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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.
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Solution
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.
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