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.