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The probabilities of three events A,B,C are P(A) = 0.6, P(B) = 0.4 and P( C) = 0.5.If P(A`cup`B) = 0.8, P(A `cap`C) = 0.3 , P(A`cap B cap`C) = 0.2 and P(A`cup Bcup C`)`ge`0.85, show that, 0.2`leP(BcapC) le` 0.35.

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CHHAYA PUBLICATION-PROBABILITY-SAMPLE QUESTIONS FOR COMPETITIVE EXAMS
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