 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 functionThusThe variance is $1/(\alpha +\beta {)}^2$, obtained from the standard first two exponential moments, $EM=1/(\alpha +\beta )$ and $E{M}^2=2/(\alpha +\beta {)}^2$.

For these competing exponential clocks, ties have probability zero. Integrating over Alice's firing time givesLet $H$ 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 givesThe Bayes' theorem therefore yields

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 $Z\sim \mathrm{Exp}(1)$ and set $A=Z/\alpha$, $B=Z/\beta$. These have the required marginals, but $M\sim \mathrm{Exp}(\max (\alpha ,\beta ))$; 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.