A six rounds pistol has three bullets and three blank shots. A die is thrown and each time the dice shows a multiple of two, the gun is fired. What is the probability that a man whose fate depends on a random throw of the dice will be shot?

The gun fires when the die shows a multiple of 2, which means outcomes 2, 4, or 6 out of 6 possible outcomes. So P(fire) = 3/6 = 1/2. Given the gun fires, it has 3 bullets and 3 blanks in 6 chambers, so P(shot | fire) = 3/6 = 1/2. Therefore P(shot) = P(fire) × P(shot | fire) = 1/2 × 1/2 = 1/4. This is a conditional probability problem where both events must occur: the die must trigger the gun (probability 1/2) and the chamber must contain a bullet (probability 1/2).