Bayes Theorem October 6, 20051Political Science 551Bayes TheoremBuses in MadisonBuses and Pedestrians95% of buses are Madison Metro (MM)80% of the time you correctly identify typeP(B) = Probability of you identifying as MMP(A) = Probability of bus being MMP(A|B) = Probability that it was MM giventhat you say it was MMP(A∩B) = Probability that you say it was MM and it was a MMBayes Theorem October 6, 20052Combined ProbabilitiesConditional JointP(A|B) P(A∩B)P(Ac|B) P(Ac∩B)P(A|Bc)P(A∩Bc)P(Ac|Bc)P(Ac∩Bc)P(Ac∩B)P(A∩Bc)P(Ac∩Bc)P(A∩B)AAcBBcMultiplication Rule()())|( ABPAPBAP ⋅=I())|( BAPBP ⋅=Probability of saying it was MM:() ( )()cABPABPBP II +=SummaryP(Ac∩B)=.01P(B|Ac)=.2P(Ac)=.05Not MMP(A∩B)=.76P(B|A)=.8P(A)=.95Was MMJoint P that is MMP(iddingas MM)P(buses that are MM)()()())|()|(BAPBPABPAPBAP⋅=⋅=IBayes Theorem October 6, 20053Probability MM given ID’d as MM()()()()()()BAPBAPBAPBPBAPBAPcIIII+==|() ( )()BAPBAPBPcII +=Since:Bayes Theorem()()()()BAPBAPBAPBAPcIII+=|()()()()()()cccABPAPBAPABPAPBAP||==IINote:Therefore:()() ( )() ( )()( )ccABPAPABPAPABPAPBAP||||⋅+⋅⋅=P(MM given you say it’s MM)P(A) = .95 P of any bus being MMP(Ac)=.05 P of any bus not MMP(B|A) = .80 P of saying MM if it was MMP(B|Ac)=.20 P of saying MM if not MM()() ( )() ( )()( )987.01.76.76.20.05.80.95.80.95.||||=+=⋅+⋅⋅=⋅+⋅⋅=ccABPAPABPAPABPAPBAPBayes Theorem October 6, 20054Venn Diagram: Is MMP(Ac∩B)=.01P(A∩Bc)=.19P(Ac∩Bc)=.04P(A∩B)=.76AAcBBcP(not MM given you say not MM)P(A) = .95 P of any bus being MMP(Ac)=.05 P of any bus not MMP(Bc|Ac)= .80 P of saying not MM if not MMP(Bc|A)=.20 P of saying MM if not MM()()( )()( )()()ABPAPABPAPABPAPBAPccccccccc||||⋅+⋅⋅=()174.19.04.04.20.95.80.05.80.05.| =+=⋅+⋅⋅=ccBAPVenn Diagram: Is Not MMP(Ac∩B)=.01P(A∩Bc)=.19P(Ac∩Bc)=.04P(A∩B)=.76AAcBBcBayes Theorem October 6, 20055Tree TrimmingMMID-MMID-MMMMc.95.05.8.76.01ID-MMc.8.04.2ID-MMc.19.2Diagnosis Sexual AbuseProbability of a child being diagnosed as having been abusedP(B)Probability of child who was not abused being incorrectly diagnosed as abusedP(B|Ac) = .10Probability of child who was abused being correctly diagnosed as abusedP(B|A) = .90Probability of any random child not having been abusedP(Ac) = .90Probability of any random child having been abusedP(A) = .10Probability of having been abused given positive diagnosisP(A|B) = ?()() ( )() ( )()( )50.90.10.10.90.10.90.|||| =⋅+⋅⋅=⋅+⋅⋅=ccABPAPABPAPABPAPBAP95% Accurate DiagnosisProbability of a child being diagnosed as having been abusedP(B)Probability of child who was not abused being incorrectly diagnosed as abusedP(B|Ac) = .05Probability of child who was abused being correctly diagnosed as abusedP(B|A) = .95Probability of any random child not having been abusedP(Ac) = .90Probability of any random child having been abusedP(A) = .10Probability of having been abused given positive diagnosisP(A|B) = ?()() ( )() ( )()( )68.90.05.10.95.10.95.|||| =⋅+⋅⋅=⋅+⋅⋅=ccABPAPABPAPABPAPBAPBayes Theorem October 6, 2005620% Abuse RateProbability of a child being diagnosed as having been abusedP(B)Probability of child who was not abused being incorrectly diagnosed as abusedP(B|Ac) = .05Probability of child who was abused being correctly diagnosed as abusedP(B|A) = .95Probability of any random child not having been abusedP(Ac) = .80Probability of any random child having been abusedP(A) = .20Probability of having been abused given positive diagnosisP(A|B) = ?()() ( )() ( )()( )81.80.05.20.95.20.95.|||| =⋅+⋅⋅=⋅+⋅⋅=ccABPAPABPAPABPAPBAPDifferent Error RatesProbability of a child being diagnosed as having been abusedP(B)Probability of child who was not abused being incorrectly diagnosed as abused (80% accurate)P(B|Ac) = .20Probability of child who was abused being correctly diagnosed as abused (10% error rate)P(B|A) = .90Probability of any random child not having been abusedP(Ac) = .90Probability of any random child having been abusedP(A) = .10Probability of having been abused given positive diagnosisP(A|B) = ?()() ( )() ( )()( )33.90.20.10.90.10.90.|||| =⋅+⋅⋅=⋅+⋅⋅=ccABPAPABPAPABPAPBAP50% False Positive RateProbability of a child being diagnosed as having been abusedP(B)Probability of child who was not abused being incorrectly diagnosed as abused (50% accurate)P(B|Ac) = .50Probability of child who was abused being correctly diagnosed as abused (10% error rate)P(B|A) = .90Probability of any random child not having been abusedP(Ac) = .90Probability of any random child having been abusedP(A) = .10Probability of having been abused given positive diagnosisP(A|B) = ?()() ( )() ( )()( )167.90.50.10.90.10.90.|||| =⋅+⋅⋅=⋅+⋅⋅=ccABPAPABPAPABPAPBAPBayes Theorem October 6, 20057Sampling ExampleP(M) = .75 P(Mc)=.25P(A|M) = .12 P(Ac|M) = .88P(A|Mc) = .20 P(Ac|Mc) = .8064.05.09.09.20.25.12.75.12.75.=+=⋅+⋅⋅=()()( )()( )()( )ccMAPMPMAPMPMAPMPAMP||||⋅+⋅⋅=As Venn DiagramMMcAcAAc∩Mc=P(Mc)·P(Ac|Mc)=.25·.80 = .20A∩Mc=P(Mc)·P(A|Mc)=.25·.20 = .05Ac∩M=P(M)·P(Ac|M)=.75·.88 = .66A∩M=P(M)·P(A|M)=.75·.12 = .09As Tree TrimmingMAAMc.75.25.12.09.05Ac.8.20.2Ac.19.88Bayes Theorem October 6, 20058Woman Instead of ManP(M) = .75 P(Mc)=.25P(A|M) = .12 P(Ac|M) = .88P(A|Mc) = .20 P(Ac|Mc) = .80()AMP |136.20.25.12.25.12.25.−==⋅+⋅⋅=()()( )()( )()( )cccccMAPMPMAPMPMAPMPAMP||||⋅+⋅⋅=Another Diagnosis ProblemIn a clinic, 10% of patients have disease A, 20% have disease B, and 70% have neither.Of those with A, 90% have headaches (H); of those with B, 50% have headaches; of those with neither, 5% have headaches?What is the probability of having a headache?Probabilities in Headache QuestionP(A) = .10 P(Ac) = .90 P(B) = .20P(A∩B) = 0 P(AUB) =.30 P(Ac∩Bc) = .70P(H|A) = .90 P(H|B)=.50 P(H|Ac∩Bc) = .05() ( )() ( )()()()225.035.100.090.05.70.20.50.10.90.|||=++=⋅+⋅+⋅=++=ccccBAPBAHPBPBHPAPAHPHP II()()()()()()40.225.10.90.|| =⋅=⋅==HPAPAHPHPHAPHAPIProbability of a headache:Probability of having A if has headache:Bayes Theorem October 6, 20059More Headache ProbabilitiesP(A) = .10 P(Ac) = .90 P(B) = .20P(A∩B) = 0 P(AUB) =.30 P(Ac∩Bc) = .70P(H|A) = .90 P(H|B)=.50 P(H|Ac∩Bc) = .05()()()()()()444.225.20.50|| =⋅=⋅==HPBPBHPHPHBPHBPIProbability of having B if has headache:Probability of having neither if has headache:()()()()156.225.70.05||