Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability resit / 8 / c / i / Solution

The two captures form a uniform ordered sample of distinct birds, that is, sampling without replacement. By exchangeability, each of the ten birds has the same chance of occupying the second position, so the probability is $4/10=2/5$. Equivalently, conditioning on the first capture gives $(4/10)(3/9)+(6/10)(4/9)=2/5$.