Let A and B be two players who are playing the game to hit the target. The probabilities of hitting the target by A and B is and , respectively. What is the probability that exactly one of them hit the target?

Question

Question

This question was previously asked in

SSC CGL 2024 Tier-II Official Paper-I (Held On: 18 Jan, 2025)

Answer 1

Given:

Probability of A hitting the target, P(A) = 2/3

Probability of B hitting the target, P(B) = 3/4

Formula used:

Probability of A not hitting the target, P(A') = 1 - P(A)

Probability of B not hitting the target, P(B') = 1 - P(B)

Probability of exactly one hitting = P(A)×P(B') + P(A')×P(B)

Calculations:

P(A') = 1 - 2/3 = 1/3

P(B') = 1 - 3/4 = 1/4

Probability of exactly one hitting = (2/3)×(1/4) + (1/3)×(3/4)

⇒ 2/12 + 3/12 = 5/12

∴ The probability that exactly one of them hits the target is 5/12.