Question

The probability of simultaneous occurrence of at least one of two events A and B is p. If the probability that exactly one of A, B occurs is q, then prove that P(A′) + P(B′) = 2 – 2p + q.

Answer 1

Since P(exactly one of A, B occurs) = q .....(Given)

We get P(A ∪ B) – P(A ∩ B) = q

⇒ p – P(A ∩ B) = q

⇒ P(A ∩ B) = p – q

⇒ 1 – P(A′ ∪ B′) = p – q

⇒ P(A′ ∪ B′) = 1 – p + q

⇒ P(A′) + P(B′) – P(A′ ∩ B′) = 1 – p + q

⇒ P(A′) + P(B′) = (1 – p + q) + P(A′ ∩ B′)

= (1 – p + q) + (1 – P(A ∪ B))

= (1 – p + q) + (1 – p)

= 2 – 2p + q.