Answer
$0.091$
Work Step by Step
The normalized wavefunction $\psi_n(x)$ for an electron in an infinite, one-dimensional potential well with length $L$ along an $x$ axis are given by
$\psi_n(x)=\sqrt {\frac{2}{L}}\sin\Big(\frac{n\pi}{L}x\Big),\;\;\;\text{for}\;n=1,2,3....,$
where $n$ is the quantum number.
Now, the probability that the electron will be detected in any finite section of the well—say, between point $x_1$ and point $x_2$— is given by
$p(x)=\int_{x_1}^{x_2}\psi^2_n(x)dx$
or, $p(x)=\int_{x_1}^{x_2}{\frac{2}{L}}\sin^2\Big(\frac{n\pi}{L}x\Big)dx$
In our case, the electron trapped in the potential well of length $L$ is in its ground state $(n=1)$.
Therefore, the probability of finding electron in an interval $x=0.75L$ and $x=L$ is
$p(x)=\int_{0.75L}^{L}{\frac{2}{L}}\sin^2\Big(\frac{\pi}{L}x\Big)dx$
We can simplify the indicated integration by changing the variable from $x$ to the dimensionless variable $y$, where
$y=\frac{\pi x}{L}$
or, $dx=\frac{L}{\pi}dy$
With the change of the variable, we must also change the
integration limits: when $x =0.75L$, $y=\frac{3\pi}{4}$ and when $x=L$, $y=\pi$
Thus,
$p(x)=\int_{\frac{3\pi}{4}}^\pi{\frac{2}{L}}\frac{L}{\pi}\sin^2(y)dy$
$p(x)=\frac{2}{\pi}\Big[\frac{y}{2}-\frac{\sin2y}{4}\Big]_\frac{3\pi}{4}^\pi$
$p(x)\approx 0.091$
