PROBABILITYConceptual FrameworkExample: Outcomes of 500 interviews to study opinions of residents of a certain city about building a prison.Area ofCityFor Against EmpiricalProbabilityTotalA 100 25 0.80 125B 115 10 0.92 125D 50 75 0.40 125E 35 90 0.28 125Total 300 200 500Def: Sample Space—set of all possible outcomesDef: Event—Outcome of interestProbability of an Event—Relative frequency of the event over a large number of trialsExample: Probability of an adult male age 40-49 years having a heart attack in next 6 years is 0.003Properties:1. Let E = some event0 ≤ Pr(E) ≤ 12. If A and B are two events that cannot occur at the same time,Pr(A or B) = Pr(A) + Pr(B)Therefore, events A and B are Mutually ExclusiveExample: Let A = Type A BloodLet B = Type O BloodPr(A) = 0.3Pr(B) = 0.4Pr(A or B) = 0.3 + 0.4 = 0.7Def: Union—A B = Probability A or B or both occursDef: Intersection—A ∩ B = Probability A and B both occurDef: Complement--A= Probability A does not occurPr(A)1)APr( Def: Independent—Two events are independent if Pr(A∩B) = Pr(A) * Pr(B)Def: Dependent—Two events are dependent ifPr(A∩B)  Pr(A) * Pr(B)Example: Two doctors test all patients in a VD clinic for syphilisA+ = {doctor A makes a positive diagnosis}B+ = {doctor B makes a positive diagnosis}Pr(A+) = 0.1Pr(B+) = 0.17Pr(A+∩B+) = 0.08Are A+, B+ independent?Pr(A+∩B+) = 0.08  Pr(A+) * Pr(B+) = 0.1 * 0.17 = 0.017Multiplication Law:If A1, A2, …, Ak are mutually independent events, thenPr (A1∩A2∩…∩Ak) = Pr(A1)*Pr(A2)*…Pr(Ak)Def: If A, B are mutually exclusive events,Pr(AB) = Pr(A) + Pr(B)Addition Law:If A, B are any two events,Pr(AB) = Pr(A) + Pr(B) – Pr(A∩B)STD Example: If either doctor A or doctor B makes a positive diagnosis, a person will need further evaluation. What is the probability that a patient will need further evaluation?Pr(A+B+) = Pr(A+) + Pr(B+) – Pr(A+∩B+)=0.1 + 0.17 – 0.08=0.19Note: If A, B are Mutually ExclusivePr(AB) = Pr(A) + Pr(B)Addition Rule for Independent EventsIf A, B are independent,Pr(AB) = Pr(A) + Pr(B) – Pr(A∩B)= Pr(A) + Pr(B) – Pr(A) * Pr(B)= Pr(A) + Pr(B)*[1-Pr(A)]Example: For two genetically unrelated individuals, supposePr(A) = 0.4 and Pr(B) = 0.4Pr(AB) = Pr(A) + Pr(B)*[1-Pr(A)]=0.4 +0.4*[1-0.4]=0.4 + 0.24=0.64Addition Law can be expanded. Let A, B, C be three events.Pr(ABC) = Pr(A) + Pr(B) + Pr(C) – Pr(A∩B) – Pr(A∩C) – Pr(B∩C) + Pr(A∩B∩C)Conditional ProbabilityLet A be a diagnostic test, say PSA for Prostate CancerA = {PSA+}B= {PC}What is the probability of prostate cancer, given the PSA test is positive?Def: Pr(B|A) = Pr(A∩B) / Pr(A)Note: If A, B are independent, thenPr(A∩B) = Pr(A) * Pr(B)Therefore, Pr(B|A) = [Pr(A) * Pr(B)] / Pr(A)= Pr(B)Also:Pr(B))]APr(B)]/Pr(*)A[Pr()A|Pr(B Def: If A, B are independent,)A|Pr(BPr(B)A)|Pr(B Def: If A, B are dependent,)A|Pr(BPr(B)A)|Pr(B  and Pr(A∩B)  Pr(A) * Pr(B)Def: Relative Risk = )A|A)/Pr(B|Pr(BIf A, B independent, RR = 1STD Example:Pr(B+|A+) = [Pr(B+∩A+)] / Pr(A+) = 0.08/0.1 = 0.8Pr(B+|A-) = [Pr(B+∩A-) / Pr(A-) = [Pr(B+∩A-)] / 0.9Need Pr(B+∩A-)Well, Pr(B+) = Pr (B+∩A+) + Pr(B+∩A-)Therefore, 0.17 = 0.08 + Pr(B+∩A-)or Pr(B+∩A-) = 0.09Therefore, Pr(B+|A-) = 0.09 / 0.9 = 0.1If doctor A says the patient does not have syphilis, doctor B will contradict 10% of the time.RR = [Pr(B+|A+)] / [Pr(B+|A-)]= 0.8 / 0.1 = 8Doctor B is 8 times as likely to diagnose as positivewhen doctor A diagnoses as positive than when doctor A diagnoses as negative.Therefore, in general:)APr(BA)Pr(BPr(B)  But,)A|Pr(B*)APr()APr(BandA)|Pr(B*Pr(A)A)Pr(BTherefore, for any two events, A and B)APr(*)A|Pr(BPr(A)*A)|Pr(BPr(B) Total Probability Rule:Let A1, A2, …, Ak be a set of mutually exclusive and exhaustive events.Pr(B) = a weighted average of the conditional probabilities of