BITSAT 2012 Paper

Since, P (exactly one of A, B occurs) = q.
∴ 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) +[1- P(A ∪ B)]
= (1 – p + q) + (1 – p) = 2 – 2p + q