Laws of probability

Laws of probability

1. LAWS OF PROBABILITY Presenter Dr. Brijesh Kumar JR Department of community Medicine, PGIMS, Rohtak

2. Contents Definition Types of probability Bayes ’ theorem Binomial Probability Law Laws of probability References

3. Probability Relative frequency or probable chances of occurrence with which an event is expected to occur on an average. Odds with which an event is expected to occur in a long run. OR

4. Usually expressed as symbol ‘p’ Probability ‘p’ ranges from 0 to 1 P=0 means ‘ no chance of an event happening’ P=1 means ‘100% chances of an event happening’

5. p = no. of events occurring total no. of trials p + q = 1 Probability If probability of an event happening is ‘p’ and probability of not happening is ‘q’ If an experiment is repeated ‘n’ times and an event A is observed ‘f’ times p(A) = f/n

6. Laws of Probability Addition law of Probability Multiplication law of Probability Binomial law of Probability

7. Addition law of Probability If one event excludes the possibility of occurrence of the other specified event or events, the events are called mutually exclusive . e.g. Getting head excludes the possibility of getting tail in coin flip e.g. Birth of a male child excludes the possibility of birth a female

8. Addition law of Probability These mutually exclusive events follow the addition law of probability. If the no. of mutually exclusive events are n and p1 is the individual probability Total prob. P = p1+ p2 + ………….+ pn = 1 P(A or B) = P(A) + P(B)

9. Addition law of Probability P(2) = 1/6 P(5) = 1/6 P(2 or 5) = 1/6 + 1/6 = 1/3 P(aces of any colour ) = 1/52 +1/52 + 1/52 + 1/52 = 1/13

10. Multiplication law of Probability It is applied when two or more events are occurring together but they are independent of each other If the no. of events are n and p1 is the individual probability Total prob. P = p1 * p2 * ………….* pn P(A and B) = P(A) * P(B)

11. P(2) = 1/6 P(5) = 1/6 P(2 and 5) = 1/6 * 1/6 = 1/36 Multiplication law of Probability

12. Multiplication law of Probability If we assume twin pregnancy will occur once in 80 preg . And child with Rh-ve blood group will be born once in 10 births. prob. of male child,p1 = ½ prob. of child being Rh+ve , p2 = 9/10 prob. of single birth, p3 = 79/80 prob. of male, Rh+ve and single birth = ½*9/10*79/80 = 711/1600

13. Binomial law of Probability It is used when events are occurring one after the other. e.g. M M F F M F M M and F F F and and and M M F 1/2*1/2=1/4 1/2*1/2=1/4 1/2*1/2=1/4 1/2*1/2=1/4

14. M M F F M F M M and F F F and and and M M F 1/2*1/2=1/4 1/2*1/2=1/4 1/2*1/2=1/4 1/2*1/2=1/4 Binomial law of Probability Chances of getting two males = ¼ = 25%

15. Binomial law of Probability M M F F M F M M and F F F and and and M M F 1/2*1/2=1/4 1/2*1/2=1/4 1/2*1/2=1/4 1/2*1/2=1/4 Chances of getting two females = ¼ = 25%

16. Binomial law of Probability M M F F M F M M and F F F and and and M M F 1/2*1/2=1/4 1/2*1/2=1/4 1/2*1/2=1/4 1/2*1/2=1/4 Chances of getting one of either sex = 1/4 + 1/4 = ½ = 50%

17. Binomial law of Probability M M F F M F M F F M M M F F e.g. MMM MFM MMF MFF FMM FMF FFM FFF 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8

18. Binomial law of Probability M M F F M F M F F M M M F F MMM MFM MMF MFF FMM FMF FFM FFF 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 Chances of getting 3 males = 1/8 = 12.5%

19. Binomial law of Probability M M F F M F M F F M M M F F MMM MFM MMF MFF FMM FMF FFM FFF 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 Chances of getting 3 Females = 1/8 = 12.5%

20. Binomial law of Probability M M F F M F M F F M M M F F MMM MFM MMF MFF FMM FMF FFM FFF 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 Chances of getting 2 males and 1 female = 1/8+1/8+1/8 = 37.5%

21. Binomial law of Probability M M F F M F M F F M M M F F MMM MFM MMF MFF FMM FMF FFM FFF 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 1/2*1/2*1/2=1/8 Chances of getting 1 male and 2 females = 1/8+1/8+1/8 = 37.5%

22. Binomial law of Probability Binomial law of probability distribution is formed by terms of the expansion of binomial expression (p +q) n n= number of events p= probability of success q= probability of failure

23. Binomial law of Probability when n = 2 ( p + q ) 2 = p 2 + q 2 + 2pq when n = 3 ( p + q ) 3 = p 3 + q 3 + 3pq 2 + 3p 2 q when n = 4 ( p + q ) 4 = p 4 + 4p 3 q+ 6p 2 q 2 + 4pq 3 + q 4

24. If p = chance of getting male child q = chance of getting female child ( p + q ) 2 = p 2 + q 2 + 2pq = 0.5 2 + 0.5 2 + 2*0.5*0.5 = 0.25 + 0.25 + 0.50 Binomial law of Probability ( p + q ) 3 = p 3 + q 3 + 3pq 2 + 3p 2 q = 0.5 3 + 0.5 3 + 3*0.5*0.5 2 + 3*0.5 2 *0.5 = 0.125 + 0.125 + 0.375 + 0.375

25. Binomial law of Probability e.g. In 4 ICDS projects, 20% of children under 6 yrs of age were found to be severely malnourished. Only 4 children were selected at random from the 4 projects. What is the prob. Of 4, 3, 2, 1, 0 being severely malnourished?

26. Binomial law of Probability p = 0.2 q = 0.8 n = 4 ( p + q ) 4 = p 4 + 4p 3 q+ 6p 2 q 2 + 4pq 3 + q 4 = 0.2 4 + 4*0.2 3 *0.8+ 6*0.2 2 *0.8 2 + 4*0.2*0.8 3 + 0.8 4 = 0.0016 + 0.0256 + 0.1536 + 0.4096 + 0.4096 Prob. of all 4 children being malnourished = 0.16% Prob. of 3 children being malnourished = 2.56% Prob. of 2 children being malnourished = 15.36% Prob. of 1 children being malnourished = 40.96% Prob. of no children being malnourished = 40.96%

27. Types of Probability Marginal Probability Joint Probability Conditional Probability

28. Marginal Probability Marginal probability is the probability of a single event without consideration of any other event. Written as P( A and B) or P( A B ) Joint Probability The probability of the intersection of two events is called their joint probability.

29. Conditional Probability Conditional probability is the probability that an event will occur given that another has already occurred. If A and B are two events, then prob. Of A given that B has already occurred will be : P(A|B) = P(A and B) / P(B) P(B|A) = P(A and B) / P(A)

30. Types of Probability Heart disease cancer diabetes total Family History 88 44 38 170 No family history 92 76 62 230 total 180 120 100 400 e.g.

31. Heart disease cancer diabetes total Family History 88 44 38 170 No family history 92 76 62 230 total 180 120 100 400 Types of Probability P ( cancer ) = 120/400 = 0.30 Marginal Probability

32. Types of Probability Heart disease cancer diabetes total Family History 88 44 38 170 No family history 92 76 62 230 total 180 120 100 400 P ( Cancer and No history ) = 76/400 = 0.19 Joint Probability

33. Types of Probability Heart disease cancer diabetes total Family History 88 44 38 170 No family history 92 76 62 230 total 180 120 100 400 P ( Diabetes | History ) = P(Dia. and Hist.)/ P(Hist.) = 38/170 = 0.2235 Conditional Probability

34. Conditional Probability Test Result Disease present (D+) Disease absent (D-) Total Positive ( T+) a (TP ) b (FP) a+b Negative (T-) c (FN) d (TN) c+d Total a+c b+d n Sensitivity = prob. of positive test given the presence of disease = P ( T+ | D+ ) = P ( T+ and D+ )/ P( D+) = a/ ( a+c )

35. Conditional Probability Test Result Disease present (D+) Disease absent (D-) Total Positive ( T+) a (TP ) b (FP) a+b Negative (T-) c (FN) d (TN) c+d Total a+c b+d n Specificity = prob. of negative test given the absence of disease = P ( T- | D- ) = P ( T- and D- )/ P( D-) = d/ ( b+d )

36. Conditional Probability Test Result Disease present (D+) Disease absent (D-) Total Positive ( T+) a (TP ) b (FP) a+b Negative (T-) c (FN) d (TN) c+d Total a+c b+d n Positive predictive value = prob. of presence of disease given the positive test = P ( D+ | T+ ) = P ( T+ and D+ )/ P( T+) = a/ ( a+b )

37. Conditional Probability Test Result Disease present (D+) Disease absent (D-) Total Positive ( T+) a (TP ) b (FP) a+b Negative (T-) c (FN) d (TN) c+d Total a+c b+d n Negative predictive value = prob. of absence of disease given the negative test = P ( D- | T- ) = P ( T- and D- )/ P( T-) = d/ ( c+d )

38. Likelihood Ratios There are two types of Likelihood Ratios one for positive test ( LR+) and another negative test (LR-). LR is the ratio of probability of a particular test result for a individual with disease of interest to probability of a particular test result for a individual without disease of interest.

39. New Diagnostic test Western Blot pos. (D+) Western Blot neg. (D-) Total HIV + ( T+) 85 20 105 HIV - (T-) 15 180 195 Total 100 200 300 Likelihood Ratios e.g. P(T+|D+)=P(T+ and D+)/P(D+) = 85/100 P(T+|D-)=P(T+ and D-)/P(D-) = 20/200

40. P(T+|D+) LR+ = ------------- P(T+|D-) P(T-|D+) LR- = ------------- P(T-|D-) D+ denotes that the individual is having a disease by Gold Standard. D- denotes that the individual is not having a disease by Gold Standard. T+ denotes that the individual is having a disease by New diagnostic kit. T- denotes that the individual is not having a disease by New diagnostic kit. Likelihood Ratios

41. Likelihood Ratios LR+ = P(T+ | D+ ) / P(T + | D-) LR+ = (85/100) / (20/200) P(T+|D+) = 8.5 times P(T+|D-) LR+ = 8.5

42. Bayes ’ Theorem The concept of conditional probability can be revised based on new information and to determine the probability that a particular effect was due to specific causes. The procedures for revising these probabilities is known as Bayes ’ theorem.

43. P (A|B) = P(A and B ) / P(B) P(A and B) = P(A|B) * P(B) Bayes ’ Theorem P (B|A) = P(A and B ) / P(A) P(A and B) = P(B|A) * P(A) P (B|A) * P(A) = P(A|B) * P(B) P(B|A) = P(A|B) * P(B) / P(A)

44. Bayes ’ Theorem e.g. Statistics show that 60% of all men have some form of prostate cancer. PSA test will show positive 80%of the time for inflicted men and 5% for healthy men. What is the probability that a man who tested positive does have prostate cancer?????

45. Bayes ’ Theorem P(C) = 0.60 P(C’) = 0.40 P(Test positive|Cancer ) = P(T|C) = 0.80 P(Test positive|No Cancer)= P(T|C’) = 0.05 Calculate P(C|T) ????

46. Bayes ’ Theorem P(C|T) = P(T|C) * P(C) / P(T) P(T) = P(C and T) + P(C’ and T) = P(C) * P(T|C) + P(C’) * P(T|C’) = 0.6*0.8 + 0.4*0.05 = o.48 + 0.02 = 0.50 P(C|T) = 0.8 * 0.6 / 0.5 = 0.96 Chance of prostate cancer if tested positive is 96%

47. Binomial Probability Law Probability of exactly ‘r’ successes in ‘n’ independent trials is given by P(r) = n! / r!(n-r)! * p r q (n-r) where p = prob. of success in each trial q = prob. of failure in each trial

48. Binomial Probability Law Trials are identical i.e. all the repetitions are performed under identical conditions. Trials are independent i.e. outcome of one trial does not affect the outcome of another trial. Each trial results in two mutually exclusive outcomes , success (p) and failure (q). Prob. of ‘p’ and ‘q’ remains constant for each trial.

49. Binomial Probability Law e.g. In a district survey, %age of NVD = 68% i.e. P(V) %age of del. conducted in health inst.=93% i.e P(H) If a sample of 10 del. cases are randomly selected, find the prob. that 6 cases were having NVD. And find the prob. of 6 cases delivered in health inst.

50. Binomial Probability Law p= 0.68 P(V) q= 0.32 P(r) = n! / r!(n-r)! * p r q (n-r) P(6) = 10! / 6! (10-6)! * 0.68 6 0.32 4 = 0.2177

51. p= 0.93 P(H) q= 0.07 P(r) = n! / r!(n-r)! * p r q (n-r) P(6) = 10! / 6! (10-6)! * 0.93 6 0.07 4 = 0.0033 Binomial Probability Law

52. Further readings…. Mahajan’s Methods in Biostatistics Handbook of statistics : Dr. Basnnar Basic and Clinical Biostatistics : Dawson and Trapp S.P.Gupta’s textbook of statistics

53. Thank You