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

Solution

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

Probability that it was the first man

$P = \dfrac{6}{31}$ ← answer

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