Question

一道证明题!求数学高手!

A.B.C是三个随机事件,证明|P(AB)-P(BC)|<=1-P(AC)

Answer 1

根据概率的加法公式有P(AB)+P(AC)-P(ABC)=P(A(BUC))<=1
也即P(AB)-P(ABC)<=1-P(AC)
而P(ABC)<=P(BC)，故P(AB)-P(BC)<=P(AB)-P(ABC)<=1-P(AC)
同理可证P(BC)-P(AB)<=1-P(AC)
故有|P(AB)-P(BC)|<=1-P(AC)

Answer 2

嗯 我看看