 Solution

A syndrome determines a phase-error set only up to multiplication by $\overline{X}=Z_1\cdots Z_n$: an error of weight $w$ and its complement of weight $n-w$ have the same error syndrome. Their probability ratio isFor $p<1/2$, maximum likelihood therefore chooses the representative of smaller weight. This is exactly majority-vote decoding of a repetition code: correction succeeds when fewer than half the qubits are flipped, with a random tie-break at $w=n/2$ when $n$ is even.

If $W\sim \mathrm{Bin}(n,p)$, then $\mathbb{E}W=np$ and $\mathrm{sd}(W)=\sqrt{np(1-p)}$. For every fixed $p<1/2$, the distance from the mean to the decision boundary in standard deviations isThus the logical failure probability $\mathrm{Pr}(W>n/2)$ tends to zero, in fact exponentially by a Chernoff bound. At $p=1/2$ the two representatives are equiprobable and decoding cannot improve with $n$. Hence