= Solution
Under <independence>, there are two disjoint ways to score two goals in total. One game can contain both, with <probability>
$$
10\left(e^{-\lambda}\frac{\lambda^2}{2}\right)(e^{-\lambda})^9
=5\lambda^2e^{-10\lambda}.
$$
Alternatively, two different games can contain one goal each, with <probability>
$$
\binom{10}{2}(\lambda e^{-\lambda})^2(e^{-\lambda})^8
=45\lambda^2e^{-10\lambda}.
$$
Adding gives
$$
\boxed{P(\text{two goals in the season})=50\lambda^2e^{-10\lambda}}.
$$
This also follows from the <Sum of independent Poisson random variables>: the season total is Poisson with parameter $10\lambda$, whose <probability> at two is $e^{-10\lambda}(10\lambda)^2/2!$.
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