作业帮 > 综合 > 作业

时间独立性的概率题A,B,C三个事件相互独立,P(A)=0.4,P(B)=0.5 ,P(C)=0.5,则P{A-C|AB

来源:学生作业帮 编辑:作业帮 分类:综合作业 时间:2024/06/06 12:57:10
时间独立性的概率题
A,B,C三个事件相互独立,P(A)=0.4,P(B)=0.5 ,P(C)=0.5,则P{A-C|AB+C}=?
理解为求P[(A-C)/(AB+C)]
有求
P[(A-C)/(AB+C)]=
={P[(A-C)(AB+C)]}/P(AB+C)
={P[(A(C逆))(AB+C)]}/P[(AB)+P(C)-P(ABC)]
={P[(A(C逆)AB+A(C逆)C}/P[(AB)+P(C)-P(ABC)]
={P[(A(C逆))AB+空]}/P[(AB)+P(C)-P(ABC)]
={P[(AB(C逆))]}/P[(AB)+P(C)-P(ABC)]
={P(A)*P(B)*P(C逆)}/P[(AB)+P(C)-P(ABC)]
=[0.4*0.5*(1-0.5)]/P[(AB)+P(C)-P(ABC)]
=0.1/[P(A)*P(B)+P(C)-P(A)*P(B)*P(C)]
=0.1/[0.4*0.5+0.5-0.4*0.5*0.5]
=0.1/0.6
=1/6.