| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Probability of ordered outcome between components |
| Difficulty | Standard +0.8 This question requires understanding of independent Poisson processes, conditional probability with enumeration of cases for part (i)(b), and normal approximation to Poisson for part (ii). The competitive scenario in (i)(b) requiring consideration of winning scores (3-0, 3-1, 3-2) adds problem-solving complexity beyond standard Poisson calculations, placing it moderately above average difficulty. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(e^{-2.1} \times \frac{2.1^3}{3!}\) alone | M1 | Allow any \(\lambda\). Allow sum of 3 or 4 relevant products, e.g. \(P(3,0)\) |
| \(= 0.189\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(e^{-1.2} \times \frac{1.2^3}{3!} \times e^{-0.9}\) | M1 | \(P(\text{Fem}=3) \times P(\text{Opp}=0)\) or \(P(\text{Fem}=2) \times P(\text{Opp}=1)\) |
| \(+ e^{-1.2} \times \frac{1.2^2}{2!} \times e^{-0.9} \times 0.9\) | M1 | \(P(3,0) + P(2,1)\) |
| \(= 0.115\) | A1 | [3] As final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(N(30, 30)\) | B1 | Seen or implied |
| \(\frac{34.5 - 30}{\sqrt{30}}\ (= 0.8216)\) | M1 | Standardising with their \(N(\lambda, \lambda)\) |
| \(1 - \Phi(0.822)\) | M1 | Allow with no or incorrect cc or no \(\sqrt{\phantom{x}}\) |
| \(= 0.206\) (3 sf) | A1 | [4] Area consistent with their working |
## Question 4(i)(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $e^{-2.1} \times \frac{2.1^3}{3!}$ alone | M1 | Allow any $\lambda$. Allow sum of 3 or 4 relevant products, e.g. $P(3,0)$ |
| $= 0.189$ | A1 | **[2]** |
## Question 4(i)(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $e^{-1.2} \times \frac{1.2^3}{3!} \times e^{-0.9}$ | M1 | $P(\text{Fem}=3) \times P(\text{Opp}=0)$ or $P(\text{Fem}=2) \times P(\text{Opp}=1)$ |
| $+ e^{-1.2} \times \frac{1.2^2}{2!} \times e^{-0.9} \times 0.9$ | M1 | $P(3,0) + P(2,1)$ |
| $= 0.115$ | A1 | **[3]** As final answer |
## Question 4(ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $N(30, 30)$ | B1 | Seen or implied |
| $\frac{34.5 - 30}{\sqrt{30}}\ (= 0.8216)$ | M1 | Standardising with their $N(\lambda, \lambda)$ |
| $1 - \Phi(0.822)$ | M1 | Allow with no or incorrect cc or no $\sqrt{\phantom{x}}$ |
| $= 0.206$ (3 sf) | A1 | **[4]** Area consistent with their working |
---4 Goals scored by Femchester United occur at random with a constant average of 1.2 goals per match. Goals scored against Femchester United occur independently and at random with a constant average of 0.9 goals per match.\\
(i) Find the probability that in a randomly chosen match involving Femchester,
\begin{enumerate}[label=(\alph*)]
\item a total of 3 goals are scored,
\item a total of 3 goals are scored and Femchester wins.
The manager promises the Femchester players a bonus if they score at least 35 goals in the next 25 matches.\\
(ii) Find the probability that the players receive the bonus.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2013 Q4 [9]}}