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?