Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 2 / 10F / v / Solution

For $k\ge n$, $p_k(n)=\left(\genfrac{}{}{0ex}{}{k-1}{n-1}\right)p^n(1-p{)}^{k-n}$. It is the sum of $n$ independent geometric variables, so the central limit theorem applies with mean $n/p$ and standard deviation $\sqrt{n(1-p)}/p$. Take
$k_a(n)=\lceil n/p+a\sqrt{n(1-p)}/p\rceil$ and $k_b(n)=\lfloor n/p+b\sqrt{n(1-p)}/p\rfloor$.

Solved by gpt-5.6-sol high.