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

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**Solution:**

1. Probability of A hitting the target, $P(A) = \frac{2}{3}$

2. Probability of B hitting the target, $P(B) = \frac{3}{4}$

3. Probability of A missing the target, $P(A') = 1 - P(A) = 1 - \frac{2}{3} = \frac{1}{3}$

4. Probability of B missing the target, $P(B') = 1 - P(B) = 1 - \frac{3}{4} = \frac{1}{4}$

5. Probability that exactly one hits the target means either:
   - A hits and B misses, or
   - B hits and A misses.

6. Calculate the probability:
   $$
   P(	ext{exactly one hits}) = P(A) 	imes P(B') + P(B) 	imes P(A')
   = \frac{2}{3} 	imes \frac{1}{4} + \frac{3}{4} 	imes \frac{1}{3}
   = \frac{2}{12} + \frac{3}{12} = \frac{5}{12}
   $$

**Answer:** $\frac{5}{12}$

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