= Solution
Let $N_A$ be the independent avocado count. There are three disjoint ways to have fewer than two total trees: no trees, only one avocado, or only one lemon. Their <probabilities> give
$$
\boxed{\begin{aligned}
\mathbb P(N_A+N_L\geq2)=1
&-(1-x)^5(1-y)^5\\
&-5x(1-x)^4(1-y)^5\\
&-5y(1-y)^4(1-x)^5.
\end{aligned}}
$$
The sum is generally a <Poisson binomial distribution>, rather than a single <binomial distribution>, because the two success <probabilities> may differ.
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