= Solution
For a one-dimensional wavefunction, the <probability current> is
$$
j=\frac{\hbar}{2mi}
\left(\psi^*\frac{d\psi}{dx}
-\psi\frac{d\psi^*}{dx}\right).
$$
A plane wave $De^{ikx}$ carries current $(\hbar k/m)|D|^2$ to the right, while $De^{-ikx}$ carries the negative of this current. Thus the asymptotic terms are interpreted as an incident wave of amplitude $A$, a reflected wave of amplitude $B$, and a transmitted wave of amplitude $C$. Their currents are
$$
j_{\rm in}=\frac{\hbar k}{m}|A|^2,
\qquad
j_{\rm ref}=-\frac{\hbar k}{m}|B|^2,
\qquad
j_{\rm tr}=\frac{\hbar k}{m}|C|^2.
$$
Current conservation for a real potential gives $j_{\rm in}+j_{\rm ref}=j_{\rm tr}$, and therefore
$$
\boxed{P_{\rm ref}=\frac{|B|^2}{|A|^2},
\qquad
P_{\rm tr}=\frac{|C|^2}{|A|^2}},
$$
with $P_{\rm ref}+P_{\rm tr}=1$.
A normalizable <bound state> has $E<0$. Writing
$$
\kappa=\frac{\sqrt{-2mE}}{\hbar}>0,
$$
its decaying asymptotic behaviour is
$$
\boxed{\psi(x)\sim D_-e^{\kappa x}\quad(x\to-\infty),
\qquad
\psi(x)\sim D_+e^{-\kappa x}\quad(x\to+\infty)}.
$$
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