(girl | test says girl)

44. Ka-fu Wong © 2003 Chap 5- 44 General Multiplication Rule The general rule of multiplication is used to find the joint probability that two events will occur. It states that for two events A and B, the joint probability that both events will happen is found by multiplying the probability that event A will happen by the conditional probability of B given that A has occurred.

45. Ka-fu Wong © 2003 Chap 5- 45 General Multiplication Rule The joint probability, P(A and B) is given by the following formula: P(A and B) = P(A)P(B/A) or P(A and B) = P(B)P(A/B) P(test says girl and girl) = P(girls) * P(test says girls | girls) P(test says boy and boy) = P(boys) * P(test says boys | boys)

46. Ka-fu Wong © 2003 Chap 5- 46 EXAMPLE 7 The Dean of the School of Business at Owens University collected the following information about undergraduate students in her college:

47. Ka-fu Wong © 2003 Chap 5- 47 EXAMPLE 7 continued If a student is selected at random, what is the probability that the student is a female (F) accounting major (A) P(A and F) = 110/1000. Given that the student is a female, what is the probability that she is an accounting major? Approach 1: P(A|F) = P(A and F)/P(F) = [110/1000]/[400/1000] =.275 Approach 2: P(A|F) = 110/400 =.275

48. Ka-fu Wong © 2003 Chap 5- 48 Tree Diagrams A tree diagram is useful for portraying conditional and joint probabilities. It is particularly useful for analyzing business decisions involving several stages. EXAMPLE 8: In a bag containing 7 red chips and 5 blue chips you select 2 chips one after the other without replacement. Construct a tree diagram showing this information.

49. Ka-fu Wong © 2003 Chap 5- 49 EXAMPLE 8 continued R1 B1 R2 B2 R2 B2 7/12 5/12 6/11 5/11 7/11 4/11

50. Ka-fu Wong © 2003 Chap 5- 50 EXAMPLE 8 continued R1 B1 R2 B2 R2 B2 7/12 5/12 6/11 5/11 7/11 4/11 The tree diagram is very illustrative about the relation between joint probability and conditional probability Let A (B) be the event of a red chip in the first (second) draw. P(B|A) = 6/11 P(A) = 7/12 P(A and B) = P(A)*P(B|A) = 6/11 * 7/12 7/12 6/11

51. Ka-fu Wong © 2003 Chap 5- 51 Bayes ’ Theorem Bayes ’ Theorem is a method for revising a probability given additional information. It is computed using the following formula: Bayes Theorem is an essential tool to understand options and real options.

52. Ka-fu Wong © 2003 Chap 5- 52 Bayes ’ Theorem Bayes ’ Theorem can be derived based on simple manipulation of the general multiplication rule. P(A 1 |B) = P(A 1 & B) /P(B) = [P(A 1 ) P(B|A 1 )] / P(B) = [P(A 1 ) P(B|A 1 )] / [P(A 1 & B) + P(A 2 & B)] = [P(A1) P(B|A 1 ) ]/ [P(A 1 ) P(B|A 1 ) + P(A 2 ) P(B|A 2 )

53. Ka-fu Wong © 2003 Chap 5- 53 EXAMPLE 9 Duff Cola Company recently received several complaints that their bottles are under-filled. A complaint was received today but the production manager is unable to identify which of the two Springfield plants (A or B) filled this bottle. What is the probability that the under-filled bottle came from plant A? The following table summarizes the Duff production experience.

54. Ka-fu Wong © 2003 Chap 5- 54 Example 9 continued The likelihood the bottle was filled in Plant A is reduced from.55 to.4783. Without the information about U, the manager will say the under-filled bottle is likely from plant A. With the additional information about U, the manager will say the under-filled bottle is likely from plant B. What is the probability that the under-filled bottle came from plant A?

55. Ka-fu Wong © 2003 Chap 5- 55 Some Principles of Counting The multiplication formula indicates that if there are m ways of doing one thing and n ways of doing another thing, there are m x n ways of doing both. Example 10: Dr. Delong has 10 shirts and 8 ties. How many shirt and tie outfits does he have? (10)(8) = 80

56. Ka-fu Wong © 2003 Chap 5- 56 Some Principles of Counting A permutation is any arrangement of r objects selected from n possible objects. Note: The order of arrangement is important in permutations.

57. Ka-fu Wong © 2003 Chap 5- 57 Some Principles of Counting A combination is the number of ways to choose r objects from a group of n objects without regard to order.

58. Ka-fu Wong © 2003 Chap 5- 58 EXAMPLE 11 There are 12 players on the Carolina Forest High School basketball team. Coach Thompson must pick five players among the twelve on the team to comprise the starting lineup. How many different groups are possible?

59. Ka-fu Wong © 2003 Chap 5- 59 Example 11 continued Suppose that in addition to selecting the group, he must also rank each of the players in that starting lineup according to their ability.

60. Ka-fu Wong © 2003 Chap 5- 60 - END - Chapter Five A Survey of Probability Concepts

61. Ka-fu Wong © 2003 Chap 5- 61 Probabilities for n Dice Suppose we roll n regular balanced six-sided dice. If the event X is the sum of the n values which appear, what are the probabilities associated for each value of X, for the possible values X = n,..., 6n? In the case n =1, these probabilities are all 1/6. For two dice, it is easiset to consider a table of possible outcomes:

62. Ka-fu Wong © 2003 Chap 5- 62 Probabilities for n Dice For two dice, it is easiest to consider a table of possible outcomes: (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)

63. Ka-fu Wong © 2003 Chap 5- 63 Probabilities for n Dice Next, we can consider the sums associated with these outcomes:: 234567 345678 456789 5678910 6789 11 789101112

64. Ka-fu Wong © 2003 Chap 5- 64 Probabilities for n Dice Since there are 36 outcomes, each equally likely, we see that the probabilities are: XCountP(X)XCountP(X) 211/36855/36 322/36944/36 433/361033/36 544/361122/36 655/361211/36 766/36

65. Ka-fu Wong © 2003 Chap 5- 65 Probabilities for n Dice If there are more than two dice, it is difficult to make tables such as these. The number of outcomes will be 6n, so to calculate the probabilities it is sufficient to count the number of times each sum occurs among the 6n possible outcomes. This is easily done by considering the generating function for the number of times each sum appears: f(x) = (x + x 2 + x 3 + x 4 + x 5 +x 6 ) n In our example, n=2 (x + x 2 + x 3 + x 4 + x 5 +x 6 ) 2 =x 2 + 2x 3 +3x 4 + 4x 5 +5x 6 +6x 7 +5x 8 +4x 9 +3x 10 +2x 11 +x 12 The coefficient of x 3 is number of times the sum of 3 occurs.