验证:设A,B,C是任意三事件,则P(AB)+P(AC)-P(BC)<=P(A),(提示:ABUAC属于A)

2025-05-11 00:07:44
推荐回答(1个)
回答1:

已知,A、B、C是任意事件,那么他们相互独立。则
P(AB)+P(AC)-P(BC)
= P(A)[P(B)+P(C)]-P(B)P(C); 相互独立,故P(AB)=P(A)P(B)
<=P(A)[P(B)+P(C)-P(B)P(C)]; P(B)P(C)>=P(A)P(B)P(C)
= P(A)[P(B)(1-P(C))+P(C)]; 合并同类项
<=P(A)[1-P(C)+P(C)]; P(B)(1-P(C))<=1-P(C)
= P(A)
原命题得证。