= Solution
The <channel capacity> is the supremum of reliable communication rates: a rate $R$ bits per use is achievable if block encoders and decoders using $n$ channel uses can transmit $M_n$ messages with error <probability> tending to zero and $\liminf n^{-1}\log_2M_n\ge R$.
A <discrete memoryless channel> has fixed transition <probabilities> $W(y|x)$ on finite input and output alphabets, and
$$
W(y_1,\ldots,y_n|x_1,\ldots,x_n)=\prod_{j=1}^nW(y_j|x_j).
$$
Its single-letter capacity formula is
$$
\boxed{C=\max_{P_X}I(X;Y)=\max_{P_X}\bigl(H(Y)-H(Y|X)\bigr),}
$$
with <logarithms> to base two. The <Shannon second coding theorem> identifies this maximum with the operational reliable rate.
For the <binary erasure channel>, use output symbols $0,1,e$, where $e$ is a recognizable erasure flag. Given input $x$, the output is $x$ with <probability> $1-p$ and $e$ with <probability> $p$. If $\Pr(X=1)=a$, the output <probabilities> are $(1-p)(1-a),(1-p)a,p$. Thus, writing $h_2$ for <binary entropy>,
$$
H(Y)=h_2(p)+(1-p)h_2(a),\qquad H(Y|X)=h_2(p),
$$
and $I(X;Y)=(1-p)h_2(a)$. Its maximum is attained by the uniform input $a=1/2$, giving
$$
\boxed{C=1-p\text{ bits per channel use},\qquad0\le p\le1.}
$$
In particular the noiseless and complete-erasure endpoints have capacities one and zero. The erasure flag reveals which positions were lost; it does not reveal the erased input values.
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