 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2021/discrete-mathematics-continuous-mathematics-and-probability/8/a/ii/solution

Every position in the random permutation has the same marginal colour probability, so $P (second red) = m / (m + n)$ . One can verify this with the law of total probability, conditioning on the first card:

\frac{m}{N} \frac{m - 1}{N - 1} + \frac{n}{N} \frac{m}{N - 1} = \frac{m ( m - 1 + n )}{N ( N - 1 )} = \frac{m}{N} .

Here

N \geq 2

because both colour counts are positive.

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