Q66 of 108 Page 271

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

We have,


Now, We know that


P(A|B) × P(B) = P(A ∩ B)


[Property of Conditional Probability]




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)




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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