 Solution

Let the probability of the first input be $t$. The two equal output probabilities are $(2+t)/6$, and the third is $s=(1-t)/3$. Writing $h_2$ for the binary entropy function, the output entropy is $h_2(s)+(1-s)$, while the conditional entropy is $t+(1-t){\log }_{2}3$. ThereforeThe derivative with respect to $s$ isand the second derivative is $-1/(\ln 2s(1-s))<0$ in the interior. Thus the unique stationary point is the global maximum. It occurs at $s=4/31$, soThe capacity itself is $\boxed{C={\log }_2(31/27)}$. The maximization is over $t$, not over arbitrary output distributions; $s=(1-t)/3$ keeps the feasible outputs explicit.