Question

概率论第五题

Answer 1

A B C相互独立
P(AB)=P(A)P(B)=0.4*0.5=0.2
P(A|AB)=1
P(ABUC)=P(AB)+P(C)-P(ABC)=0.2+0.5-0.2*0.5=0.6
P((A-C)|(ABUC)=P((A-C)(ABUC))/P(ABUC)
=(P((A-C)AB)+P((A-C)C)/P(ABUC) (A-C)与C独立

=(P((A-C)AB)+0)/P(ABUC)
=P(((A-C)A)B)/P(ABUC)
=P((A-C)B)/P(ABUC)
=P((A-AC)B)/P(ABUC)
=P(A-AC)*P(B)/P(ABUC)
=(P(A)-P(AC))*P(B)/P(ABUC)
=(0.4-0.4*0.5)*0.5/0.6
=1/6

或者用你的分解方式算
P(A|ABUC)
=P(A(ABUC))/P(ABUC)
=P(A(AB)UAC)/P(ABUC)
=P(ABUAC)/P(ABUC)
=P(A(BUC))/P(ABUC)
=P(A)P(BUC)/P(ABUC)
=0.4*(0.5+0.5-0.5*0.5)/0.6
=0.3/0.6=0.5
P(AC|ABUC)

=P(AC(ABUC))/P(ABUC)
=P(ACABUAC)/P(ABUC)
=P(ABCUAC)/P(ABUC)
=P(AC)/P(ABUC)
=0.4*0.5/0.6
=0.2/0.6
P(A-C|ABUC)=0.5-0.2/0.6=0.1/0.6=1/6