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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)=0.4, P(C)=0.8,P(A nn B)=0.09, P(Ann C) = 0.28, P(A nn B nn C) 0.08 and P(Auu B uu C) leq 0.75`.Show that `P(B nn C)` lies in the interval `[0.21, 0.46]`.

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The correct Answer is:
`0.16`

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