Three men, P, Q and R aim at a target, the probabilities that P, Q and R hit the target are \(\frac{1}{2}\), \(\frac{1}{3}\) and \(\frac{3}{4}\) respectivel...

Answer Details

To find the probability that exactly 2 of them hit the target, we need to consider all possible combinations of two men hitting the target while the other misses. There are three such combinations: PQ, QR, and PR. For the combination PQ, the probability that P hits the target and Q misses is \(\frac{1}{2} \times \frac{2}{3} = \frac{1}{3}\). The probability that P misses and Q hits is \(\frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\). Therefore, the probability that exactly PQ hit the target is \(\frac{1}{3} + \frac{1}{6} = \frac{1}{2}\). Similarly, for QR, the probability that Q hits and R misses is \(\frac{1}{3} \times \frac{1}{4} = \frac{1}{12}\). The probability that Q misses and R hits is \(\frac{2}{3} \times \frac{3}{4} = \frac{1}{2}\). Therefore, the probability that exactly QR hit the target is \(\frac{1}{12} + \frac{1}{2} = \frac{7}{12}\). Finally, for PR, the probability that P hits and R misses is \(\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\). The probability that P misses and R hits is \(\frac{1}{2} \times \frac{3}{4} = \frac{3}{8}\). Therefore, the probability that exactly PR hit the target is \(\frac{1}{8} + \frac{3}{8} = \frac{1}{2}\). The total probability that exactly 2 of them hit the target is the sum of the probabilities for each combination, which is \(\frac{1}{2} + \frac{7}{12} + \frac{1}{2} = \frac{5}{6}\). Therefore, the answer is (C) \(\frac{5}{12}\).