Question

You are trying to hit a target composed of three points, if the probability that you hit the 1st point is 0.15, the probability that you hit the second is

Answer 1

Split it into disjoint events, and then add up their probabilities:

The probability that the target is hit by the 1st shooter only is (0.3)×(1−0.4)×(1−0.5)(0.3)×(1−0.4)×(1−0.5)

The probability that the target is hit by the 2nd shooter only is (1−0.3)×(0.4)×(1−0.5)(1−0.3)×(0.4)×(1−0.5)

The probability that the target is hit by the 3rd shooter only is (1−0.3)×(1−0.4)×(0.5)(1−0.3)×(1−0.4)×(0.5)

The overall probability is therefore:

(0.3)×(1−0.4)×(1−0.5)+(1−0.3)×(0.4)×(1−0.5)+(1−0.3)×(1−0.4)×(0.5)=0.44