Represent $Y_n$ as a sum of $n$ independent $\operatorname{Poisson}(1)$ variables. The <central limit theorem> gives
$$
\boxed{\frac{Y_n-n}{\sqrt n}\xrightarrow{\mathrm d}N(0,1).}
$$
Since zero is a continuity point of the standard normal distribution,
$$
e^{-n}\sum_{k=0}^n\frac{n^k}{k!}
=\mathbb P(Y_n\leq n)\longrightarrow\boxed{\frac12}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 2 / 11F / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ia/paper-2/11f/b/solution