= Solution
For a finite step and equal masses, continuity of the wavefunction and its derivative gives
$$
A+B=C,\qquad k_1(A-B)=k_2C.
$$
Hence the reflected and transmitted amplitude ratios are
$$
r=\frac BA=\frac{k_1-k_2}{k_1+k_2},\qquad
t=\frac CA=\frac{2k_1}{k_1+k_2}.
$$
A travelling wave has <probability current> $j=(\hbar k/m)|A|^2$, so the flux coefficients are
$$
\boxed{R=|r|^2=\left(\frac{k_1-k_2}{k_1+k_2}\right)^2},\qquad
\boxed{T=\frac{k_2}{k_1}|t|^2=\frac{4k_1k_2}{(k_1+k_2)^2}}.
$$
They satisfy $R+T=1$. Using $|t|^2$ alone as the transmission probability would omit the different transmitted velocity.
For $E=2\,\mathrm{eV}$ and a downward step of $2\,\mathrm{eV}$, the transmitted kinetic energy is $4\,\mathrm{eV}$ and $k_2/k_1=\sqrt2$. Therefore
$$
\boxed{R=0.0294\;(2.94\%),\qquad T=0.9706\;(97.1\%).}
$$
The nonzero reflection is compatible with current conservation even though the potential drops.
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