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If P(B) = 3|5, P(A|B) = 1|2 and P(A ∪ B) = 4|5, then P(A ∪ B)′ + P(A′ ∪ B) = A. 1|5 B. 4|5 C. 1|2 D. 1

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If P(B) = 3|5, P(A|B) = 1|2 and P(A ∪ B) = 4|5, then P(A ∪ B)′ + P(A′ ∪ B) = A. 1|5 B. 4|5 C. 1|2 D. 1 Select the correct answer from above options

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Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) [Additive Law of Probability] ∴ P(A ∪ B)’ = P[A’ ∩ B’] = 1 – P(A ∪ B) = 1 - 4/5 = 1/5 and P(A’ ∪ B) = 1 – P(A’ ∩ B) = 1 – [P(A) – P(A ∩ B)] = 1 Hence, the correct option is D

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