Probability

1. Probability

2. Lotto I am offered two lotto cards: • Card 1: has numbers • Card 2: has numbers Which card should I take so that I have the greatest chance of winning lotto?

3. Roulette In the casino I wait at the roulette wheel until I see a run of at least five reds in a row. I then bet heavily on a black. I am now more likely to win.

4. Coin Tossing I am about to toss a coin 20 times. What do you expect to happen? Suppose that the first four tosses have been heads and there are no tails so far. What do you expect will have happened by the end of the 20 tosses ?

5. Coin Tossing • Option A • Still expect to get 10 heads and 10 tails. Since there are already 4 heads, now expect to get 6 heads from the remaining 16 tosses. In the next few tosses, expect to get more tails than heads. • Option B • There are 16 tosses to go. For these 16 tosses I expect 8 heads and 8 tails. Now expect to get 12 heads and 8 tails for the 20 throws.

6. TV Game Show • In a TV game show, a car will be given away. • 3 keys are put on the table, with only one of them being the right key. The 3 finalists are given a chance to choose one key and the one who chooses the right key will take the car. • If you were one of the finalists, would you prefer to be the 1st, 2nd or last to choose a key?

7. Let’s Make a Deal Game Show • You pick one of three doors • two have booby prizes behind them • one has lots of money behind it • The game show host then shows you a booby prize behind one of the other doors • Then he asks you “Do you want to change doors?” • Should you??! (Does it matter??!) • See the following website: • http://www.stat.sc.edu/~west/javahtml/LetsMakeaDeal.html

8. Game Show Dilemma Suppose you choose door A. In which case Monty Hall will show you either door B or C depending upon what is behind each. No Switch Strategy ~ here is what happens P(WIN) = 1/3

9. Game Show Dilemma Suppose you choose door A, but ultimately switch. Again Monty Hall will show you either door B or C depending upon what is behind each. Switch Strategy ~ here is what happens Monty will show either B or C. You switch to the one not shown and lose. Monty will show door C, you switch to B and win. Monty will show door B, you switch to C and win. P(WIN) = 2/3 !!!!

10. Matching Birthdays • In a room with 23 people what is the probability that at least two of them will have the same birthday? • Answer: .5073 or 50.73% chance!!!!! • How about 30? • .7063 or 71% chance! • How about 40? • .8912 or 89% chance! • How about 50? • .9704 or 97% chance!

11. Probability What is Chapter 6 trying to do? • Introduce us to basic ideas about probabilities: • what they are and where they come from • simple probability models (genetics) • conditional probabilities • independent events • Baye’s Rule Teach us how to calculate probabilities: • tables of counts and using properties of probabilities such as independence.

12. Probability I toss a fair coin (where fair means ‘equally likely outcomes’) What are the possible outcomes? Head and tail ~ This is called a “dichotomous experiment” because it has only two possible outcomes. S = {H,T}. What is the probability it will turn up heads? 1/2 I choose a patient at random and observe whether they are successfully treated. What are the possible outcomes? “Success” and “Failure” What is the probability of successful treatment? ????? What factors influence this probability? ?????

13. What are Probabilities? • A probability is a number between 0 & 1 that quantifies uncertainty. • A probability of 0 identifies impossibility • A probability of 1 identifies certainty

14. Where do probabilities come from? • Probabilities from models: The probability of getting a four when a fair dice is rolled is 1/6 (0.1667 or 16.7% chance)

15. Where do probabilities come from? • Probabilities from data or Empirical probabilities What is the probability that a randomly selected patient is successfully treated? • In a clinical trial n = 67 patients are “randomly” selected. • 40 of these patients are successfully treated. • The estimated probability that a randomly chosen patient will have a successful outcome is 40/67 (0.597 or 59.7% chance)

16. Where do probabilities come from? • Subjective Probabilities • The probability that there will be another outbreak of ebola in Africa within the next year is 0.1. • The probability of rain in the next 24 hours is very high. Perhaps the weather forecaster might say a there is a 70% chance of rain. • A doctor may state your chance of successful treatment.

17. For equally likely outcomes, and a given event A: Number of outcomes in A P(A) = Total number of outcomes Simple Probability Models “The probability that an event A occurs” is written in shorthand as P(A).

18. Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 1. Heart Disease In 1996, 6631 Minnesotans died from coronary heart disease. The numbers of deaths classified by age and gender are:

19. Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 1. Heart Disease Let A be the event of being under 45 B be the event of being male C be the event of being over 64

20. Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 1. Heart Disease Find the probability that a randomly chosen member of this population at the time of death was: • under 45 • P(A) = 92/6631 = 0.0139

21. Conditional Probability • We wish to find the probability of an event occuring given information about occurrence of another event. For example, what is probability of developing lung cancer given that we know the person smoked a pack of cigarettes a day for the past 30 years. • Key words that indicate conditional probability are: “given that”, “of those”, “if …”, “assuming that”

22. Conditional Probability “The probability of event A occurring given that event B has already occurred” is written in shorthand as P(A|B)

23. Conditional Probability and Independence P(A|B) =__________ , P(B) > 0 P(A and B) P(B) Two events A and B are said to be independent if P(A|B) = P(A) and P(B|A) = P(B) i.e. knowing the occurrence of one of the events tells you nothing about the occurrence of the other.

24. Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 1. Heart Disease Find the probability that a randomly chosen member of this population at the time of death was: • male assuming that the person was younger than 45.

25. Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 Heart Disease Find the probability that a randomly chosen member of this population at the time of death was: • male given that the person was younger than 45. • P(B|A) = 79/92 = 0.8587 P(B|A) = P(A and B)/P(A) = (79/6631)/(92/6631) = 79/92

26. Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 1. Heart Disease Find the probability that a randomly chosen member of this population at the time of death was: c) male and was over 64. • P(B and C)=(1081+1795)/6631=2876/6631=.434

27. Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 1. Heart Disease Find the probability that a randomly chosen member of this population at the time of death was: d) over 64 given they were female (not B).

28. Sex Age Male Female Total < 45 79 13 92 45 - 64 772 216 988 65 - 74 1081 499 1580 > 74 1795 2176 3971 Total 3727 2904 6631 1. Heart Disease Find the probability that a randomly chosen member of this population at the time of death was: d) over 64 given they were female (not B). • P(C|not B) = (499+2176)/2904 = .9211

29. 2. Hodgkin’s Disease Response to Treatment

30. 2. Hodgkin’s Disease

31. 2. Hodgkin’s Disease Response to Treatment • Had positive response to treatment P(pos) = 314/538 = .584 or 58.4% chance

32. 2. Hodgkin’s Disease Response to Treatment • Had at least some response to treatment P(par or pos) = (98 + 314)/538 = 412/538 = .766 or 76.6% chance

33. 2. Hodgkin’s Disease Response to Treatment • Had LP and positive response to treatment P(LP and pos) = 74/538 = .138 or 13.8%

34. 2. Hodgkin’s Disease Response to Treatment • Had LP or NS as there histological type. P(LP or NS) = (104 + 96)/538 = .372 or 37.2% chance

35. 2. Hodgkin’s Disease Response to Treatment What conditional probabilities would be of interest? EXAMPLES IN NOTES

36. 3. Right Heart Catheterization and 30-day Mortality (Conners, et al. 1996) Died within 30 days? RHC = patient had catheter put in No RHC = patient did not have catheter YES = Died within 30 days NO = Survived 30 days What is the probability that a heart patient in this study died? P(YES) = 1918 / 5735 = .3344 or 33.44%

37. 3. Right Heart Catheterization and 30-day Mortality (Conners, et al. 1996) Died within 30 days? RHC = patient had catheter put in No RHC = patient did not have catheter YES = Died within 30 days NO = Survived 30 days What is the probability that a heart patient had a right heart catheter put in during treatment? P(RHC) = 2184 / 5735 = .3808 or 38.08%

38. 3. Right Heart Catheterization and 30-day Mortality (Conners, et al. 1996) Died within 30 days? RHC = patient had catheter put in No RHC = patient did not have catheter YES = Died within 30 days NO = Survived 30 days What is the probability that a patient would die within 30 days given that they had a right heart catheter put in? P(YES | RHC) = 830 / 2184 = .3800 or 38.00%

39. 3. Right Heart Catheterization and 30-day Mortality (Conners, et al. 1996) Died within 30 days? RHC = patient had catheter put in No RHC = patient did not have catheter YES = Died within 30 days NO = Survived 30 days What is the probability that a patient would die within 30 days given that they did not have a right heart catheter put in? P(YES | No RHC) = 1088 / 3551 = .3064 or 30.64%

40. 3. Right Heart Catheterization and 30-day Mortality (Conners, et al. 1996) How many times more likely is a patient who had a right heart catheter put in to die within 30 days than patient who did not have a Swan-Ganz line put in? P(YES | RHC) = .3800 P(YES | No RHC) = .3064 .3800/ .3064 = 1.24 times more likely. This is called the relative risk or risk ratio (denoted RR).Risk of death is 24% greater for those that had a Swan-Ganz line put in.

41. 3. Right Heart Catheterization and 30-day Mortality (Conners, et al. 1996) The shading for 30-day mortality is 1.24 times higher for the RHC group than for the No RHC group (recall RR = 1.24). Patients having a Swan-Ganz line put in have 1.24 times higher risk of death within 30-days of initial treatment.

42. Building a Contingency Table from a Story 4. HIV Example A European study on the transmission of the HIV virus involved 470 heterosexual couples. Originally only one of the partners in each couple was infected with the virus. There were 293 couples that always used condoms. From this group, 3 of the non-infected partners became infected with the virus. Of the 177 couples who did not always use a condom, 20 of the non-infected partners became infected with the virus.

43. I NI 4. HIV Example Let C be the event that the couple always used condoms.(NC be the complement) Let I be the event that the non-infected partner became infected. (NI be the complement) Condom Usage Infection Status C NC Total Total

44. I NI 4. HIV Example A European study on the transmission of the HIV virus involved 470 heterosexual couples. Originally only one of the partners in each couple was infected with the virus. There were 293 couples that always used condoms. From this group, 3 of the non-infected partners became infected with the virus. Condom Usage Infection Status C NC Total 3 Total 293 470

45. I NI 4. HIV Example Of the 177 couples who did not always use a condom, 20 of the non-infected partners became infected with the virus. Condom Usage Infection Status C NC Total 3 20 23 290 157 447 Total 293 177 470

46. Condom Usage Infection Status C NC Total I 3 20 23 290 157 447 NI Total 293 177 470 4. HIV Example • What proportion of the couples in this study always used condoms? P(C)

47. Condom Usage Infection Status C NC Total I 3 20 23 290 157 447 NI Total 293 177 470 4. HIV Example • What proportion of the couples in this study always used condoms? P(C) = 293/470 (= 0.623)

48. Condom Usage Infection Status C NC Total I 3 20 23 290 157 447 NI Total 293 177 470 4. HIV Example • If a non-infected partner became infected, what is the probability that he/she was one of a couple that always used condoms? P(C|I) = 3/23 = 0.130

49. 4. HIV Example c) In what percentage of couples did the non-HIV partner become infected amongst those that did not use condoms? P(I|NC) = 20/177 = .113 or 11.3% • Amongst those that did where condoms? P(I|C) = 3/293 = .0102 or 1.02% • What is relative risk of infection associated with not wearing a condom? RR = P(I|NC) / P(I|C) = 11.08 times more likely to become infected.

50. 4. HIV Example The percentage of couples where the non-HIV partner became infected in the non-condom user group is 11 times higher than that for condom group.