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 2/3 and 3/4 , respectively. What is the probability that exactly one of them hit the target?

Let A and B be two players who are playing the game to hit…

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 $\frac 2 3$ and $\frac 3 4$ , respectively. What is the probability that exactly one of them hit the target?

Show answer & explanation

Correct answer: C

Explanation

The correct answer is option 3 i.e. $\frac{5}{12}$ .

P(A) = 2/3

P(B) = 3/4

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

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

There are two scenarios:

A hits and B misses (A and B') and A misses and B hits (A' and B)

P(A and B') = P(A) × P(B') = (2/3) × (1/4) = 2/12 = 1/6

P(A' and B) = P(A') × P(B) = (1/3) × (3/4) = 3/12 = 1/4

P(exactly one hits) = P(A and B') + P(A' and B) = (1/6) + (1/4) = 5/12

Therefore, the probability that exactly one of them hits the target is 5/12.

SSC CGL 2025 PYQ — Question 9