Pre-U Pre-U 9795/2 2016 June — Question 4 7 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2016
SessionJune
Marks7
TopicSum of Poisson processes
TypeMulti-period repeated application
DifficultyStandard +0.3 This question tests standard Poisson distribution properties: independence assumption for sum of Poissons, scaling parameter for multiple trials, and calculating specific probabilities. While it requires understanding of Poisson processes and conditional probability for part (ii)(b), the techniques are straightforward applications of textbook results with no novel problem-solving required. The multi-part structure and Further Maths context place it slightly above average difficulty.
Spec5.02i Poisson distribution: random events model5.02n Sum of Poisson variables: is Poisson

4 In a Football League match, the number of goals scored by the home team can be modelled by the distribution \(\mathrm { Po } ( 2.4 )\). The number of goals scored by the away team can be modelled by the distribution Po(1.8).
  1. State a necessary assumption for the total number of goals scored in one match to be modelled by the distribution \(\operatorname { Po } ( 4.2 )\).
  2. Assume now that this assumption holds.
    1. Write down an expression for the probability that the total number of goals scored in \(n\) randomly chosen games is less than 4 .
    2. Find the probability that the result of a randomly chosen game is either 0-0 or 1-1.

(i) Number of goals scored by home team is independent of number of goals scored by away team
- B1: Not just *goals* independent. Extras, including conditions already implied by given Poisson distributions: B0
[1]
(ii)(a) \(e^{-4.2n}\!\left(1+4.2n+\dfrac{(4.2n)^2}{2!}+\dfrac{(4.2n)^3}{3!}\right)\)
- M1: Po\((4.2n)\) implied
- A1: Correct \(\pm 1\) term
- A1: Fully correct expression, aef.
SR Po\((4.2)\): Fully correct formula B1
[3]
(b) \(e^{-2.4}e^{-1.8}(1+2.4\times1.8)\)
\(= \mathbf{0.0798}\)
- M1: Individual Poisson distributions multiplied
- A1: Correct expression \([= 0.0150 + 0.0647]\)
- A1: Answer, a.r.t. 0.080 [0.07977]
[3]
(i) Number of goals scored by home team is independent of number of goals scored by away team

- B1: Not just *goals* independent. Extras, including conditions already implied by given Poisson distributions: B0

**[1]**

(ii)(a) $e^{-4.2n}\!\left(1+4.2n+\dfrac{(4.2n)^2}{2!}+\dfrac{(4.2n)^3}{3!}\right)$

- M1: Po$(4.2n)$ implied
- A1: Correct $\pm 1$ term
- A1: Fully correct expression, aef.

SR Po$(4.2)$: Fully correct formula B1

**[3]**

(b) $e^{-2.4}e^{-1.8}(1+2.4\times1.8)$

$= \mathbf{0.0798}$

- M1: Individual Poisson distributions multiplied
- A1: Correct expression $[= 0.0150 + 0.0647]$
- A1: Answer, a.r.t. 0.080 [0.07977]

**[3]**
4 In a Football League match, the number of goals scored by the home team can be modelled by the distribution $\mathrm { Po } ( 2.4 )$. The number of goals scored by the away team can be modelled by the distribution Po(1.8).\\
(i) State a necessary assumption for the total number of goals scored in one match to be modelled by the distribution $\operatorname { Po } ( 4.2 )$.\\
(ii) Assume now that this assumption holds.
\begin{enumerate}[label=(\alph*)]
\item Write down an expression for the probability that the total number of goals scored in $n$ randomly chosen games is less than 4 .
\item Find the probability that the result of a randomly chosen game is either 0-0 or 1-1.
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q4 [7]}}