The probability of simultaneous occurrence of at least one of two even

1.	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» 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.

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»

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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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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