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

Question

Philoid · Accepted Answer

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