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The probabilities of three events A,B,C ...

The probabilities of three events A,B,C are such that P(A)=0.3, P(B)=0.04, P( C) = 0.8 `P(AcapB)=0.08`
`P(AcapC)=0.28,P(AcapBcapC)=0.09` and
`P(AcupBcupC)ge0.75.` Show that `P(BcapC)` lies in the interval [0.23,0.48]

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