Question

The probabilities that three men hit a target are 1/6, 1/4 and 1/3. Each man shoots once at the target. What is the probability that exactly one of them hits the target?

• A. 11/72
• B. 21/72
• C. 31/72
• D. 3/4

Answer 1

The correct option is C 31/72

Let the probability of A hitting the target $=\frac{1}{6}$

Let the probability of B hitting the target $=\frac{1}{4}$

Let the probability of C hitting the target $=\frac{1}{3}$

The probability that exactly one of them hits the target (it means only one will hit the target while others two cannot)

$\frac{1}{6}\times \frac{3}{4}\times \frac{2}{3}+\frac{5}{6}\times \frac{3}{4}\times \frac{1}{3}+\frac{5}{6}\times \frac{1}{4}\times \frac{2}{3}$

$=\frac{31}{72}$