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

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

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

$P = \dfrac{6}{31}$   ←   answer