P(AC)=0.0P(2C)=0.5P(A3C)=020 P(A4C)=010P(A4C)=005 P(BA)+P(CA) P(A3B) P(AC) P(43)P(4) P(A3)=P(AB)+P(A4C)=0.65 P(B4) P(A3B)0.459 P(43)0.6513
( ) P B A3 + ( ) = P C A3 ( ) ( ) 3 3 P A P A B 1 ( ) ( ) 3 3 + = P A P A C ( ) ( ) P A3 = P A3 B ( ) 0.65 +P A3 C = ( ) P B A3 . 13 9 0.65 0.45 ( ) ( ) 3 3 = = = P A P A B ( ) 0.30 1 P AC = ( ) 0.35 2 P A C = ( ) 0.20 3 P A C = ( ) 0.10 4 P A C = ( ) 0.05 5 P A C =
解二设事件A为5:47到家事件B为乘 地铁回家.由 Bayes公式 P(BAy P(AB) P(B)P(AB) P(A) P(B)P(AB)+P(B)P(AB) 0.45 0.459 0.45+-×0.20 0.6513
解二 设事件 A为 5:47 到家;事件B为乘 地铁回家. P(B A) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) P B P A B P B P A B P B P A B P A P A B + = = . 13 9 0.65 0.45 0.20 2 1 0.45 2 1 0.45 2 1 == = + = 由 Bayes 公式