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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2022

Paper 2

When $r$ red marbles remain, the next draw reduces their number with probability $r/n$. The waiting time $W_r$ for that reduction has a <geometric distribution> on $\{1,2,\ldots\}$ and therefore
$$
G_{W_r}(s)
=\frac{(r/n)s}{1-(1-r/n)s}
=\frac{rs}{n-(n-r)s}.
$$
The successive waiting times are independent, and
$$
T=W_n+W_{n-1}+\cdots+W_1.
$$
Using the product rule for probability-generating functions,
$$
\boxed{
G_T(s)=\prod_{r=1}^{n}
\frac{rs}{n-(n-r)s}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 10F / d / Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 10F / d / Solution

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Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 2 / 10F / d / Solution

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