A closed circuit surrounding the origin and passing through $(m,0)$ has diameter at least $m$. By the one-point estimate used in part (b2), its probability is at most $Ce^{-cm}$. Therefore
$$
\mathbb P(\exists m\geq n\text{ on such a circuit})
\leq\sum_{m\geq n}Ce^{-cm}
\leq C'e^{-c'n}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 209 / 2 / b3 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-209/2/b3/solution