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

Answer 1

Since P (exactly one of A, B occurs) = q:

$P (A\cup B) -P ( A \cap B) = q\\p -P (A \cap B) = q\\P (A \cap B) = p-q\\1 - P (A' \cup B') = p - q\\P (A' \cup B') = 1 -p + q\\P (A') + P (B') -P (A' \cap B') = 1 - p + q\\P (A') + P (B') = (1 - p + q) + P (A' \cap B')\\= (1 - p + q) + (1 - P (A \cup B))\\= (1 -p + q) + (1 - p)\\= 2 -2p + q$