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Prove that
(i) `P(A)=P(AcapB)+P(AcapbarB)`
(ii) `P(AcupB)=P(AcapB)+P(AcapbarB)+P(barAcapB)`

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1 Answers

(i) `because P(A)=P(AcapB)+P(AcapbarB)`
`therefore RHS=P(AcapB)+P(AcapbarB)`
`-P(A)cdotP(B)+P(A)cdotP(barB)`
`=P(A)[P(B)+P(barB)]`
`=P(A)[P(B)+1-P(B)]` `[becauseP(barB)=1-P(B)]`
=P(A)=LHS Hence proved
(ii) `becauseP(AcupB)=P(AcapB)+P(AcapbarB)+P(barAcapB)`
`therefore RHS=P(A)cdotP(B)+P(A)cdotP(barB)+P(barA)cdotP(B)`
`=P(A)cdotP(B)+P(A)cdot[1-P(B)]+[1-P(A)]P(B)`
`=P(A)cdotP(B)+P(A)-P(A)-P(A)cdotP(B)+P(B)-P(A)cdotP(B)`
`=P(A)+P(B)-P(A)cdotP(B)`
`=P(A)+P(B)-P(AcapB)`
`=P(AcupB)=LHS`

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