# Conditional probability involving 3 people shooting a target

Suppose that $3$ people will shoot at a target, and that each will hit it with probability $0.9$, independently of whether or not any of the others hit the target. If Gary is one of the $3$ shooters, what is the probability that Gary will hit the target given that at least one of the three shooters will hit it?

Attempted Solution:

P(Gary hits | $1$ hit) = $1\over3$

P(Gary hits | $2$ hit) = $2\over3$

P(Gary hits | $3$ hit) = $3\over3$

P($1$ hit) = $3\choose1$$(.9)(.1^2) = .027 P(2 hit) = 3\choose2$$(.9^2)(.1) = .243$

P($3$ hit) = $3\choose3$$(.9^3) = .729 Then P(Gary hits | \geq 1 hit) = 1\over{3}$$(.027)$+$2\over3$$(.243)+3\over3$$(.729)$ = $.9$

Is this a valid solution?

• The unconditional probability that Gary hits is $0.9$. The conditional probability given that someone hits it should be higher. So I think your solution is incorrect. Sep 24, 2017 at 2:07
• Yeah I was worried about that. I think it should be slightly higher than $0.9$ since you're guaranteed that it's not the case that everyone misses their shot.
– Remy
Sep 24, 2017 at 2:07

The probability that Gary hits and at least one hits is 0.9, since if Gary hits he counts as one of those who hit. The probability that at least one hits is the complement of the probability that nobody hits: $1-(1-0.9)^3=0.999$. Thus the probability that Gary hits given that at least one hits is $$0.9/0.999=100/111=0.900900\dots$$