1. Probability

2. 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? Lotto

3. 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. Roulette

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. 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? TV Game Show

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 Result A B C WinCarGoat LoseGoatCarGoat LoseGoat Car 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 Result A B C LoseCarGoat WinGoatCarGoat WinGoat Car 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 In this presentation we will examine basic ideas about probabilities: what they are and where they come from simple probability models 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. 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) Where do probabilities come from?

16. 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. Where do probabilities come from?

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

18. 1. Heart Disease In 1996, 6631 Minnesotans died from coronary heart disease. The numbers of deaths classified by age and gender are: Sex AgeMaleFemaleTotal < 45791392 45 - 64772216988 65 - 7410814991580 > 74179521763971 Total372729046631

19. Let A be the event of being under 45 B be the event of being male C be the event of being over 64 1. Heart Disease Sex AgeMaleFemaleTotal < 45791392 45 - 64772216988 65 - 7410814991580 > 74179521763971 Total372729046631

20. Find the probability that a randomly chosen member of this population at the time of death was: a)under 45 P(A) = 92/6631 = 0.0139 1. Heart Disease Sex AgeMaleFemaleTotal < 45791392 45 - 64772216988 65 - 7410814991580 > 74179521763971 Total372729046631

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

23. P(A|B) =__________, P(B) > 0 Conditional Probability and Independence 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. 1. Heart Disease Sex AgeMaleFemaleTotal < 45791392 45 - 64772216988 65 - 7410814991580 > 74179521763971 Total372729046631 Find the probability that a randomly chosen member of this population at the time of death was: b)male assuming that the person was younger than 45.

25. Sex AgeMaleFemaleTotal < 45791392 45 - 64772216988 65 - 7410814991580 > 74179521763971 Total372729046631 Find the probability that a randomly chosen member of this population at the time of death was: b)male given that the person was younger than 45. P(B|A) = 79/92 = 0.8587 2.Heart Disease P(B|A) = P(A and B)/P(A) = (79/6631)/(92/6631) = 79/92

26. Sex AgeMaleFemaleTotal < 45791392 45 - 64772216988 65 - 7410814991580 > 74179521763971 Total372729046631 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 1. Heart Disease

27. Sex AgeMaleFemaleTotal < 45791392 45 - 64772216988 65 - 7410814991580 > 74179521763971 Total372729046631 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). 1. Heart Disease

28. Sex AgeMaleFemaleTotal < 45791392 45 - 64772216988 65 - 7410814991580 > 74179521763971 Total372729046631 P(C|not B) = (499+2176)/2904 =.9211 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).

29. 2. Hodgkin’s Disease Type NonePartialPositive Row Totals LD441018 72 LP121874 104 MC5854154 266 NS121668 96 Column Totals 12698314n = 538 Response to Treatment

30. 2. Hodgkin’s Disease Type NonePartialPositive Row Totals LD441018 72 LP121874 104 MC5854154 266 NS121668 96 Column Totals 12698314n = 538 Response to Treatment a)Had positive response to treatment P(pos) = 314/538 =.584 or 58.4% chance

31. 2. Hodgkin’s Disease Type NonePartialPositive Row Totals LD441018 72 LP121874 104 MC5854154 266 NS121668 96 Column Totals 12698314n = 538 Response to Treatment b)Had at least some response to treatment P(par or pos) = (98 + 314)/538 = 412/538 =.766 or 76.6% chance

32. 2. Hodgkin’s Disease Type NonePartialPositive Row Totals LD441018 72 LP121874 104 MC5854154 266 NS121668 96 Column Totals 12698314n = 538 Response to Treatment c)Had positive response to treatment given they have LP P(pos|LP) = 74/104 =.7115 or 71.15%

33. 2. Hodgkin’s Disease Type NonePartialPositive Row Totals LD441018 72 LP121874 104 MC5854154 266 NS121668 96 Column Totals 12698314n = 538 Response to Treatment d)Had positive response to treatment given they have LD. P(pos|LD) = 18/72=.25 or 25.0% chance

34. Mosaic Plot with Conditional Probs. These percentages are all conditional probabilities expressed as a chance (0 – 100%). They give the conditional probability of a certain response to treatment given histological type, i.e. P(response to treatment | histological type)

35. 3. Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991) Brain Injury No Brain Injury Row Totals No Helmet 9719182015 Helmet Worn 17977994 Column Totals11428953009 BI = the event the motorcyclist sustains brain injury NBI = no brain injury H = the event the motorcyclist was wearing a helmet NH = no helmet worn P(BI) = 114 / 3009 =.0379 What is the probability that a motorcyclist involved in a accident sustains brain injury?

36. 3. Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991) Brain Injury No Brain Injury Row Totals No Helmet 9719182015 Helmet Worn 17977994 Column Totals1142893009 BI = the event the motorcyclist sustains brain injury NBI = no brain injury H = the event the motorcyclist was wearing a helmet NH = no helmet worn P(H) = 994 / 3009 =.3303 What is the probability that a motorcyclist involved in a accident was wearing a helmet?

37. 3. Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991) Brain Injury No Brain Injury Row Totals No Helmet 9719182015 Helmet Worn 17977994 Column Totals11428953009 What is the probability that the cyclist sustained brain injury given they were wearing a helmet? P(BI|H) = 17 / 994 =.0171 BI = the event the motorcyclist sustains brain injury NBI = no brain injury H = the event the motorcyclist was wearing a helmet NH = no helmet worn

38. 3. Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991) Brain Injury No Brain Injury Row Totals No Helmet 9719182015 Helmet Worn 17977994 Column Totals11428953009 What is the probability that the cyclist not wearing a helmet sustained brain injury? P(BI|NH) = 97 / 2015 =.0481 BI = the event the motorcyclist sustains brain injury NBI = no brain injury H = the event the motorcyclist was wearing a helmet NH = no helmet worn

39. 3. Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991) Brain Injury No Brain Injury Row Totals No Helmet 9719182015 Helmet Worn 17977994 Column Totals11428953009 How many times more likely is a non-helmet wearer to sustain brain injury?.0481 /.0171 = 2.81 times more likely. This is called the relative risk or risk ratio (denoted RR).

40. Example 3: Helmet Use and Head Injuries in Motorcycle Accidents (Wisconsin, 1991) The shading for Brain Injury for the No Helmet group is roughly three times higher than the shading for Brain Injury for the Helmet Worn group. (recall RR = 2.81) Motorcyclists not wearing a helmet are at three times the risk of suffering brain injury.

41. 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.

42. 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) CNC NI I 4. HIV Example Total Condom Usage Infection Status

43. 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. CNC NI I 4. HIV Example Total Condom Usage Infection Status 470 293 3

44. Of the 177 couples who did not always use a condom, 20 of the non-infected partners became infected with the virus. CNC NI I 4. HIV Example Total Condom Usage Infection Status 470 293 320 177 290157 23 447

45. a)What proportion of the couples in this study always used condoms? CNC NI I Total Condom Usage Infection Status 470 293 320 177 290157 23 447 4. HIV Example P(C )

46. a)What proportion of the couples in this study always used condoms? CNC NI I Total Condom Usage Infection Status 470 293 320 177 290157 23 447 4. HIV Example P(C ) = 293/470 (= 0.623)

47. b)If a non-infected partner became infected, what is the probability that he/she was one of a couple that always used condoms? 4. HIV Example CNC NI I Total Condom Usage Infection Status 470 293 320 177 290157 23 447 P(C|I ) = 3/23 = 0.130

48. 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.

49. 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.

50. Relative Risk (RR) and Odds Ratio (OR) Example: Age at First Pregnancy and Cervical Cancer A case-control study was conducted to determine whether there was increased risk of cervical cancer amongst women who had their first child before age 25. A sample of 49 women with cervical cancer was taken of which 42 had their first child before the age of 25. From a sample of 317 “similar” women without cervical cancer it was found that 203 of them had their first child before age 25. Q: Do these data suggest that having a child at or before age 25 increases risk of cervical cancer?

51. Definition of Odds The ODDS for an event A are defined as Odds for A = _______ P(A) 1 – P(A) For example suppose we roll a single die the odds for a 3 are: Odds for 3 = P(3)/(1 – P(3)) = = (1/6)/(1 – (1/6)) = 1/5 1 three for every 5 rolls that don’t result in a six. (Odds for a 3 are 1:5 and odds against are 5:1)

52. Relative Risk (RR) and Odds Ratio (OR) The Odds Ratio (OR) for a disease associated with a risk factor is ratio of the odds for disease for those with risk factor and the odds for disease for those without the risk factor OR = _________________________ P(Disease|Risk Factor) 1 – P(Disease|Risk Factor) _____________________ P(Disease|No Risk Factor) 1 – P(Disease|No Risk Factor) _______________________ The Odds Ratio gives us the multiplicative increase in odds associated with having the “risk factor”. Odds for disease amongst those with risk factor present Odds for disease amongst those without the risk factor.

53. Relative Risk (RR) and Odds Ratio (OR) Age at 1 st Pregnancy CaseControl Row Totals Age < 2542203 245 Age > 257114 121 Column Totals49317 n = 366 Cervical Cancer a) Why can’t we calculate P(Cervical Cancer | Age < 25)? Because the number of women with disease was fixed in advance and therefore NOT RANDOM !

54. Relative Risk (RR) and Odds Ratio (OR) Age at 1 st Pregnancy CaseControl Row Totals Age < 2542203 245 Age > 257114 121 Column Totals49317 n = 366 Cervical Cancer b) What is P(risk factor|disease status) for each group? P(Age < 25|Case) = 42/49 =.857 or 85.7% P(Age < 25|Control) = 203/317 =.640 or 64.0%

55. Relative Risk (RR) and Odds Ratio (OR) Age at 1 st Pregnancy CaseControl Row Totals Age < 2542203 245 Age > 257114 121 Column Totals49317 n = 366 Cervical Cancer c) What are the odds for the risk factor amongst the cases? Amongst the controls? Odds for risk factor cases =.857/(1-.857) = 5.99 Odds for risk factor controls =.64/(1-.64) = 1.78

56. Relative Risk (RR) and Odds Ratio (OR) Age at 1 st Pregnancy CaseControl Row Totals Age < 2542203 245 Age > 257114 121 Column Totals49317 n = 366 Cervical Cancer d) What is the odds ratio for the risk factor associated with being a case? Odds Ratio (OR) = 5.99/1.78 = 3.37, the odds for having 1 st child on or before age 25 are 3.37 times higher for women who currently have cervical cancer versus those that do not have cervical cancer.

57. Relative Risk (RR) and Odds Ratio (OR) Odds Ratio The ratio of dark to light shading is 3.37 times larger for the cervical cancer group than it is for the control group.

58. e)Even though it is inappropriate to do so calculate P(disease|risk status). P(case|Age<25) = 42/245 =.171 or 17.1% P(case|Age>25) = 7/121 =.058 or 5.8% Now calculate the odds for disease given the risk factor status Odds for Disease for 1 st Preg. Age < 25 =.171/(1 -.171) =.207 Odds for Disease for 1 st Preg. Age > 25 =.058/(1 -.058) =.061 Relative Risk (RR) and Odds Ratio (OR)

59. f) Finally calculate the odds ratio for disease associated with 1 st pregnancy age < 25 years of age. Odds Ratio =.207/.061 = 3.37 This is exactly the same as the odds ratio for having the risk factor (Age < 25) associated with being in the cervical cancer group!!!! Relative Risk (RR) and Odds Ratio (OR) Final Conclusion: Women who have their first child at or before age 25 have 3.37 times the odds of developing cervical cancer when compared to women who had their first child after the age of 25.

60. Relative Risk (RR) and Odds Ratio (OR) Risk Factor Status CaseControl Risk Factor Present a b Risk Factor Absent cd Disease Status OR = _____ a X d b X c Much easier computational formula!!!

61. Relative Risk (RR) and Odd’s Ratio (OR) When the disease is fairly rare, i.e. P(disease) <.10 or 10%, then one can show that the odds ratio and relative risk are similar. i.e. OR is approximately equal to RR when P(disease) <.10 or 10% chance. In these cases we can use the phrase: “… times more likely” when interpreting the Odds Ratio.

62. Relative Risk (RR) and Odds Ratio (OR) Age at 1 st Pregnancy CaseControl Row Totals Age < 25 a 42 b 203 245 Age > 25 c7c7 d 114 121 Column Totals49317 n = 366 OR = (42 X 114)/(7 X 203) = 3.37 Because less than 10% of the population of women develop cervical cancer we can say women who have their first child at or before age 25 are 3.37 times more likely to develop cervical cancer than women who have their first child after age 25.

63. More About RR and OR The most commonly cited advantage of the RR over the OR is that the former is the more natural interpretation. The relative risk comes closer to what most people think of when they compare the relative likelihood of events. e.g. suppose there are two groups, one with a 25% chance of mortality and the other with a 50% chance of mortality. Most people would say that the latter group has it twice as bad. But the odds ratio is 3, which seems too big. Calculations RR =.50/.25 = 2.00 OR = P(death|high mortality)/P(survive|high mortality) P(death|low mortality)/P(survive|low mortality) =.50/(1 -.50) = 3.00.25/(1 -.25)

64. More About RR and OR Even more extreme examples are possible. A change from 25% to 75% mortality represents a relative risk of 3, but an odds ratio of 9. A change from 10% to 90% mortality represents a relative risk of 9 but an odds ratio of 81.

65. More About RR and OR OR’s arise as part of logistic regression which we will study later in the course. Despite their pitfalls OR’s are really the only option when case-control studies are used. Any study of risk needs to adjust for potential confounding factors which is typically done using logistic regression.

66. Baye’s Rule and Medical Screening Tests Baye’s Rule is used in medicine and epidemiology to calculate the probability that an individual has a disease, given that they test positive on a screening test. Example: Down syndrome is a variable combination of congenital malformations caused by trisomy 21. It is the most commonly recognized genetic cause of mental retardation, with an estimated prevalence of 9.2 cases per 10,000 live births in the United States. Because of the morbidity associated with Down syndrome, screening and diagnostic testing for this condition are offered as optional components of prenatal care. Prenatal diagnosis of trisomy 21 allows parents the choice of continuing or terminating an affected pregnancy.

67. Baye’s Rule and Medical Screening Tests Many studies have been conducted looking at the effectiveness of screening methods used to identify “likely” Down syndrome cases. One of these tests called “triple test” or “triple screen” is described below: Alpha-fetoprotein (AFP), unconjugated estriol and human chorionic gonadotropin (hCG) are the serum markers most widely used to screen for Down syndrome. This combination is known as the "triple test" or "triple screen." AFP is produced in the yolk sac and fetal liver. Unconjugated estriol and hCG are produced by the placenta. The maternal serum levels of each of these proteins and of steroid hormones vary with the gestational age of the pregnancy. With trisomy 21, second-trimester maternal serum levels of AFP and unconjugated estriol are about 25 percent lower than normal levels and maternal serum hCG is approximately two times higher than the normal hCG level.

68. Baye’s Rule and Medical Screening Tests One study looking at effectiveness produced

69. Baye’s Rule and Medical Screening Tests How well does this study suggest the “triple test” perform? False Positive = P(T+|D-) = 203/4072 =.0499 False Negative = P(T-|D+) = 31/118 =.2627 Sensitivity = P(T+|D+) = 87/118 =.7373 Specificity = P(T- |D-) = 3869/4072 =.9501

70. Baye’s Rule and Medical Screening Tests Now suppose you are have just been given the news the results of the “triple test” are positive for Down syndrome. What do you want to know now? You probably would like to know what the probability that your unborn child actually has Down syndrome, i.e. what is P(D+|T+)? To answer this question we need to use Baye’s Rule to “reverse the conditioning”.

71. Baye’s Rule for Medical Screening Tests Positive Predictive Value P(D+|T+) This requires “prior knowledge” of the probability of the disease/defect being present. We have this because we know that on average in the U.S. there are 9.2 Down’s cases per 10,000 live births, i.e. P(D+) =.00092 and P(D-) =.99908 So with a positive screening test result in hand, how does the probability of disease change?

72. Baye’s Rule for Medical Screening Tests Positive Predictive Value P(D+|T+) Thus the probability of Down’s given a positive test result is about 1.3 in 100 births vs. the 9.2 per 10,000 births without the test result, which is a 14.6 times higher. However, fewer than 50 positive test results actually correspond to Down’s cases.

73. Baye’s Rule for Medical Screening Tests Negative Predictive Value P(D-|T-) Thus the probability of not having Down’s given a negative test result is a 99.97% chance vs. 99.908% chance without a negative test result, which is ONLY 1.0007 times higher, i.e. a.07% increase!

74. Link to Nice Presentation on ROC Curves Review Slides 1 – 35 of the presentation below, paying particular attention to the material on ROC curves. We will discuss these curves when we examine logistic regression. http://www.rad.washington.edu/gems/outcomes /radiologylecturebryan.ppt