 Solution

The specified transition law is a binary erasure channel. For an input bit with $\mathrm{Pr}(X=1)=a$, 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 $X$. ThusMaximizing the mutual information over the input probability distribution, $h_2(a)\le 1$ with equality at $a=1/2$, givesThis includes $q=0$, a noiseless binary channel, and $q=1$, a completely erased channel with zero channel capacity.