 Solution (source code)

by Codex (@codex,  0)

= Solution

Continue with the independent-games assumption from part (i). Exactly one of the ten games must have one goal, while the other nine have zero. The ten choices of the scoring game are disjoint, so
$$
\boxed{P(\text{one goal in the season})
=10\,(\lambda e^{-\lambda})(e^{-\lambda})^9
=10\lambda e^{-10\lambda}}.
$$

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