1. Probability Week 4 GT00303

2. Probability A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will occur. 4-2

3. Ways of Assigning Probability There are three ways to assign a probability to an outcome: 1.CLASSICAL APPROACH Based on the assumption that the outcomes of an experiment are equally likely. 2.RELATIVE FREQUENCY APPROACH The probability of an event happening is the fraction of the time similar events happened in the past. 3.SUBJECTIVE APPROACH The likelihood (probability) of a particular event happening that is assigned by an individual based on subjective judgment. 4-3

4. (1) Classical Approach Consider an experiment of rolling a six-sided die. What is the probability of the event “an even number of spots appear face up”? The possible outcomes are: Probability (Even Number) = 3/6 = 0.5 4-4

5. P(10) = 3/36P(6) = 5/36 P(2) = 1/36 123456 1234567 2345678 3456789 45678910 56789 11 6789101112 Consider an experiment of rolling 2 six-sided dice and observing the total. What is the probability of the event (a) “the total is 2”; (b) “the total is 6”; (c) “the total is 10”? The possible outcomes are: 4-5

6. Bits & Bytes Computer Shop tracks the number of desktop computer systems it sells over a month (30 days): Desktops Sold# of Days 01 12 210 312 45 (2) Relative Frequency Approach From this past record/ historical data, we can construct the probabilities of an event (i.e. the number of desktop sold on a given day). 4-6

7. “There is a 40% chance Bits & Bytes will sell 3 desktops on any given day” Desktops Sold# of DaysDesktops Sold 011/30 =.03 122/30 =.07 21010/30 =.33 31212/30 =.40 455/30 =.17 ∑ = 1.00 4-7

8. (3) Subjective Approach If there is little or no past experience or information on which to base a probability, it may be arrived at subjectively. For example: Estimating the likelihood the New England Patriots will play in the Super Bowl next year. Estimating the likelihood you will be married before the age of 30. Estimating the likelihood the U.S. budget deficit will be reduced by half in the next 10 years. 4-8

9. Complement of Event Intersection of Events Union of Events Mutually Exclusive Events A A’A’ AB AB AB Basic Relationships of Probability 4-9

10. The complement of event A is defined to be the event consisting of all sample points that are “not in A”. Complement of A is denoted by A ’ The Venn diagram below illustrates the concept of a complement. P(A) + P(A ’ ) = 1 OR P(A) = 1 - P(A ’ ) (1) Complement of An Event A A’A’ 4-10

11. Illustration: An automatic Shaw machine fills plastic bags with a mixture of beans, broccoli, and other vegetables. Most of the bags contain the correct weight, but because of the variation in the size of the beans and other vegetables, a package might be underweight or overweight. Use the complement rule to show the probability of a satisfactory bag is 0.90. P(B) = 1 – P(B ’ ) = 1 – P(A or C) = 1 – [P(A) + P(C)] = 1 – [0.025 + 0.075] = 1 - 0.10 = 0.90 4-11

12. The intersection of events A and B is the set of all sample points that are in both A and B. The joint probability of A and B is the probability of the intersection of A and B, i.e. P(A and B) OR (2) Intersection of Two Events AB 4-12

13. A = tosses where first toss is 1 ={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)} B = tosses where the second toss is 5 = {(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)} The intersection of events A and B contains those points where the first toss is 1 and the second toss is 5. The intersection is {(1,5)}. Illustration: 4-13

14. The union of two events A and B, is the event containing all sample points that are in A or B or both: Union of A and B is denoted: A or B. P (A or B) OR (3) Union of Two Events AB 4-14

15. A = tosses where first toss is 1 ={(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)} B = tosses where the second toss is 5 = {(1,5), (2,5), (3,5), (4,5), (5,5), (6,5)} The union of events A and B contains those points where the first toss is 1 or the second toss is 5 or both. The union is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,5), (3,5), (4,5), (5,5), (6,5)}. Illustration: 4-15

16. Events are mutually exclusive if the occurrence of any one event means that none of the others can occur at the same time. Hence, their joint probability is 0. AB (4) Mutually Exclusive Events 4-16

17. Why are some mutual fund managers more successful than others? One possible factor is where the manager earned his or her MBA. The following table compares mutual fund performance against the ranking of the school where the fund manager earned their MBA: Mutual fund outperforms the market Mutual fund doesn’t outperform the market Top 20 MBA program 0.110.29 Not top 20 MBA program 0.060.54 E.g. This is the probability that a mutual fund outperforms AND the manager was in a top- 20 MBA program; it’s a joint probability. Illustration: 4-17

18. Alternatively, we could introduce shorthand notation to represent the events: A = Fund manager graduated from a top-20 MBA program A ’ = Fund manager did not graduate from a top-20 MBA program B = Fund outperforms the market B ’ = Fund does not outperform the market BB’ A 0.110.29 A’ 0.060.54 B and B ’ are mutually exclusive events. 4-18

19. BB’ A 0.110.29 A’ 0.060.54 4-19 What’s the probability that a fund outperforms the market or the manager graduated from a top-20 MBA program? ✔✔ ✔ OR Return

20. Types of Probability JOINT PROBABILITY A probability that measures the likelihood two or more events will happen concurrently. 4-20 MARGINAL PROBABILITY A probability that measures the likelihood that a specific event will happen. CONDITIONAL PROBABILITY The probability of a particular event occurring, given that another event has occurred.

21. 4-21 Illustration: BB’ Marginal Probability A 0.110.29P(A)=0.40 A’ 0.060.54P(A’)=0.60 Marginal Probability P(B)=0.17P(B’)=0.831.00 What is the probability a fund manager isn’t from a top school? What is the probability a fund outperforms the market? Conditional probability is not directly observable from contingency table.

22. 4-22 BB’ Marginal Probability A 0.110.29P(A)=0.40 A’ 0.060.54P(A’)=0.60 Marginal Probability P(B)=0.17P(B’)=0.831.00 What’s the probability that a fund will outperform the market given that the manager graduated from a top-20 MBA program?

23. Thus, there is a 27.5% chance that that a fund will outperform the market given that the manager graduated from a top-20 MBA program. BB’ Marginal Probability A 0.110.29P(A)=0.40 A’ 0.060.54P(A’)=0.60 Marginal Probability P(B)=0.17P(B’)=0.831.00 4-23

24. One of the objectives of calculating conditional probability is to determine whether two events are related. In particular, we would like to know whether they are independent, that is, if the probability of one event is not affected by the occurrence of the other event. Two events A and B are said to be independent if: P(A|B) = P(A) or P(B|A) = P(B) Independence of Events 4-24

25. 4-25 Illustration: BB’ Marginal Probability A 0.110.29P(A)=0.40 A’ 0.060.54P(A’)=0.60 Marginal Probability P(B)=0.17P(B’)=0.831.00 Are the two events A and B independent? Since P(A|B) ≠ P(A), A and B are not independent events.

26. Since P(B|A) ≠ P(B), A and B are not independent events. Stated another way, A and B are dependent. That is, the probability of one event (B) depends on the occurrence of the other event (A). 4-26

27. Probability Rules 4-27 (1) Complement Rule

28. 4-28 (2) Multiplication Rule OR This applies when events A and B are dependent. What if they are independent?

29. 4-29 If events A and B are independent:

30. A graduate statistics course has 7 male and 3 female students. The professor wants to select two students at random to help her conduct a research project. What is the probability that the two students chosen are female? Illustration: Let A = The first student is female B = The second student is female P(A) = 3/10 = 0.3 P(B|A) = 2/9 = 0.22 4-30 Return

31. 4-31 What is the probability that the two students chosen are female? Since events A and B are dependent: There is a 6.7% chance that the professor will choose 2 female students from her grad class of 10.

32. The professor who teaches the course is suffering from the flu and will be unavailable for two classes. The professor’s replacement will teach the next two classes. His style is to select one student at random and pick on him or her to answer questions during that class. What is the probability that the two students chosen are female? Illustration: Let A = The first student is female B = The second student is female P(A) = 3/10 = 0.3 P(B) = 3/10 = 0.3 4-32

33. 4-33 What is the probability that the two students chosen are female? Since events A and B are independent (the student selected in the 1 st class can still be chosen in the 2 nd class):

34. Illustration: Let A = The first shirt is white B = The second shirt is white P(A) = 9/12 = 0.75 P(B|A) = 8/11 = 0.727 A golfer has 12 golf shirts in his closet. Suppose 9 of these shirts are white and the others blue. He gets dressed in the dark, so he just grabs a shirt and puts it on. He plays golf two days in a row and does not do laundry. What is the likelihood both shirts selected are white? 4-34

35. 4-35 What is the probability that both shirts selected are white? Since events A and B are dependent: There is a 54.5% chance that the golfer will choose 2 white shirts from his closet.

36. ABAB = +– If A and B are mutually exclusive, then this term goes to zero (3) Addition Rule Confirm with Slide 19! (Click here)here 4-36

37. 4-37 If events A and B are mutually exclusive: AB

38. In a large city, two newspapers are published, the Sun and the Post. The circulation departments report that 22% of the city’s households have a subscription to the Sun and 35% subscribe to the Post. A survey reveals that 6% of all households subscribe to both newspapers. What is the probability of selecting a household at random that subscribes to the Sun or the Post or both? Illustration: Let A = Subscription to the Sun B = Subscription to the Post P(A) = 0.22 P(B) = 0.35 4-38

39. 4-39 What is the probability of selecting a household at random that subscribes to the Sun or the Post or both? There is a 51% probability that a randomly selected household subscribes to one or the other or both papers

40. Illustration: An automatic Shaw machine fills plastic bags with a mixture of beans, broccoli, and other vegetables. Most of the bags contain the correct weight, but because of the variation in the size of the beans and other vegetables, a package might be underweight or overweight. What is the probability that a particular package will be either underweight or overweight? 4-40

41. 4-41 BB’ Marginal Probability A 0.110.29P(A)=0.40 A’ 0.060.54P(A’)=0.60 Marginal Probability P(B)=0.17P(B’)=0.831.00 Contingency Tables Contingency Tables should be used when marginal and joint probabilities are given. Conditional probability is not needed!

42. This is P(F|F), the probability of selecting a second female student, given that a female was already chosen first First selectionSecond selection P(F) = 3/10 P( M) = 7/10 P(F|M) = 3/9 P(F|F) = 2/9 P( M|M) = 6/9 P( M|F) = 7/9 This is P(F), the probability of selecting the first female student Probability Trees When marginal and conditional probabilities are given, it is best to use Probability Trees. 4-42 See Slide 30!Slide 30

43. At the ends of the “branches”, we calculate joint probabilities as the product of the individual probabilities on the preceding branches. First selectionSecond selection P(F) = 3/10 P( M) = 7/10 P(F|M) = 3/9 P(F|F) = 2/9 P( M|M) = 6/9 P( M|F) = 7/9 P(F F)=(3/10)(2/9) P(F M)=(3/10)(7/9) P(M F)=(7/10)(3/9) P(M M)=(7/10)(6/9) Joint probabilities 4-43

44. 3/9 + 6/9 = 9/9 = 1 2/9 + 7/9 = 9/9 = 1 3/10 + 7/10 = 10/10 = 1 The probabilities associated with any set of branches from one “node” must add up to 1.00… First selectionSecond selection P(F) = 3/10 P( M) = 7/10 P(F|M) = 3/9 P(F|F) = 2/9 P( M|M) = 6/9 P( M|F) = 7/9 Handy way to check your work ! 4-44

45. Suppose we have our grad class of 10 students again, but make the student sampling independent, that is “with replacement” – a student could be picked first and picked again in the second round. F MFMF M FMFM P(F) = 3/10 P( M) = 7/10 P(F|M) = 3/10 P(F|F) = 3/10 P( M|M) =7/10 P( M|F) = 7/10 P(F F)=(3/10)(3/10) P(F M)=(3/10)(7/10) P(M F)=(7/10)(3/10) P(M M)=(7/10)(7/10) Illustration: 4-45

46. Law school grads must pass a bar exam. Suppose pass rate for first-time test takers is 72%. They can re-write if they fail and 88% pass their second attempt. What is the probability that a randomly grad passes the bar? P(Pass) = 0.72 P(Fail and Pass) =( 0.28)(0.88) =0.2464 P(Fail and Fail) = (0.28)(0.12) = 0.0336 First exam P(Pass) = 0.72 P( Fail) = 0.28 Second exam P(Pass|Fail) = 0.88 P( Fail|Fail) = 0.12 Illustration: 4-46

47. What is the probability that a randomly grad passes the bar? P(Pass) = P(Pass 1 st ) + P(Fail 1 st and Pass 2 nd ) = 0.7200 + 0.2464 = 0.9664 P(Pass) = 0.72 P(Fail and Pass) = (0.28)(0.88)=0.2464 P(Fail and Fail) = (0.28)(0.12) = 0.0336 First exam P(Pass) = 0.72 P( Fail) = 0.28 Second exam P(Pass|Fail) = 0.88 P( Fail|Fail) =.12 There is a 96.64% chance they will pass the bar 4-47

48. Illustration: 4-48