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Show that for any events Eand F,

P(EEF)P(EF)

Hint: Compute P(EEF)by conditioning on whether F occurs.

Short Answer

Expert verified

The result is that P(EEF)is weighted average betweenP(EF)and1

Step by step solution

01

Prove

Demonstrate as all eventualities E,F

P(EEF)=P[EF(EF)]P(FEF)+PEFc(EF)PFcEF

If it's not clear, evaluate that equations via enlarging the correct hand side by a probability density statement.

F(EF)=F

Fc(EF)=EFc

PEFc(EF)=PEEFc=1

02

Weighted Average

That's also sufficient to ascertain the disparity, as

P(EEF)=P(EF)P(FEF)+1PFcEF

P(FEF)+PFcEF=1

P(EF)1

thus localid="1649659759663" P(EEF)is that the weighted combination of such constants localid="1649659766251" P(EF)and localid="1649659771667" 1, so just between that two integers,

localid="1649659823011" P(EF)P(EEF)1

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Let S = {1, 2, . . . , n} and suppose that A and B are, independently, equally likely to be any of the 2n subsets (including the null set and S itself) of S.

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