# Three Men Shoot and Only One of Them Hits the Target. Find the Probability that it was the First Man

**Problem**

The probabilities that three men hit a target are 1/6, 1/4, and 1/3, respectively. Each shoot once at the target. If only one of them hits the target, find the probability that it was the first man.

**Answer Key**

6/31

**Solution**

Probability that one of them hits the target
$P = \dfrac{P_1}{P_1 + P_2 + P_3} = \dfrac{1/12}{31/72}$

- The 1
^{st}man hit the target

$P_1 = \dfrac{1}{6} \times \dfrac{3}{4} \times \dfrac{2}{3} = \dfrac{1}{12}$ - The 2
^{nd}man hit the target

$P_2 = \dfrac{5}{6} \times \dfrac{1}{4} \times \dfrac{2}{3} = \dfrac{5}{36}$ - The 3
^{rd}man hit the target

$P_3 = \dfrac{5}{6} \times \dfrac{3}{4} \times \dfrac{1}{3} = \dfrac{5}{24}$

$P_1 + P_2 + P_3 = \dfrac{31}{72}$

Probability that it was the first man

$P = \dfrac{6}{31}$ ← *answer*