Sec1.6 Bays' Theorem

Sec1.6 Bays’ TheoremExample 1.6-1Bowl B 1, contains two red and four white chips;B 2 ,bowl contains one red and two white chips; andbowl contains five red and four white chips.B 3The probabilities for selecting the bowls are given byP( B 1 ) =13P( B 2 ) =1612, , and P(B 3 ) = .1

Ex1.6-1(1)The experiment consists of selecting a bowl withthese probabilities and then drawing a chip at randomfrom that bowl. Find the probability, P(R), of event R,drawing a red chip.P(R)= P(B I R)+ P(B I R)+ P(B I=P(B11) P(R|B1)+2P(B23) P(R | B2)R)+P(B3) P(R|B3)2

(2)Suppose that the outcome of the experiment is a red chip,but we do not know from which bowl it was drawn. Wecompute the conditional probability that the chip was drawnfrom bowl B1 namely, P( B 1| R).P(B1I R)P( B1| R)=P(R)=P(B1) P(R|B1) +P(B1I R)P(B ) P(R | B22) +P(B3) P(R|B3)=( 1 )( 2 )3 6( 1 )( 2 ) + ( 1 )( 1 ) + ( 1 )( 5 )366329=28≠P(B 1).3

Similarly,P( 1 )( 1 )6( 4 )3P(B )1| )2 ∩ RB2 R = = = ≠ P(BP(R)89( 2)and( 1 )( 5 )2( 4 )9P(B )5( | )3 ∩ RP B3 R = = = ≠ P(B3)P(R)894

Prior probabilities ()&Posterior probabilities ()The original probabilities P( B i ), i = 1,2,3, are calledprior probabilities, and,the conditional probabilities P( B i | R),i = 1,2,3 arecalled the posterior probabilities.5

GeneralizationDef:Let { B1 , B2,...,Bm}be a collection of subsets of thesample space S. We say that { B1 , B2,...,Bm}ispartition of S if the following hold.(1) P( B i) > 0, i = 1,..., m.(2) S = B1U B2U ... U(3) B I B = φ,i ≠ j .ijB m6

Remark(1)Let A be an event, thenA = ( B1 I A)U ( B2I A)U...U ( Bm I A).andm∑P( A)= P(B Ii=1iA)=m∑i=1P(Bi) P(A |Bi)(2) If P(A) > 0, we have thatP(BkI A)P( Bk | A)= , k = 1,2,..., m.P(A)7