 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-2/14/b/solution

There are three possible count pairs conditional on two trees. Their probabilities, from the two independent binomial distributions, are

p_{20} = 10 x^{2} (1 - x)^{3} (1 - y)^{5}, p_{02} = 10 y^{2} (1 - y)^{3} (1 - x)^{5},

p_{11} = 25 x y (1 - x)^{4} (1 - y)^{4} .

The first two correspond to matching kinds, so the conditional probability is

\frac{p _{20} + p _{02}}{p _{20} + p _{02} + p _{11}} .

This form remains valid at endpoints whenever the conditioning event has positive probability; if the denominator vanishes, the conditional probability is undefined. As a check, when

x = y \in (0, 1)

the result is

20/45 = 4/9

, the fraction of two-element subsets containing matching kinds.

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