Slide 1Slide 2Slide 3Consider the following scenario…Consider the following scenario…Consider the following scenario…Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Reconsider the previous scenario…Reconsider the previous scenario…The Debate Continues…NEXT UP…CHAPTER 3Probability Theory•3.1 - Basic Definitions and Properties•3.2 - Conditional Probability and Independence•3.3 - Bayes’ Formula•3.4 - Applications (biomedical)P(B) P(B )Bayes’ FormulaBayes’ FormulaExactly how does one event A affect the probability of another event B?2AP(B)prior probabilityposterior probabilityP(B ∩ A)P(A)But what if the numerator and denominator are not explicitly given?B BCAP(A ⋂ B) P(A⋂ BC)P(A)ACP(AC ⋂ B) P(AC ⋂ BC)P(AC)P(B) P(BC) 1.0In general, for any two events A and B, there are 4 disjoint intersections:A BA ⋂ BA ⋂ BcAc ⋂ B “A only” “B only”Ac ⋂ Bc “Neither A nor B”“A and B”Probability TableI Ic( ) ( )P(A)= P A B + P A BRecall…) )=+P PP P P P(Overcast | Rain) (Rain)(Overcast | Rain) (Rain (Overcast | No Rain) (No RainConsider the following scenario…4•Weather Forecast: P(Rain) = .50 prior probability•Track Record:P(Overcast | Rain) = .65P(Overcast | No Rain) = .15•Want: P(Rain | Overcast) = ??? posterior probabilityBy def, P(Rain | Overcast) = ))I(Rain Overcast(OvercastPP ) ) )=+II IPP P(Rain Overcast(Rain Overcast (No Rain OvercastAlso recall…IP A B = P A | B P B( ) ( ) ( )Let A = Overcast B = Rain(.65)(.5)(.65)(.5) (.15)(1 .5)=+ -= 0.81250.8125 ))IPP(Rain Overcast(OvercastI Ic( ) ( )P(A)= P A B + P A B(.65)(.5)(.65)(.5) (.15)(1 .5)=+ -(.65)( )(.65)( ) (.15)(1 )=+ -.7.7 .7Consider the following scenario…5•Weather Forecast: P(Rain) = .70 prior probability•Track Record:P(Overcast | Rain) = .65P(Overcast | No Rain) = .15•Want: P(Rain | Overcast) = ??? posterior probabilityBy def, P(Rain | Overcast) = ) )=+P PP P P P(Overcast | Rain) (Rain)(Overcast | Rain) (Rain (Overcast | No Rain) (No Rain= 0.8125= 0.910.81250.91 ))IPP(Rain Overcast(Overcast ) ) ) =+II IPP P(Rain Overcast(Rain Overcast (No Rain OvercastAlso recall…IP A B = P A | B P B( ) ( ) ( )Let A = Overcast B = RainI Ic( ) ( )P(A)= P A B + P A B) )=+P PP P P P(Overcast | Rain) (Rain)(Overcast | Rain) (Rain (Overcast | No Rain) (No Rain(.65)(.5)(.65)(.5) (.15)(1 .5)=+ -(.65)( )(.65)( ) (.15)(1 )=+ -.7.7 .7Consider the following scenario…6•Weather Forecast: P(Rain) = .70 prior probability•Track Record:P(Overcast | Rain) = .65P(Overcast | No Rain) = .15•Want: P(Rain | Not Overcast) = ??? posterior probabilityBy def, P(Rain | Overcast) = = 0.8125= 0.91Exercise ))IPP(Rain Overcast(OvercastHint: Use probability tableAlso recall…IP A B = P A | B P B( ) ( ) ( )Let A = Overcast B = RainI Ic( ) ( )P(A)= P A B + P A B ) ) ) =+II IPP P(Rain Overcast(Rain Overcast (No Rain Overcast )) ) =+I IIP PP(Rain Overcast (No Rain(RaOin OvercasvercasttP(Rain) = .70Example: Vitamin B-complex deficiency among general population B1ThiamineB2RiboflavinB3 NiacinB4 No B deficiencyAssume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive.“10% of pop is B1-deficient (only), 20% is B2-deficient (only), and 30% is B3-deficient (only). The remaining 40% is not B-deficient.” Given:Example: Vitamin B-complex deficiency among general population B1ThiamineB2RiboflavinB3 NiacinB4 No B deficiencyAssume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive.P(B2) = .20 P(B3) = .30P(B1) = .10P(B4) = .40A = AlcoholicAc = Not AlcoholicGiven:P(A ∩ B1)To find these intersection probabilities, we need more information!Prior probs1.00P(A ∩ B2) P(A ∩ B3)P(A ∩ B4)P(Ac ∩ B1)P(Ac ∩ B2) P(Ac ∩ B3) P(Ac ∩ B4)Example: Vitamin B-complex deficiency among general population B1ThiamineB2RiboflavinB3 NiacinB4 No B deficiencyAssume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive.P(B2) = .20 P(B3) = .30P(B1) = .10P(B4) = .40Alsogiven…“Alcoholics comprise 35%, 30%, 25%, and 20% of the B1, B2, B3, B4 groups, respectively.” Given:A = AlcoholicAc = Not AlcoholicPrior probs1.00P(A ∩ B1) P(A ∩ B2) P(A ∩ B3)P(A ∩ B4)P(Ac ∩ B1)P(Ac ∩ B2) P(Ac ∩ B3) P(Ac ∩ B4)Example: Vitamin B-complex deficiency among general population B1ThiamineB2RiboflavinB3 NiacinB4 No B deficiencyAssume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive.P(B2) = .20 P(B3) = .30P(B1) = .10P(B4) = .40Alsogiven…P(A | B1) = .35 P(A | B2) = .30 P(A | B3) = .25 P(A | B4) = .20Prior probs1.00Given:A = AlcoholicAc = Not AlcoholicP(A ∩ B1) P(A ∩ B2) P(A ∩ B3)P(A ∩ B4)P(Ac ∩ B1)P(Ac ∩ B2) P(Ac ∩ B3) P(Ac ∩ B4)Example: Vitamin B-complex deficiency among general population B1ThiamineB2RiboflavinB3 NiacinB4 No B deficiencyAssume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive.Alsogiven…Prior probs1.00Given:A = AlcoholicAc = Not AlcoholicP(A ∩ B1) P(A ∩ B2) P(A ∩ B3)P(A ∩ B4)P(Ac ∩ B1)P(Ac ∩ B2) P(Ac ∩ B3) P(Ac ∩ B4)P(A | B1) = .35 P(A | B2) = .30 P(A | B3) = .25 P(A | B4) = .20P(B2) = .20 P(B3) = .30P(B1) = .10P(B4) = .40P(A ∩ B) = P(A | B) P(B) Recall:Example: Vitamin B-complex deficiency among general population B1ThiamineB2RiboflavinB3 NiacinB4 No B deficiencyAssume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive.Alsogiven…Prior probs1.00Given:A = AlcoholicAc = Not AlcoholicP(Ac ∩ B1)P(Ac ∩ B2) P(Ac ∩ B3) P(Ac ∩ B4)P(A ∩ B) = P(A | B) P(B) Recall: P(A | B1) = .35 P(A | B2) = .30 P(A | B3) = .25 P(A | B4) = .20.10  .35 .20  .30 .30  .25 .40  .20 .035 .060 .075 .080 P(B2) = .20 P(B3) = .30P(B1) = .10P(B4) = .40Example: Vitamin B-complex deficiency among general population B1ThiamineB2RiboflavinB3 NiacinB4 No B deficiencyAssume B1, B2, B3, B4 “partition” the population, i.e., they are disjoint and exhaustive.Prior probs1.00Given:A = AlcoholicAc = Not Alcoholic.10  .35 .20  .30 .30  .25 .40  .20 .035 .060 .075 .080 .065.065.140.140.225.225.320.320P(B2) = .20 P(B3) = .30P(B1) = .10P(B4) = .40P(B1 | A) = ? P(B2 | A) = ? P(B3 | A) = ?P(B4 | A) = ?Posterior