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 $\mathbb{P}(\text{second red})=m/(m+n)$. One can verify this with the law of total probability, conditioning on the first card:Here $N\ge 2$ because both colour counts are positive.