= Solution
For an incident particle from the left, take $E>0$ and $E>V_0$, and define
$$
k=\frac{\sqrt{2mE}}\hbar,\qquad q=\frac{\sqrt{2m(E-V_0)}}\hbar.
$$
The <Time-independent Schrödinger equation> gives an incident plus reflected <plane wave> on the left and an outgoing transmitted <plane wave> on the right:
$$
\psi(x)=e^{ikx}+re^{-ikx}\quad(x<0),\qquad \psi(x)=te^{iqx}\quad(x>0).
$$
<Continuity> of the <wavefunction> and its <derivative> at a finite potential jump yields $1+r=t$ and $k(1-r)=qt$. Hence $r=(k-q)/(k+q)$ and $t=2k/(k+q)$. The incident and reflected <probability currents> have equal wave-number magnitude, so
$$
\boxed{P=|r|^2=\left(\frac{\sqrt E-\sqrt{E-V_0}}{\sqrt E+\sqrt{E-V_0}}\right)^2}.
$$
As a check, the transmitted current fraction is $(q/k)|t|^2=4kq/(k+q)^2$, and the two probabilities sum to one.
For a positive step, $E\downarrow V_0$ makes $q\downarrow0$, so \b[$P\to1$]. For a negative step with fixed positive incident energy, $V_0\to-\infty$ makes $q\to\infty$, so again \b[$P\to1$]. The latter is quantum reflection from an increasingly large wave-number mismatch, even though the step is downward.
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