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 $\frac{3}{4}$ respectively. Find the probability that exactly 2 of them hit the target.

$1$

$\frac{1}{2}$

$\frac{5}{12}$

$\frac{1}{12}$

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Correct answer is C

$p(P) = \frac{1}{2}, p(P') = \frac{1}{2}$

$p(Q) = \frac{1}{3}, p(Q') = \frac{2}{3}$

$p(R) = \frac{3}{4}, p(R') = \frac{1}{4}$

p(exactly two hit the target) = p(P and Q and R') + p(P and R and Q') + p(Q and R and P')

= $(\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}) + (\frac{1}{2} \times \frac{3}{4} \times \frac{2}{3}) + (\frac{1}{3} \times \frac{3}{4} \times \frac{1}{2})$

= $\frac{1}{24} + \frac{6}{24} + \frac{3}{24}$

= $\frac{5}{12}$

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