 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-2/14/c/iii/solution

Under independence, there are two disjoint ways to score two goals in total. One game can contain both, with probability

10 (e^{- λ} \frac{λ ^{2}}{2}) (e^{- λ})^{9} = 5 λ^{2} e^{- 10 λ} .

Alternatively, two different games can contain one goal each, with probability

(2 10) (λ e^{- λ})^{2} (e^{- λ})^{8} = 45 λ^{2} e^{- 10 λ} .

Adding gives

P (two goals in the season) = 50 λ^{2} e^{- 10 λ} .

10 λ

, whose probability at two is

e^{- 10 λ} (10 λ)^{2} /2!

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