Uncertainty and probability

1. Uncertainty and probability Using probabilities Using decision trees Probability revision

2. Today’s agenda • Important terms • Simple review (objective, subjective, marginal, joint, and conditional probabilities) • Examples: outcomes, expected values, risk attitudes • Examples: action choices, decision trees

3. Vocabulary • A probability is a number between zero and one representing the likelihood of the occurrence of some event. • Probability • objective vs. subjective • marginal vs. joint • joint vs. conditional • prior vs. posterior • likelihood vs. posterior

4. Vocabulary continued • Outcomes or payoffs (mutually exclusive) • Action choices • States of nature • Decision tree • Expected value • Risk

5. Probability Imagine an urn containing 1500 red, pink, yellow, blue and white marbles. Take one ball from the urn. What is: P(black) = 0 ~ = NOT P(~black) = 1 Probabilities are all greater than or equal to zero and lessthan or equal to one.

6. Same urn: Suppose the number of balls is as follows: Red 400 Pink 100 Yellow 400 Blue 500 White 100 Total 1500 What is: P(Red) = 400/1500 = .267 P(Pink) = 100/1500 = .067 P(Yellow) = 400/1500 = .267 P(Blue) = 500/1500 = .333 P(White) = 100/1500 = .067 Total = 1

7. Joint probabilities and independence Define A as the event “draw a red or a pink marble.” We know 500 marbles are either red or pink. What are: P(A) = = .33 (1 - P(A)) = .67 P(~A) =

8. Joint probabilities and independence (we’re getting there) Define B as the event, “draw a pink or white marble.” We know 200 marbles are pink or white. What are: P(B) = .133 P(~B) = .867

9. Joint probabilities and independence Define A as the event “draw a red or a pink marble.” Define B as the event “draw a pink or white marble.” What is: P(A, B) = P(A  B) This is the joint probability of A and B. What color is the marble? Pink P(A, B) = P(pink) = = .0667

10. Joint probabilities and independence Are A and B independent? Note that P(A, B)  P(A) * P(B) = .33 * .13 = .0429 Are A and B mutually exclusive? What is the probability of A or B? P(A or B) = P(A  B) = P(A) + P(B) - P(A  B) = .40

11. Joint probabilities and independence Suppose we draw one marble from the urn andreplace it. Then, we draw a second marble. What is: P(Red, Red) = = .071 P(Red, Blue) = .088 Are (Red, Red) independent? Are (Red, Blue) independent?

12. Joint and marginal probabilities What are: = .133 P(B) = P(~B) = .867 1500 P(A, B) = P(A or B) =

13. Conditional probabilities The probability that a particularevent will occur, given we alreadyknow that another event hasoccurred. What is P(A | B) = We have information to bringto bear on the base rate probability of the event P(~A | ~B) = 1500 P(A | ~B) = P(~B | A) =

14. Definition of independence P(B | A) = Events A and B are independent if P(B | A) = P(B) Here P(B | A) = P(B) P(B) = 1500

15. Marginal and joint probability table: The joint probabilities are in the box. The marginalsare outside. How do you compute conditionals from this? A ~A P(B | A) = B ~B

16. Joint probability tables A ~A What are: B P(A, B) = .0667 ~B P(~A, ~B) = .6 P(~A) = .667 P(~B) = .867 = .8 P(~B | A)

17. Outcomes or payoffs Example: Win $1,000 if you draw a pink marble, win $0 otherwise. Outcomes: $1,000 or $0 Events: A pink marble or a marble of another color Probabilities: P(Pink) = 1/15 P(~Pink) = 14/15 The expected value of this gamble: E(gamble) = (1/15)*($1,000) + (14/15)*($0) = $66.67

18. Example We expect to sell 10,000 computers if the market isgood and to sell 1,000 computers if the market isbad. The marketing department’s best estimate ofthe likelihood of a good market is .5. Outcomes: 10,000 computers sold or 1,000 computers sold. Probabilities: P(good market) = .5 P(bad market) = .5 Are these objective or subjective probabilities? What are expected computer sales? 5,500

19. More expected values With discrete outcomes, an expected value is theprobability-weighted sum of the outcomes for thedecision of interest. Here are some two-outcome lotteries. Compute theexpected values. L1: Win $1,000 with probability .5 or lose $500 with probability .5. E(L1) = $250 L2: Win $2,000 with probability .5 or lose $1,000 with probability .5. E(L2) = $500

20. More lotteries Here are some two-outcome lotteries. Compute theexpected values. L3: Win $1,500 with probability 1/3 or lose $750 with probability 2/3. E(L3) = $0 L4: Win $750 with probability 2/3 or lose $750 with probability 1/3. E(L4) = $250 L5: Win $300 with probability .5 or lose $200 with probability .5. E(L5) = $50 L6: Win $10,000 with probability 9/10 or lose $85,000 with probability 1/10. E(L6) = $500

21. L1: Win $1,000 with probability .5 or lose $500 with probability .5. E(L1) = $250 L2: Win $2,000 with probability .5 or lose $1,000 with probability .5. E(L2) = $500 L3: Win $1,500 with probability 1/3 or lose $750 with probability 2/3. E(L3) = $0 L4: Win $750 with probability 2/3 or lose $750 with probability 1/3. E(L4) = $250 L5: Win $300 with probability .5 or lose $200 with probability .5. E(L5) = $50 L6: Win $10,000 with probability 9/10 or lose $85,000 with probability 1/10. E(L6) = $500

22. Action choices If I build a large hotel (cost = $5,000,000) and tourismis high (P(high) = 2/3), I will make $15,000,000 in revenue, but if it is low, I will make $2,000,000. If I build a small hotel (cost = $2,000,000) and tourismis high, I will make $5,000,000, but if tourism is low, I will make $2,000,000. I can also choose to do nothing.

23. Action choices If I build a large hotel (cost = $5,000,000) and tourism is high (P(high) = 2/3), I will make $15,000,000 in revenue, but if it is low, I will make $2,000,000. If I build a small hotel (cost = $2,000,000) and tourism is high, I will make $5,000,000, but if tourism is low, I will make $2,000,000. I can also choose to do nothing. Action choices: Do nothing, build large, build small Outcomes: $15,000,000; $2,000,000; $5,000,000, $0 States of nature: high tourism, low tourism Probabilities: P(high) = 2/3; P(low) = 1/3

24. Decision trees Suppose I need to decide whether to invest $10,000 in the market or leave it in the bank to earn interest. If I invest, there is a 50% chance that the market will increase 20% over the coming year and a 50% chancethat the market will be stagnant (no change). If I leave the money in the bank, there is an 80% chance that interest rates will increase to 10% and a 20% chance that interest rates will remain at 5%. What should I do? Use a decision tree.

25. A decision by an individual is required Nature makes these decisions $2,000 or EV = $1,000 Nature decides .5 (Good) (Stagnant) .5 I decide StockMarket $0 or EV = $0 $1,000 $1,000 orEV = $800 .8 Bank (Increase) (Same) .2 $500 or EV = $100 $900

26. Homework assignment Problems 5-15 and 16 Urn 1 Urn 2 Consider two urns: Red balls 7 4 Black balls 3 6 P(R1) = Probability of red on first draw P(R2) = Probability of red on second draw P(B1) = Probability of black on first draw P(B2) = Probability of black on second draw a(1) Take one ball from urn 1, replace it, and take a second ball. What is the probability of two reds beingdrawn? P(R1, R2) = .7 x .7 = .49

27. Homework a(2) What is the probability of a red on the seconddraw if a red is drawn on the first draw? P(R2 | R1) = = a(3) What is the probability of a red on the second draw if a black is drawn on the first draw? P(R2 | B1) =

28. Homework b(1) Take a ball from urn 1; replace it. Take a ball from urn 2 if the first ball was black; otherwise, draw a ball from urn 1. What is the probability of two reds being drawn? P(R1, R2) = .7 x .7 = .49 b(2) What is the probability of a red on the second draw if a red is drawn on the first draw? P(R2 | R1) =

29. Homework b(3) What is the probability of a red on the second draw if a black is drawn on the first draw? P(R2 | B1) = What is the unconditional probability of red on thesecond draw? P(R2) = P(B1, R2) + P(R1, R2) = (.3 x .4) + (.7 x .7) = .61

30. Homework 5-16. Draw a tree diagram for Problem 5-15a P(R1, R2) = .49 Red Draw 2 .7 Red .3 Draw 1 .7 Black P(R1, B2) = .21 P(B1, R2) = .21 Red .3 .7 Black .3 P(B1, B2) = .09 Black

31. Homework 5-17. Draw a tree diagram for Problem 5-15b P(R1, R2) = .49 Red Draw 2 .7 Red .3 Draw 1 .7 Black P(R1, B2) = .21 P(B1, R2) = .12 Red .3 .4 Black .6 P(B1, B2) = .18 Black

32. Homework 5-29. This is the survey problem involving home- ownership and income levels. The results can be summarized by the table below

33. Survey A. Suppose a reader of this magazine is selected atrandom and you are told that the person is a home- owner. What is the probability that the person has income in excess of $25,000? P(>$25,000 | homeowner) =

34. Survey b. Are home ownership and income (measured only as above or below $25,000) independent factors for this group? They are NOT independent. If yes, then P(>$25 | home) = P(>$25) But, P(>$25) = .7 and P(>$25 | home) = .75

35. Homework 5-38. The president of a large electric utility has to decide whether to purchase one large generator (Big Jim) or four smaller generators (Little Arnies) to attain a given amount of electric generating capacity. On any given summer day, the probability of a generator being in service is 0.95 (the generators are equally reliable). Equivalently, there is a 0.05 probability of a failure.

36. Homework 5-38. a. What is the probability of Big Jim’s being out of service on a given day? Let P(out) = the probability of any generator being out of service = .05 If P(BJout) = the probability of Big Jim’s being out of service. Then P(out) = P(BJout) = .05

37. Homework 5-38. b. What is the probability of either zero or one of the four Arnies being out? (At least three will be running.) Since the probability of a failure (f) for one Arnie is .05, We want P(f  1|n=4, p=.05) P(f = 0 | n=4, p=.05) = P(f = 1 | n=4, p=.05) = P(f  1|n=4, p=.05) = .8145 + .1715 = .9860

38. Homework 5-38. c. If five Little Arnies are purchased, what is the probability of at least four operating? P(f  1 | n=5, p=.05) =

39. Homework 5-38. d. If six Little Arnies are purchased, what is the probability of at least four operating? P(f  2 | n=6, p=.05) =

40. Homework 5-40. Newspaper articles frequently cite the fact that in any one year, a small percentage (say, 10%) of all drivers are responsible for all automobile accidents. The conclusion is often reached that if only we could single out these accident-prone drivers and either retrain them or remove them from the roads, we could drastically reduce auto accidents. You are told that of 100,000 drivers who were involved in one or more accidents in one year, 11,000 of them were involved in one or more accidents in the next year. A. Given the above information, complete the entries in the joint probability table in Table 5-21.

41. Accidents: Joint probability table A1 = accident in year 1, A2 = accident in year 2 11,000/100,000 = .11 Given: P(A2 | A1) = P(A2 | A1)= P(A1, A2) = .089 .011 Therefore, .11 P(A1, A2) = P(A2 | A1)xP(A1) = .089 .811 .11 x .10 = .011

42. Accidents B. Do you think searching for accident-prone drivers is an effective way to reduce auto accidents? Why? If the information in the problem is representative, then searching for accident-prone drivers will not be very helpful, since having had an accident in Year 1 has only a minor effect on the probability of an accident in Year 2.

43. Uncertainty continued . . . Probability revisions Continue decision trees

44. Today’s agenda • Finish the homework problems • Work through a decision tree example that • Uses no information • Uses perfect information • Uses imperfect information • Briefly discuss Freemark Abbey Winery • Group problem solving

45. Homework 5-42. A safety commissioner for a certain city performed a study of the pedestrian fatalities at intersections. He noted that only 6 of the 19 fatalities were pedestrians who were crossing the intersection against the light (i.e., in disregard of the proper signal), whereas the remaining 13 were crossing with the light. He was puzzled because the figures seemed to show that it was roughly twice as safe for a pedestrian to cross against the light as with it. Can you explain this apparent contradiction to the commissioner?

46. 5-42. Homework The commissioner is looking at the wrong conditional frequencies (probabilities). P(?) = 6/19 It’s a conditional probability P(crossing against the light | killed at intersection) = 6/19 It is not the probability of being killed if you cross against the light. Further, 13/19 is not the probability of being killed if you cross with the light.

47. 5-42. Homework The relevant probabilities are: P(killed | crossed with light) and P(killed | crossed against light) The deaths must be considered relative to the number of pedestrians who cross with and against the light. As an extreme possibility, it may be that the only six persons who crossed against the light were killed, a fatality rate of 100%; whereas, 1 million crossed with the light, a fatality rate of .000013. It isn’t likely that this is the case, but the commissioner’s data do not rule it out.

48. 5-43. Homework Probability revision Suppose a new test is available to test for drug addiction. The test is 95 percent accurate “each way”; that is, if the person is an addict, there is a 95 percent chance the test will indicate “yes”; if the person is not an addict, then 95 percent of the time the test will indicate “no.” Suppose it is known that the incidence of drug addiction in urban populations is about 1 out of 1,000. Given a positive (yes) test result, what are the chances that the person being tested is addicted?

49. Homework: Probability revision 5-43. We were given P(“yes” | addicted) = .95 and P(“no” | not addicted) = .95 and the prior P(addicted) = .001 We want P(addicted | “yes”) and P(not addicted | “no”) This is a different conditional probability called a“revised” or “posterior” probability.

50. Test for drug addiction We know: P(addicted | yes) = and P(yes | addicted) = = .95 P(addicted) = .001 We can solve for the joint.