# Sheldon Ross 10: Exercise 3.66 Question: The opponents of soccer team A are of two types: either they are a class 1 or a class 2 team. The number of goals team A scores against a class-i opponent is a Poisson random variable with mean λi, where λ1 = 2, λ2 = 3. This weekend the team has two games against teams they are not very familiar with. Assuming that the first team they play is a class 1 team with probability 0.6  and the second is, independently of the class of the first team, a class 1 team with probability 0.3, determine the following

1. expected number of goals team A will score this weekend.
2. the probability that team A will score a total of five goals.

Analytical Solution

1. E[ X ] = E[Xw/Team 1+Xw/Team 2] = E[Xw/Team 1] + E[Xw/Team 2]
• ⇒ E[ X ] = ∑ E[Xw/Team 1 | class = i ]  P{ class=i } +  ∑ E[Xw/Team 2 | class = i ]  P{ class=i }
• ⇒ E[ X ] = (2*0.6 + 3*0.4) + (2*0.3 + 3*0.7) = 5.1
2. Part 2 can be observed from the simulation below

Simulation Solution

The simulation pretty much explains the distribution of the outcomes upon multiple iterations. This chart shows the simulation of about 10000 games. Code

```Module[{teams = {{0.6, 0.4} -> {2, 3}, {0.3, 0.7} -> {2, 3}}, goals,
games, outcomes, barChart, distributionChart,
iterations = Power[10, 4]},
goals :=
Plus @@ (RandomVariate[PoissonDistribution[RandomChoice[#]]] & /@
teams);
outcomes = Table[goals, iterations];
games = KeySort[Counts[outcomes]];

barChart =
BarChart[games, ImageSize -> 788,
Frame -> {{True, True}, {False, True}},
FrameTicks -> {Range@Max@Keys@games, Automatic},
ChartLabels ->
Placed[{N[Values@games / iterations], Values@games}, {Above,
Below}], AspectRatio -> 0.5, ChartStyle -> LightBlue,
PlotLabel -> Style["Distribution of the outcomes", Black, 13]];
distributionChart =
DistributionChart[{{outcomes, Mean /@ Partition[outcomes, 10]}},
ImageSize -> 788, ChartElementFunction -> "PointDensity",
ChartLegends ->
Placed[{"Actual Outcomes", "Batched Means (Batch size = 10)"},
Below], AspectRatio -> 0.5,
PlotLabel ->
Style["Distribution Chart of the outcomes and the batch means",
13, Black]];
Column[{barChart, "\n\n", distributionChart}]
]```

End of the post 😉

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