解A.B.C相互独立,则P(AB)=P(A)P(B),P(AC)=P(A)P(C),P(BC)=P(B)P(C),P(ABC)=P(A)P(B)P(C)∴P(AB)P(C)=P(A)P(B)P(C)=P(ABC),AB与C独立;P((A-B)C)=P(AC-BC)=P(AC)-P(ABC)=P(A)P(C)-P(A)P(B)P(C)=(P(A)-P(A)P(B))P(C)=P(A-B)P(C),A-B与C独立。
解A.B.C相互独立,则P(AB)=P(A)P(B),P(AC)=P(A)P(C),P(BC)=P(B)P(C),P(ABC)=P(A)P(B)P(C)∴P(AB)P(C)=P(A)P(B)P(C)=P(ABC),AB与C独立;P((A-B)C)=P(AC-BC)=P(AC)-P(ABC)=P(A)P(C)-P(A)P(B)P(C)=(P(A)-P(A)P(B))P(C)=P(A-B)P(C),A-B与C独立。
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