1 / 48

Probability

Probability. Week 4 GT00303. Probability. A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will occur. 4- 2. Ways of Assigning Probability. There are three ways to assign a probability to an outcome: 1. CLASSICAL APPROACH

snowy
Download Presentation

Probability

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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. 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”? P(6) = 5/36 P(10) = 3/36 P(2) = 1/36 The possible outcomes are: 4-5

  6. (2) Relative Frequency Approach Bits & Bytes Computer Shop tracks the number of desktop computer systems it sells over a month (30 days): 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” 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. A A’ A B A A B B Basic Relationships of Probability Complement of Event Union of Events Intersection of Events Mutually Exclusive Events 4-9

  10. A A’ (1) Complement of An Event 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’ ) 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. A B (2) 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 joint probabilityof A and B is the probability of the intersection of A and B, i.e. P(A and B) OR 4-12

  13. Illustration: 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)}. 4-13

  14. A B (3) Union of Two Events 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 4-14

  15. Illustration: 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)}. 4-15

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

  17. Illustration: 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: 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. 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 B and B’ are mutually exclusive events. 4-18

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

  20. Types of Probability 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. JOINT PROBABILITY A probability that measures the likelihood two or more events will happen concurrently. 4-20

  21. Illustration: What is the probability a fund manager isn’t from a top school? Whatis the probability a fund outperforms the market? Conditional probability is not directly observable from contingency table. 4-21

  22. What’s the probability that a fund will outperform the market given that the manager graduated from a top-20 MBA program? 4-22

  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. 4-23

  24. Independence of Events 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 independentif: P(A|B) = P(A) or P(B|A) = P(B) 4-24

  25. Illustration: Are the two events A and B independent? Since P(A|B) ≠ P(A), A and B are not independent events. 4-25

  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 onthe occurrence of the other event (A). 4-26

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

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

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

  30. Illustration: 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? 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 Return 4-30

  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. 4-31

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

  34. Illustration: 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? 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 4-34

  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. 4-35

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

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

  38. Illustration: 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? Let A = Subscription to the Sun B = Subscription to the Post P(A) = 0.22 P(B) = 0.35 4-38

  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 4-39

  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. Contingency Tables Contingency Tables should be used when marginal and joint probabilities are given. Conditional probability is not needed! 4-41

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

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

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

  45. P(F F)=(3/10)(3/10) P(F|F) = 3/10 P(F) = 3/10 P( M|F) = 7/10 P(F M)=(3/10)(7/10) P(M F)=(7/10)(3/10) P(F|M) = 3/10 P( M) = 7/10 P( M|M) =7/10 P(M M)=(7/10)(7/10) Illustration: 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. FF FM MF MM 4-45

  46. P(Pass) = 0.72 P(Fail and Pass) =( 0.28)(0.88) =0.2464 P(Fail and Fail) = (0.28)(0.12) = 0.0336 Illustration: 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? First exam P(Pass) = 0.72 Second exam P(Pass|Fail) = 0.88 P( Fail) = 0.28 P( Fail|Fail) = 0.12 4-46

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

  48. Illustration: 4-48

More Related