At distance $r$, inclusion requires $L\geq4\pi r^2f_{\min}$. Thus
$$
\boxed{
\mathbb P(I=1\mid r)
=1-\Phi\left(\frac{4\pi r^2f_{\min}-L_0}{\sigma_L}\right)
=\Phi\left(\frac{L_0-4\pi r^2f_{\min}}{\sigma_L}\right)}.
$$
It equals $1/2$ when
$$
r=\sqrt{\frac{L_0}{4\pi f_{\min}}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 219 / 1 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-219/1/e/solution