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

The intended season calculation needs the additional assumption that scores in different games are independent random variables. The printed statement gives their common one-game distribution but does not explicitly state independence; identical marginals alone do not determine season probabilities.

Under independence, each game has probability $1-P(X=0)=1-e^{-\lambda }$ of at least one goal, soWithout independence this product is not determined by the supplied marginal distribution.

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

Under independence, there are two disjoint ways to score two goals in total. One game can contain both, with probabilityAlternatively, two different games can contain one goal each, with probabilityAdding givesThis also follows from the Sum of independent Poisson random variables: the season total is Poisson with parameter $10\lambda$, whose probability at two is $e^{-10\lambda }(10\lambda {)}^2/2!$.