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.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option C
Explanation:
\(p(P) = \frac{1}{2}, p(P') = \frac{1}{2}\)
\(p(Q) = \frac{1}{3}, p(Q') = \frac{2}{3}\)
\(p(R) = \frac{3}{4}, p(R') = \frac{1}{4}\)
p(exactly two hit the target) = p(P and Q and R') + p(P and R and Q') + p(Q and R and P')
= \((\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}) + (\frac{1}{2} \times \frac{3}{4} \times \frac{2}{3}) + (\frac{1}{3} \times \frac{3}{4} \times \frac{1}{2})\)
= \(\frac{1}{24} + \frac{6}{24} + \frac{3}{24}\)
= \(\frac{5}{12}\)
\(p(P) = \frac{1}{2}, p(P') = \frac{1}{2}\)
\(p(Q) = \frac{1}{3}, p(Q') = \frac{2}{3}\)
\(p(R) = \frac{3}{4}, p(R') = \frac{1}{4}\)
p(exactly two hit the target) = p(P and Q and R') + p(P and R and Q') + p(Q and R and P')
= \((\frac{1}{2} \times \frac{1}{3} \times \frac{1}{4}) + (\frac{1}{2} \times \frac{3}{4} \times \frac{2}{3}) + (\frac{1}{3} \times \frac{3}{4} \times \frac{1}{2})\)
= \(\frac{1}{24} + \frac{6}{24} + \frac{3}{24}\)
= \(\frac{5}{12}\)