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2013-04-05
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1°证明: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∪B)=P(A)+P(B)-P(AB) P(A∪B)*P(C)=P(A)*P(C)+P(B)P(C)-P(AB)P(C)=P(AC)+P(BC)-P(ABC)=P[(A∪B)C]
P(A-B)*P(C)=P(A)P(C)-P(AB)P(C)=P(AC)-P(ABC)=P(AC-ABC)=P((A-B)C)
P(ABC)=P(A)*P(B)*P(C)=P(AB)*P(C)
P(A∪B)=P(A)+P(B)-P(AB) P(A∪B)*P(C)=P(A)*P(C)+P(B)P(C)-P(AB)P(C)=P(AC)+P(BC)-P(ABC)=P[(A∪B)C]
P(A-B)*P(C)=P(A)P(C)-P(AB)P(C)=P(AC)-P(ABC)=P(AC-ABC)=P((A-B)C)
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