Lesson 3:

1. Lesson 3: A Survey of Probability Concepts

2. Outline

3. Definition: Probability 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. Definition: Classical probability • The classical definition applies when there are nequally likely outcomes. • For example, in the tossing of a single perfectly cubical die, made of completely homogeneous material, the equally likely events are the appearance of any of the specific number of dots (from 1 to 6) on its upper face.

5. Definition: Empirical or statistical probability • The empirical or statistical definition applies when the number of times the event happens is divided by the number of observations, based on data. 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 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)

6. Definition: Subjective probability • Subjective probability is based on whatever information is available, based on subjective feelings. • 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.

7. Example 1(to be used to illustrate the definitions) • 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.

8. Definitions: Experiment and outcome • An experimentis the observation of some activity or the act of taking some measurement. • The experiment is rolling the one die in the first example, and rolling one die twice in the second example. • An outcome is the particular result of an experiment. • The possible outcomes are the numbers 1, 2, 3, 4, 5, and 6 in the first example. • The possible outcomes are number pairs (1,1), (1,2), …, (6,6), in the second example.

9. Definition: Event • 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. • For John: the occurrence of (2,1), (2,2), (2,3),…, (2,6), (4,1),…,(4,6), (6,1),…,(6,6) [John does not care about the result of the second roll].

10. Learning exercise 1: University Demographics • Current enrollments by college and by sex appear in the following table. • If we select a student at random, what is the probability that the student is : • A female or male, i.e., P(Female or Male). • Not from Agricultural and Forestry, i.e., P(not-Ag-For) • A female given that the student is known to be from BusEcon, i.e., P(Female |BusEcon). • A female and from BusEcon, i.e., P(Female and BusEcon). • From BusEcon, i.e., P(BusEcon).

11. Learning exercise 1: University Demographics 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(Female and BusEcon) = 400 /10000 = 0.04 P(BusEcon) = 900 /10000 = 0.09 P(Female and BusEcon) = P(BusEcon) P(Female | Bus Econ)

12. Definition: Mutually Exclusive 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.

13. Definition: 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.

14. 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)

15. 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, i.e., applying the test to 100 baby girls, 90 of the tests will indicate girls. and • the number of boys correctly identified is 3 out of 4. i.e., applying the test to 100 baby boys, 75 of the tests will indicate boys. • 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).

16. 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 562 babies, 432 are girls. • Thus P(girl | test says girl ) = 432/562 = 0.769

17. Learning exercise 2: Predicting Sex of Babies With the information given, we can fill in the following table in sequence from (1) to (9), with the initial assumption of 1000 babies in total. For the question at hand, i.e., P(girl | test says girl ), we only need to fill in the cells from (1) to (6). P(girl | test says girl ) = 432/562 = 0.769

18. 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) 1000*P(boys)*P(test says girls | boys) 1000*[P(girls)*P(test says girls|girls) + 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)]

19. Learning exercise 3: Putting in Extra Trunk Lines • 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 ).

20. Learning exercise 3: Putting in Extra Trunk Lines • 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.

21. Learning exercise 3: Putting in Extra Trunk Lines • P( individual line works ) = 50/100 • Find the number of lines we must install so that P(at least one of the lines works ) is larger than 0.9. Suppose we install k independent lines P( at least one of the k lines works ) = 1 – P(all the k lines fail) = 1 – P(Line 1 fails)*P(line 2 fails)*… *P(line k fails) = 1 – 0.5k We want to find the smallest integer k such that 1 – 0.5k >0.9 or 0.1> 0.5k Log 0.1< k log(0.5) or k>log 0.1 / log(0.5) = 3.32 Hence k=4.

22. 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)

23. EXAMPLE 3 • New England Commuter Airways recently supplied the following information on their commuter flights from Boston to New York:

24. EXAMPLE 3 • If A is the event that a flight arrives early, thenP(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.

25. 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).

26. The Complement Rule continued • A Venn diagram illustrating the complement rule would appear as: A ~A

27. 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

28. EXAMPLE 4 continued • P(A or B) = 1 - P(C or D) = 1 - [.8 +.025] =.175 D .025 C .8 ~(C or D) = (A or B) .175

29. 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) B A and B A

30. 75 220 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: TV 175 Both 100 Stereo 320

31. EXAMPLE 5 continued • In a sample of 500 students, 320 said they had a stereo, 175 said they had a TV, and 100 said they had both. • 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(student has a stereo) = 320/500 = .64. P(student has a TV) = 175/500 = .35. P(student has both a stereo and TV) = 100/500 = .20. P(student as only a stereo) = 220/500 = .44. P(student has only a TV) = 75/500 = .15. P(student as only a stereo) = P(student has a stereo) - P(student has both a stereo and TV) P(student has only a TV) = P(student has a TV) - P(student has both a stereo and TV)

32. EXAMPLE 5 continued • 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. • 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(either S or T) = P(only S) + P (only T) = [P(S) - P(S and T)] +[P(T) - P(S and T)] = .64 - .20+.35 - .20 = .59.

33. 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. • In an experiment of throwing only one die, let x denote the outcome, P(x=5, and x=6) =0 because the two outcomes are mutually exclusive.

34. Special Rule of Multiplication • The special rule of multiplication requires that two events A and B are independent. • Two events A and B are independentif 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)

35. 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.

36. EXAMPLE 6 continued • 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. • 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 2: P(at least one) = P(GE) + P(IBM) – P(both) = 0.5 + 0.7 – 0.35 = .85. Approach 1: P(at least one) = P(GE only) + P(IBM only) + P(both) = (.5)(.3) + (.5)(.7) +(.7)(.5) = .85. Approach 3: P(at least one) = 1- P(both GE and IBM do not increase) = 1 – 0.15 = .85.

37. 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.

38. 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) • Examples: • 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)

39. Independence -- an illustration • Consider whether the decision of a young man going to party depends on whether his girlfriend goes to the same party. • Assume the probability of the young man going to party is 0.7 (i.e., he goes to 70 out of 100 parties on average). • If he tends to go to whichever party his girlfriend goes, his party behavior depends on his girlfriend’s. That is, the probability of going to a party conditional on his girlfriend’s presence is larger than 0.7 (extreme case being 1.0). • If he tends to avoid going to whichever party his girlfriend goes, his party behavior also depends on his girlfriend’s. That is, the probability of going to a party conditional on his girlfriend’s presence is less than 0.7 (extreme case being 0.0). • If in making the party decision, he never considers whether his girlfriend is going to a party, his party behavior does not depends on his girlfriend’s. That means, the probability of going to a party conditional on his girlfriend’s presence is 0.7.

40. Independence -- an illustration • Define events: • A: a young man goes to a party • B: his girlfriend goes to the same party. • Assume P(A) =0.7 • His party behavior does not depend on his girlfriend’s only if P(A|B) =P(A) = 0.7. And, event A is said to be independent of event B. • P(the young man and his girlfriends shows up in a party) = P(A & B) = P(B)*P(A|B). • If he always goes to whichever party his girlfriend goes, P(A|B) = 1. Hence, P(A & B) = P(B)*P(A|B) = P(B). • If he always avoid to whichever party his girlfriend goes, P(A|B) = 0. Hence, P(A & B) = P(B)*P(A|B) = 0. • If in making the party decision, he never considers whether his girlfriend is going to a party, P(A|B) = 0.7. Hence, P(A & B) = P(B)*P(A|B) = P(B)*P(A) = 0.7.

41. Independence • Event A is independent of B • P(A|B) = P (A) • Event B is independent of A • P(B|A) = P (B) • P(B|A) = P (B) implies P(A&B) = P(B|A) * P(A) = P(B) * P(A) • P(A|B) = P (A) impliesP(A&B) = P(A|B) * P(B) = P(A) * P(B) • Thus, if P(A&B) = P(B) * P(A), we must have either P(A|B) = P (A) or P(B|A) = P (B).

42. EXAMPLE 7 The Dean of the School of Business at Owens University collected the following information about undergraduate students in her college:

43. 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 2: P(A|F) = 110/400 = .275 Approach 1: P(A|F) = P(A and F)/P(F) = [110/1000]/[400/1000] = .275

44. 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.

45. EXAMPLE 8 continued 6/11 R2 R1 7/12 B2 5/11 R2 7/11 B1 5/12 4/11 B2

46. EXAMPLE 8 continued 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 6/11 6/11 R2 P(A) = 7/12 R1 7/12 7/12 P(A and B) = P(A)*P(B|A) = 6/11 * 7/12 B2 5/11 R2 7/11 B1 5/12 4/11 B2

47. Bayes’ Theorem • Bayes’ Theorem is a method for revising a probability given additional information. • If A1 and A2 are mutually exclusive and exhaustive, and A1 and A2 together covers event B, we have : Bayes Theorem is an essential tool to understand options and real options in finance.

48. Bayes’ Theorem • A1 and A2 are mutually exclusive means P (A1 or A2 ) = P(A1) + P(A2 ) P (A1 and A2 ) = 0 • A1 and A2 are exhaustive means P (A1 or A2 ) = P(A1) + P(A2 ) = 1 • A1 and A2 together covers event B means P(B) = P(A1 & B) + P(A2 & B) (see the Venn diagram illustration)

49. Venn Diagram illustration of Bayes’ theorem A1 A2 B A1 & B A2 & B

50. Bayes’ Theorem • Bayes’ Theorem can be derived based on simple manipulation of the general multiplication rule. P(A1|B) = P(A1 & B) /P(B) = [P(A1) P(B|A1)] / P(B) = [P(A1) P(B|A1)] / [P(A1 & B) + P(A2 & B)] = [P(A1) P(B|A1) ]/ [P(A1) P(B|A1) + P(A2) P(B|A2)]