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 P (A′) + P (B′) =?

It is given that

The probability of simultaneous occurrence of at least one of two events A and B is p

P(A ∪ B) = p

Due to the probability that just one of A, B occurs = q

So we get

P(A ∪ B) - P(A ∩ B) = q

p - P(A ∩ B) = q

P(A ∩ B) = p - q

We know that,

P(A ∪ B) = P(A) + P(B) - P(A ∩ B)

p = P(A) + P(B) - (p - q)

P(A) + P(B) = p + (p - q)

P(A) + P(B) = 2p - q

Similarly

P (A′) + P (B′) = 1 - P(A) + 1 - P(B)

= 2 - [P(A) + P(B)]

= 2 - (2p - q)

= 2 - 2p + q.

Summary:

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 P (A′) + P (B′) =?

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