= Solution
The specified transition law is a <binary erasure channel>. For an input <bit> with $\Pr(X=1)=a$, its <Shannon entropy> is the <binary entropy> $h_2(a)$. A non-erased output determines $X$, whereas the erasure symbol leaves its <probability distribution> unchanged because the erasure event is <independent> of $X$. Thus
$$
H(X\mid Y)=qH(X),\qquad I(X;Y)=(1-q)H(X)=(1-q)h_2(a).
$$
Maximizing the <mutual information> over the input <probability distribution>, $h_2(a)\leq1$ with equality at $a=1/2$, gives
$$
\boxed{C=1-q\quad\text{bits per channel use}.}
$$
This includes $q=0$, a noiseless binary <channel>, and $q=1$, a completely erased <channel> with zero <channel capacity>.
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