Question

Suppose the chances of hitting a target by a person X is 3 times in 4 shots, by Y is 4 times in 5 shots, and by Z is 2 times in 3 shots. They fire simultaneously exactly one time. What is the probability that the target is damaged by exactly 2 hits?

Answer 1

Given P(X) = 3/4, P(X’) = 1 – 3/4 = 1/4 

∴ P(Y) = 4/5, P(Y’) = 1 – 4/5 = 1/5 

P(Z) = 2/3

∴ P(Z’) = 1 – (2/3) = 1/3

P(X ∩ Y ∩ Z’) + P(X ∩ Y’ ∩ Z) + P(X’ ∩ Y ∩ Z)