At distance $r_s$, detection requires $L_s\geq4\pi r_s^2f_{\min}$. Hence
$$
\mathbb P(I_s=1\mid r_s)
=1-\Phi\left(\frac{4\pi r_s^2f_{\min}-L_0}{\sigma_L}\right)
=\Phi\left(\frac{L_0-4\pi r_s^2f_{\min}}{\sigma_L}\right).
$$
The probability is $1/2$ when its normal quantile is zero, namely at
$$
r_s=\sqrt{\frac{L_0}{4\pi f_{\min}}}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 219 / 1 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-219/1/d/solution