One of the basic tenets of quantum mechanics is that
$\begin{align*}\frac{\int_{V} |\psi|^2}{\int_{\mbox{all space}} |\psi|^2}\end{align*}$
is the probability that the particle is in the volume $V$ (assuming that $\psi$ is normalizable). In $1$ dimension, this becomes

$\begin{align*}\frac{\int_a^b |\psi|^2 dx }{\int_{-\infty}^{\infty} |\psi|^2 dx}\end{align*}$
In our case,
$\begin{align*}\int_a^b |\psi|^2 dx = \int_2^4 |\psi|^2 dx = 4 + 9 = 13\end{align*}$
and
$\begin{align*}\int_{-\infty}^{\infty} |\psi|^2 dx = 1 + 1 + 4 + 9 + 1 = 16\end{align*}$
Therefore, the answer is $\boxed{13/16}$ . So, answer (E) is correct.

Solution to 1996 Problem 17