= Solution
Use the intended model of <independent random variables> $A,B$ with <exponential distributions> of rates $\alpha,\beta>0$. For $M=\min(A,B)$, independence gives the <survival function>
$$
P(M>t)=P(A>t)P(B>t)=e^{-(\alpha+\beta)t},\qquad t\geq0.
$$
Thus
$$
\boxed{M\sim\operatorname{Exp}(\alpha+\beta),\qquad
f_M(t)=(\alpha+\beta)e^{-(\alpha+\beta)t}\ (t>0),\qquad
\operatorname{sd}(M)=\frac1{\alpha+\beta}.}
$$
The <variance> is $1/(\alpha+\beta)^2$, obtained from the standard first two exponential moments, $E M=1/(\alpha+\beta)$ and $E M^2=2/(\alpha+\beta)^2$.
For these <competing exponential clocks>, ties have <probability> zero. Integrating over Alice's firing time gives
$$
\boxed{P(A<B)=\int_0^\infty\alpha e^{-\alpha t}e^{-\beta t}\,dt
=\frac{\alpha}{\alpha+\beta}.}
$$
Let $H$ denote the event that the next shot hits, and use the stated success probability as $P(H\mid A<B)=1/2$. Bill's corresponding conditional success probability is one. The <law of total probability> gives
$$
P(H)=\frac12\frac{\alpha}{\alpha+\beta}+\frac{\beta}{\alpha+\beta}.
$$
The <Bayes' theorem> therefore yields
$$
\boxed{P(A<B\mid H)=\frac{\alpha}{\alpha+2\beta}.}
$$
The printed assumptions specify the two marginal <exponential distributions> but do not explicitly specify their independence. The answers above require that extra assumption; marginals alone do not determine the race. For a concrete counterexample, take one $Z\sim\operatorname{Exp}(1)$ and set $A=Z/\alpha$, $B=Z/\beta$. These have the required marginals, but $M\sim\operatorname{Exp}(\max(\alpha,\beta))$; if $\alpha>\beta$, Alice always fires first and $P(A<B\mid H)=1$. Thus the independence qualification changes actual answers, rather than merely their derivation.
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