= Solution
There are two possible messages and three channel uses, giving
$$
\boxed{R=\frac{\log_2 2}{3}=\frac13\ \text{bit per use}.}
$$
Conditional on either transmitted message, a decoding error occurs exactly when at least two of the three <bits> are flipped. The <discrete memoryless channel> makes these flips <independent>, each with <probability> $\varepsilon=1/10$. Their count therefore has a <binomial distribution> with parameters $3,\varepsilon$. <Majority-vote decoding of a repetition code> gives
$$
\begin{aligned}
\mathbb P(\text{error}\mid W=a)
&=\binom32\varepsilon^2(1-\varepsilon)+\varepsilon^3\\
&=3\left(\frac1{10}\right)^2\frac9{10}+\left(\frac1{10}\right)^3
=\frac7{250}=0.028.
\end{aligned}
$$
Interchanging the two <bit> values leaves the <binary symmetric channel> and majority decision unchanged, so \b[$\mathbb P(\text{error}\mid W=b)=7/250$ as well]. Both the maximum conditional error and the average error, for any message prior, are therefore $0.028$. The <probability> of correct decoding for each message is $243/250=0.972$. These are finite-block error <probabilities>; a channel-capacity statement alone would not determine them.
Back to article page
