Engineering
Mathematics
Conditional Probability
Question
A, B, C are three events for which P(A) = 0.6, P(B) = 0.4, P(C) = 0.5, P(AUB) = 0.8, P(A⋂C) = 0.3 and P(A⋂B⋂C) = 0.2. If P(AUBUC) ≥ 0.85 then the interval of values of P(B⋂C) is
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Solution
We know that P(AUB) = P(A) + P(B) – P(A⋂B)
Hence, P(A⋂B) = 0.6 + 0.4 – 0.8 = 0.2
We also know that
P(AUBUC) = P(A) + P(B) + P(C) – P(A⋂B) – P(A⋂C) – P(B⋂C) + P(A⋂B⋂C)
Now, 1 ≥ P(AUBUC) ≥ 0.85
Thus, 1 ≥ 0.6 + 0.4 + 0.5 – 0.2 – P(BUC) – 0.3 + 0.2 ≥ 0.85
⇒ 0.35 ≥ P(B⋂C) ≥ 0.2
Hence, (a) is correct.
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