= Solution
Use the binary <Hamming code> $C=\ker H\subset\mathbb F_2^{15}$, where the $j$th column $h_j$ of the $4\times15$ <parity-check matrix> $H$ is the four-digit binary expansion of $j$, for $1\le j\le15$. Thus the columns are precisely the nonzero vectors of $\mathbb F_2^4$. They include the four coordinate vectors, so $\operatorname{rank}H=4$ and $\dim C=15-4=11$. Its <code rate> is therefore $\boxed{11/15}$.
For a received word $y$, compute its <syndrome> $s=Hy$. If $s=0$, report no error. Otherwise identify the unique column $h_j=s$ and flip bit $j$. Indeed, if $y=c+e_j$ with $c\in C$, then $Hy=Hc+He_j=h_j$, so this procedure recovers $c$ exactly. No column is zero and no two columns coincide; hence no nonzero <codeword> has <Hamming weight> one or two. Some three columns sum to zero, for example those indexed by $1,2,3$, so the <minimum distance> is exactly three. Thus \b[every single-bit error is corrected]. The assertion requires at most one error; arbitrary multiple errors need not be detected by this procedure.
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