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 \b[the <probability> is $4/10=2/5$]. Equivalently, conditioning on the first capture gives $(4/10)(3/9)+(6/10)(4/9)=2/5$.

Solution

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

 Solution (source code)