= Solution
There are three possible count pairs conditional on two trees. Their <probabilities>, from the two independent <binomial distributions>, are
$$
p_{20}=10x^2(1-x)^3(1-y)^5,\qquad
p_{02}=10y^2(1-y)^3(1-x)^5,
$$
$$
p_{11}=25xy(1-x)^4(1-y)^4.
$$
The first two correspond to matching kinds, so the <conditional probability> is
$$
\boxed{\frac{p_{20}+p_{02}}{p_{20}+p_{02}+p_{11}}}.
$$
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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