Conditional probability involving 3 people shooting a target

Question

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?

Answer 1

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 $$0.9/0.999=100/111=0.900900\dots$$

Answer 2

According to the total probability rule,

P(Gary hits)=1\over{3}(.027)+2\over{3}(.243)+3\over{3}(.729)=.9.

P(Gary hits|\geq 1 hit)=P("Gary hits" intersect "\geq 1 hit")/P(\geq 1 hit)=.9/.999=100/111