大学概率论：设A,B相互独立，P（A）=0.7，P(A∪B)=0.8，求P(B)，P(AB),，P(A-B)。请给过程，谢谢~

Question

如题所述

Answer 1

我再来提供个不用楼上那种多个式子加减的方法吧~
P(A+B)=P(A)+P(B)-P(AB)
因为A，B相互独立，所以
=P(A)+P(B)-P(A)*P(B)
带入，0.8=0.7+P(B)-0.7P(B)
解得P(B)=1/3
P(AB)=P(A)*P(B)
带入，得到P(AB)=(7/10)*(1/3)=7/30
P(A-B)=P(A)-P(AB)
带入，得到P(A-B)=(7/10)-(7/30)=7/15

Answer 2

A,B相互独立 => P(AB) = P(A)*P(B) => P(AB) = 0.7P(B) (1)
P(A) = P(A-B) + P(AB) => 0.7 = P(A-B) + P(AB) (2)
P(A∪B)= P(A-B)+P(B) => 0.8= P(A-B)+P(B) (3)
(1)(2)(3) =>
P(B) =1/3
P(AB) = 7/30
P(A-B) = 7/15本回答被提问者采纳