 Solution

Each of the three input values has two possible outputs, so their union has at most six elements. The preceding union formula also gives this bound. Choose noise values $0,3$, whose two translated input alphabets are disjoint. With columns ordered $0,1,2,3,4,5$ the transition matrix isThus $|\mathcal{Y}|=6$. Every output determines its input, explicitly $X=Y\mathrm{mod}3$, so $H(X\mid Y)=0$. Consequently $I(X;Y)=H(X)\le {\log }_{2}3$, with equality for a uniform input. ThereforeThe extra uncertainty of which noise value occurred does not reduce information about the input: the two possible outputs for each input remain distinguishable from those of every other input.