Stat 321 – Lecture 8Last Time - Conditional ProbabilityLast Time - IndependenceExample 3: ELISA TestTable SolutionExample 1: Governator VotesSlide 7Slide 8Law of Total ProbabilitySlide 10General ResultExample 2: Randomized ResponseSlide 13Slide 14For FridayStat 321 – Lecture 8Law of Total Probability (2.4)Last Time - Conditional ProbabilityP(A|B) = probability of event A occurring given that event B has occurredTo calculate:restrict sample space to outcomes in B (if equally likely)P(A|B) = P(AB)/P(B)PropertiesP(A|B) vs. P(B|A)P(A'|B) = 1-P(A|B)Multiplication rule: P(AB)=P(B)P(A|B)=P(A)P(B|A)Last Time - IndependenceEvents A and B are independent if P(A|B) = P(A) P(B|A) = P(B)Multiplication Rule for independent eventsP(AB) = P(A|B)P(B) = P(A)P(B) = P(B|A)P(A) = P(A)P(B)To “check” independenceSee if P(A|B) = P(A) or if P(B|A) = P(B) or if P(AB) = P(A)P(B)ABExample 3: ELISA TestGivenIf has AIDS, test says AIDS 97.7% of timeIf doesn’t have AIDS, test says no AIDS 92.6% of timeAbout .5% of people have AIDSGiven that someone has tested positive, what is the probability they have AIDS?Table Solution Positive test Negative test TotalCarries AIDS virusDoes not carry AIDSTotal1,000,000Probability is only .062 that a person who tests positive actually has AIDS!5000995,0004885 1159213707363078515921485Of those with positive test, proportion with AIDS: 4885/78515 = .0622Most physicians think the probability is about 60-70% - confusion of the inverseP(AIDS | positive test) ≠ P(positive test | AIDS)Example 1: Governator VotesCNN.com52% of the white votes17% of the black votes31% of the Hispanic votes37% of the votes from other races. So overall (.52+.17+.31+.37)/4 = .3425 => 34% of votes?Example 1: Governator VotesGiven:.52 = P(A|W).17 = P(A|B).31 = P(A|H).37 = P(A|O).70 = P(W)…Example 1: Governator VotesWhite Black Hisp. Other TotalArnoldA'Total.70 .06 .18 .06 1.00700 60 180 60 1000.52(700)=364P(AW)=.364.0102 .0558 .0222=P(A).4522P(A|W)P(W) + P(A|B)P(B) + P(A|H)P(H) + P(A|O)P(O) = P(A)Law of Total ProbabilityP(B)=P(B|Ai)P(Ai) where {Ai}=partition.70 .18 .06White H B O.06Arnold1234AWAWAHABP(A|W)P(W)P(A|W)P(W)P(A|H)P(H)P(A|B)P(B)Law of Total ProbabilityWHBOA|WNot A|WA|HNot A|HA|BNot A|BA|ONot A|O.70.18.06.06.52.31.17.37P(AW)=(.52)(.70)=.364P(AH)=(.31)(.18)=.0558P(AB)=(.17)(.06)=.0102P(AO)=(.37)(.06)=.0222.364+.0558+.0102+.0222 =.4522P(A)?P(A'W)=(.48)(.70)=.336General ResultLaw of Total ProbabilityP(B)=P(B|A1)P(A1)+P(B|A2)P(A2)+…= (.7)(.52)+(.18)(.31)+(.17)(.06)+.37(.06)Example 2: Randomized ResponseTechnique for asking sensitive questionsRandomly decide which question respondents will answer: sensitive or boringWork backwards with probability rules to estimate proportions for sensitive questionExample 2: Randomized ResponseFlip fair coinHeads: answer sensitive questionTails: answer boring question=“does your home phone number end in even digit?”Determine proportion of “yeses”Define eventsY=“response is yes”S=“respondent answered sensitive question”Example 2: Randomized ResponseRespondents are ensured confidentialityCan still obtain estimate for P(Y|S)For FridayLab 3 dueFinish reading Ch. 2 (Section 2.4!)Will not begin a new lab but have a regular class session (in lab)Review handout available