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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2021

Paper 2

Urn $r$ has $n-r$ blue balls among $n-1$ balls. Averaging over the uniformly chosen urn,
$$
\mathbb P(\text{first blue})
=\frac1n\sum_{r=1}^n\frac{n-r}{n-1}
=\boxed{\frac12}.
$$
The probability that both removed balls are blue is
$$
\begin{aligned}
\mathbb P(B_1\cap B_2)
&=\frac1n\sum_{r=1}^n
\frac{(n-r)(n-r-1)}{(n-1)(n-2)}\\
&=\frac13.
\end{aligned}
$$
Therefore
$$
\boxed{
\mathbb P(B_2\mid B_1)
=\frac{\mathbb P(B_1\cap B_2)}{\mathbb P(B_1)}
=\frac23}.
$$
The first blue draw makes urns with more blue balls more likely, which explains why the conditional probability exceeds one half.


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 2 / 9E / a / ii / Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 2 / 9E / a / ii / Solution

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