1. In today’s lecture… Probability Counting methods- Permutations & Combinations Independence Non-independence/Bayes’ Rule

2. Example: Prostate Cancer Study Thompson et al. (2006) Prostate Specific Antigen (PSA) evaluation leads to early detection of prostate cancer Study looked at 5519 men who underwent prostate biopsy Characteristics looked at: age, race, history of prostate cancer, previous biopsies/screening

3. Table 1:Racial characteristics of study participants (n=5519) n% Race White531096.2 ~ 96 Black2093.8~ 4

4. Table 2: Number of prostate cancers and high grade prostate cancers

5. Probability: Chance of something happening (from 0-1) 0: cannot happen 1: sure to happen P(A) = probability that event “A” will occur P(PSA 0-1) = probability that PSA level is from 0-1 P(~A) = probability that event “A” will NOT occur [complement] P(~PSA 0-1)= probability that PSA level is NOT from 0-1 P(A & B) = the probability that both A and B happen [joint probability] P(PSA 0-1 & white) = the probability of being a white male with PSA 0-1 P(A|B) = the probability that A occurs, given that B occurred [conditional probability] P(PSA 0-1|white) = the probability that PSA is 0-1, given that the patient is white *Sainani K., Stanford AB P(A&B) AB P(A/B)

6. Assessing Probability 1. Theoretical/Classical probability—based on theory (a priori understanding of a phenomena) e.g.: theoretical probability of rolling a 2 on a standard die is 1/6 theoretical probability of choosing an ace from a standard deck is 4/52 theoretical probability of getting heads on a regular coin is 1/2 2. Empirical probability—based on empirical data e.g.: you toss an irregular die (probabilities unknown) 100 times and find that you get a 2 twenty-five times; empirical probability of rolling a 2 is 1/4 empirical probability of an Earthquake in Bay Area by 2032 is.62 (based on historical data) empirical probability of a lifetime smoker developing lung cancer is 15 percent (based on empirical data) *Sainani K., Stanford

7. Computing theoretical probabilities:counting methods Great for gambling! Fun to compute! If outcomes are equally likely to occur… Note: these are called “counting methods” because we have to count the number of ways A can occur and the number of total possible outcomes. *Sainani K., Stanford

8. Applying our example… P (PSA level 0-1) = (# cases with PSA 0-1) (total number of cases) = (1963)/(5519) = 0.35 P (PSA level >6) = (# cases with PSA >6) (total number of cases) = (150)/(5519) = 0.03 You randomly pick a patient to test his PSA. What’s the probability that he is white? P(white) = (# cases who are white) (total number of cases) = (5310)/(5519) = 0.96 …that he is black? P(black) = (# cases who are black) (total number of cases) = (209)/(5519) = 0.04

9. Example 2 What’s the probability that you pick two patients who are black? P(1 st patient black) = (# cases who are black) (total number of cases) = (209)/(5519) = 0.038 ~ 0.04 P(2 nd patient black) = (# cases who are black) (total number of cases) = (208)/(5518) = 0.037 ~0.04 P(black & black) = P(1 st patient black) x P(2 nd patient black) = 0.0016 This is an example of joint probability…more on this coming up!

10. Example 3 If you have 5 patients (3 white, 2 black), and you want to test PSA of two randomly chosen patients, what’s the probability that they are white (W) and black (B)? Considering order of picking, P (1B, 1W patient) = # ways to pick one B, one W pair # total patient pairs Numerator = W 1 B 1 W 1 B 2 W 2 B 1 W 2 B 2 W 3 B 1 W 3 B 2 B 1 W 1 B 2 W 1 B 1 W 2 B 2 W 2 B 1 W 3 B 2 W 3 = 12 Denominator = 5x4 = 20 P(1B, 1W) = 12/20 = 0.6 5 patients4 patients

11. Applying our PSA example, using a probability tree… B P(B=.04) W P(W=.96) First pick B P(B=.04) P(W=.96) B P(B=.04) W P(W=.96) Second pickOutcome P(WB)=0.04*0.96 = 0.038 B P(BB)=0.04*0.04 = 0.0016 B P(BW)=0.04*0.96 = 0.038 P(WW)=0.96*0.96 = 0.922 Rule of thumb: in probability, “and” means multiply, “or” means add P(B)=0.04, From our example 2 P(1B,1W) = P(BW) +P(WB) = 0.038 + 0.038 = 0.076

12. Ignoring order of picking: P (1B,1W patient) = (# ways to pick one B, one W ) (total # ways to pick 2 patients) Numerator = W 1 B 1 W 1 B 2 W 2 B 1 W 2 B 2 W 3 B 1 W 3 B 2 = 6 Denominator = (5x4)/2 P (picking a B,W patient) = 6= 12 = 0.6 (5x4)/2 20 We divide out the order, by dividing by 2 here

13. Summary of Counting Methods Counting methods for computing probabilities With replacement Without replacement Permutations— order matters! Combinations— Order doesn’t matter Without replacement *Sainani K., Stanford

14. Permutations — Order matters! A permutation is an ordered arrangement of objects. With replacement = once an event occurs, it can occur again (after you roll a 6, you can roll a 6 again on the same die). Without replacement = an event cannot repeat (after you draw an ace of spades out of a deck, there is 0 probability of getting it again). *Sainani K., Stanford

15. Permutations with replacement Sample space: the set of all possible outcomes. Example: in genetics, if both the mother and father carry one copy of a recessive disease-causing mutation (d), there are three possible outcomes (the sample space):  child is not a carrier (DD)  child is a carrier (Dd)  child has the disease (dd). Probabilities: the likelihood of each of the possible outcomes (always 0  P  1.0).  P(genotype=DD)=.25  P(genotype=Dd)=.50  P(genotype=dd)=.25. *Sainani K., Stanford

16. Summary: order matters, with replacement Formally, “order matters” and “with replacement”  use powers  Equation for total number of possible outcomes: *Sainani K., Stanford

17. Example 1: ♀ P( ♀ D=.5) ♀ P( ♀ d=.5) Mother’s allele ♂ P( ♂ D=.5) ♂ P( ♂ d=.5) ♂ P( ♂ D=.5) ♂ P( ♂ d=.5) Father’s allele ______________ 1.0 P(DD) =.5*.5 =.25 P(Dd) =.5*.5 =.25 P(dD) =.5*.5 =.25 P(dd) =.5*.5 =.25 Child’s outcome What’s the chance of having a child with the disease(dd) if both parents are heterozygote (Dd)? P(dd) = 1 way to get (dd) = 1/4 = 0.25 2 2 possible outcomes *Sainani K., Stanford

18. Permutations without replacement Example 1: Suppose you want to test PSA levels of 4 patients: A,B,C,D. How many ways can you test them? A B C D B A C D C B A D D B C A..# permutations = 4x3x2x1 = 4! = 24. OR Reminder! Factorial notation: n! =n x (n-1) x (n-2) x……….x1 A B C D A B C D D C D C So there are 4! ways of doing 4 tests for 4 patients

19. Example 2: What if you had 3 different tests and 5 people? E B A C D E A B D A B C D Test 1: 5 possible Test 2: Only 4 possible E B D Test 3: only 3 possible *Sainani K., Stanford

20. Summary: order matters, without replacement Formally, “order matters” and “without replacement”  use factorials  Note: This formula also worked for Example 1. We were picking 4 people for 4 spots. So, 4!/ (4-4)! = 4!/0! = 4! = 24 *Sainani K., Stanford

21. Recall Permutation Theory… If you want to see if there is a difference between the mean PSA scores for black ( n=209) and white patients (n=5310) in PSA example: 1. Calculate mean scores of black patients & white patients 2. Shuffle scores of 5000 random patients, 3. Number of possible permutations of shuffling are: (5519!)= A huge number of permutations (5519-5000)! 4. Compare original mean scores to mean scores of each permutation.

22. 2. Combinations—Order doesn’t matter A combination helps determine the number of ways “r” objects can be chosen from “n” larger group of objects Introduction to combination function, or “choosing” Spoken: “n choose r” Written as: *Sainani K., Stanford

23. Example of combinations If you have 3 identical tests. What are the # of ways you can choose 3 out of 5 patients, to be tested? = 5 C 3 = 5! = 5 x 4 = 10 3! (5-3)! 2

24. Example: Distinct vs. Nondistinct objects Suppose you want to calculate mean PSA scores of 4 patients (3 white, 1black): A (White), B (White), C (White), & D (Black). How many ways can we arrange the 4 patients based on race? Total number of arrangements of 4 people = 4! = 24 However, based on race, 3 of them are identical (White- A, B, C) and 1 of them is identical (Black- D). If you only consider race of the patients, there will be fewer arrangements possible …

25. For example : arrangementA B C D (W W W B) = arrangement C B A D (W W W B) In fact, the arrangement (W W W B) can be done in 6 distinct ways: A B C D A C B D B C A D B A C D C A B D C B A D = 3! permutations of white patients x 1 permutation of the black patient = 6 x1 = 6 This is one race based arrangement.

26. Similarly, the arrangement (W W B W) can be done in 3!x1!=6 ways : A B D C A C D B B C D A B A D C C A D B C B D A

27. Since we don’t care about order, 4! ways of arranging the 4 patients is reduced to: 4! = 4! = 4 3! 1! 6 Hence, number of ways of arranging n objects, of which k are white and m are black: = n! k! m! ( 1. White or black are just examples of being nondistinct 2. Can be extended to any number of nondistinct sets)

28. This is also a “choosing” problem since we are choosing 3 tests for white patients & 1 for the black patient: 4 C 3 = 4 C 1 = 4! (3!)(1!) = 4

29. Summary: combinations If r objects are taken from a set of n objects without replacement and disregarding order, how many different samples are possible? Formally, “order doesn’t matter” and “without replacement”  use choosing  *Sainani K., Stanford

30. Summary of Counting Methods Counting methods for computing probabilities With replacement: n r Permutations— order matters! Without replacement: n(n-1)(n-2)…(n-r+1)= Combinations— Order doesn’t matter Without replacement: *Sainani K., Stanford

31. Independence Formal definition: A and B are independent if and only if P(A&B)=P(A)*P(B) Going back to our Genetics example: The mother’s and father’s alleles are segregating independently. P(♂D|♀D)=.5 and P(♂D|♀d)=.5 What father’s gamete looks like is not dependent on the mother’s – doesn’t depend which branch you start on! Formally, P(DD)=.25=P(D♂)*P(D♀) Conditional Probability: Read as “the probability that the father passes a D allele given that the mother passes a d allele.” Joint Probability: The probability of two events happening simultaneously. Marginal probability: This is the probability that an event happens at all, ignoring all other outcomes. *Sainani K., Stanford

32. On the tree ♀ P( ♀ D =.5) ♀ P( ♀ d=.5) Mother’s allele ♂♀ P( ♂ D/ ♀ D )=.5 ♂ P( ♂ d=.5) ♂ P( ♂ D=.5) ♂ P( ♂ d=.5) Father’s allele ______________ 1.0 P(DD)=.5*.5=.25 P(Dd)=.5*.5=.25 P(dD)=.5*.5=.25 P(dd)=.5*.5=.25 Child’s outcome Conditional probabilityMarginal probability: motherJoint probability Marginal probability: father *Sainani K., Stanford

33. Independent  mutually exclusive Events A and ~A are mutually exclusive, but they are NOT independent. P(A&~A)= 0 P(A)*P(~A)  0 Conceptually, once A has happened, ~A is impossible; thus, they are completely dependent. *Sainani K., Stanford

34. Practice problem If HIV has a prevalence of 3% in San Francisco, and a particular HIV test has a false positive rate of.001 and a false negative rate of.01, what is the probability that a random person selected off the street will test positive?

35. Answer ______________ 1.0 P (+, test +)=.0297 P(+, test -)=.003 P(-, test +)=.00097 P(-, test -) =.96903  P(test +)=.0297+.00097=.03067 Marginal probability of carrying the virus. Joint probability of being + and testing + P(+&test+)  P(+)*P(test+).0297 .03*.03067 (=.00092)  Dependent! Marginal probability of testing positive Conditional probability: the probability of testing + given that a person is + P(+)=.03 P(-)=.97 P(test +)=.99 P(test - )=.01 P(test +) =.001 P(test -) =.999 * Sainani K., Stanford

36. Law of total probability One of these has to be true (mutually exclusive, collectively exhaustive). They sum to 1.0. *Sainani K., Stanford

37. Law of total probability Formal Rule: Marginal probability for event A= B 2 B 3 B 1 Where: A *Sainani K., Stanford

38. Non independent events/Conditional Probability When two events are not independent, the occurrence of one event depends on whether the other has occurred

39. Bayes’ Rule

40. Definition: Let A and B be two events with P(B)  0. The conditional probability of A given B is: *Sainani K., Stanford

41. Bayes’ Rule: From the “Law of Total Probability” OR *Sainani K., Stanford

42. In-Class Exercise If HIV has a prevalence of 3% in San Francisco, and a particular HIV test has a false positive rate of.001 and a false negative rate of.01, what is the probability that a random person who tests positive is actually infected (also known as “positive predictive value”)? *Sainani K., Stanford

43. Answer: using probability tree ______________ 1.0 P(test +)=.99 P(+)=.03 P(-)=.97 P(test - =.01) P(test +) =.001 P (+, test +)=.0297 P(+, test -)=.003 P(-, test +)=.00097 P(-, test -) =.96903 P(test -) =.999 A positive test places one on either of the two “test +” branches. But only the top branch also fulfills the event “true infection.” Therefore, the probability of being infected is the probability of being on the top branch given that you are on one of the two circled branches above. *Sainani K., Stanford

44. Answer: using Bayes’ rule *Sainani K., Stanford

45. Conditional probability in epidemiology: Odds and Risk (probability) Definitions: Risk = P(A) = cumulative probability (you specify the time period!) For example, what’s the probability that a person with a high sugar intake develops diabetes in 1 year, 5 years, or over a lifetime? Odds = P(A)|P(~A) For example, “the odds are 3 to 1 against a horse” means that the horse has a 25% probability of winning. Note: An odds is always higher than its corresponding probability, unless the probability is 100%. *Sainani K., Stanford

46. Introduction to the 2x2 Table Exposure (E)No Exposure (~E) Disease (D)aba+b = P(D) No Disease (~D)cdc+d = P(~D) a+c = P(E)b+d = P(~E) Marginal probability of disease Marginal probability of exposure *Sainani K., Stanford

47. Coming soon…(Applications of today’s lecture) More on odds ratios, risk ratios Patterns of categorical data/distributions Frequency tables Chi square Logistic regression Kaplan Meier, survival analysis Special thanks to Dr. Cobb for her great slides from last year!