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?

Correct option is D

Given:

Probability of A hitting the target =$$\frac{2}{3}$$​​

Probability of B hitting the target = $$\frac{3}{4}$$​​

We need to find the probability that exactly one of them hits the target.

Solution:

The probability of exactly one player hitting the target means either:

A hits and B misses, or

B hits and A misses.

Using probability rules:

Probability of A missing = $$1 - \frac{2}{3} = \frac{1}{3}$$​​

Probability of B missing = 1 $$- \frac{3}{4} = \frac{1}{4}$$​​

Thus, the probability of exactly one hitting = 

(A hits, B misses) + (B hits, A misses) = $$\left( \frac{2}{3} \times \frac{1}{4} \right) + \left( \frac{3}{4} \times \frac{1}{3} \right)$$​ 

​$$= \left( \frac{2}{12} \right) + \left( \frac{3}{12} \right) = \frac{5}{12}$$​​