= Solution
The <probability current> of the incident wave is $j_{\rm in}=\hbar k_1|A|^2/m$, and that of the reflected wave is $j_{\rm ref}=-\hbar k_1|B|^2/m$. The <quantum reflection probability> is the positive ratio of reflected to incident flux magnitudes:
$$
\boxed{R=\frac{|j_{\rm ref}|}{j_{\rm in}}=\frac{|B|^2}{|A|^2}
=\left(\frac{k_1-k_2}{k_1+k_2}\right)^2.}
$$
For comparison, the transmitted <probability current> is $j_{\rm tr}=\hbar k_2|C|^2/m$, so the <transmission probability> is
$$
T=\frac{k_2}{k_1}\left|\frac CA\right|^2
=\frac{4k_1k_2}{(k_1+k_2)^2}.
$$
These satisfy $R+T=1$, as required by conservation of <probability current> for a real, nonabsorbing <potential energy>. In particular, $|C/A|^2$ alone is not the transmitted flux fraction when $k_2\ne k_1$.
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