| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sum of Poisson processes |
| Type | Conditional probability given total |
| Difficulty | Standard +0.3 This is a standard Poisson distribution question requiring (a) direct calculation from tables, (b) sum of independent Poissons (X+Y ~ Po(5.5)), and (c) conditional probability. Part (c) requires recognizing that given X+Y=5, X follows a binomial distribution, which is slightly above routine but well-covered in S2 syllabi. Overall slightly easier than average A-level due to straightforward application of known results. |
| Spec | 2.03d Calculate conditional probability: from first principles5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(e^{-3.1}\!\left(1 + 3.1 + \frac{3.1^2}{2!} + \frac{3.1^3}{3!}\right)\) or \(e^{-3.1}(1 + 3.1 + 4.805 + 4.965)\) or \(0.0450 + 0.1397 + 0.2165 + 0.2238\) | M1 | Condone one end error. Any \(\lambda\). Accept fully correct \(\Sigma\) notation. Expression must be seen |
| \(= 0.625\) (3sf) | A1 | Correct answer with no working scores SC B1 |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([\lambda] = 5.5\) | B1 | SOI |
| \(1 - e^{-5.5}(1 + 5.5 + \frac{5.5^2}{2!} + \frac{5.5^3}{3!} + \frac{5.5^4}{4!})\) | M1 | Condone one end error. Any \(\lambda\). Accept fully correct \(\Sigma\) notation. Expression must be seen. |
| \(= 0.642\) or \(0.643\) (3sf) | A1 | Correct answer with no working scores SC B1 B1 |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X=3) \times P(Y=2) = e^{-3.1} \times \frac{3.1^3}{3!} \times e^{-2.4} \times \frac{2.4^2}{2!}\) or \(0.223676 \times 0.261267 [= 0.05844]\) | M1 | Find P(3 in first half AND 2 in second half). Must see expression. |
| \(P(\text{total } 5) = e^{-5.5} \times \frac{5.5^5}{5!}\) or \(0.17140\) | M1 | Use of 5.5 to find P(5). |
| \(P(\text{exactly 3 in 1st half} \mid \text{total } 5) = \frac{P(\text{exactly 3 in 1st half and total 5})}{P(\text{total 5})}\) | M1 | Attempt at conditional probability; numerator = *their* 0.05844 and denominator = P(total 5). Note: \((\frac{3.1^3}{3!} \times \frac{2.4^2}{2!}) \div (\frac{5.5^5}{5!})\) scores M1 M1 M1. |
| \([= \frac{0.05844}{0.17140}] = 0.341\) (3sf) | A1 | |
| 4 |
## Question 5:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $e^{-3.1}\!\left(1 + 3.1 + \frac{3.1^2}{2!} + \frac{3.1^3}{3!}\right)$ or $e^{-3.1}(1 + 3.1 + 4.805 + 4.965)$ or $0.0450 + 0.1397 + 0.2165 + 0.2238$ | M1 | Condone one end error. Any $\lambda$. Accept fully correct $\Sigma$ notation. Expression must be seen |
| $= 0.625$ (3sf) | A1 | Correct answer with no working scores **SC B1** |
| **Total: 2** | | |
## Question 5(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[\lambda] = 5.5$ | B1 | SOI |
| $1 - e^{-5.5}(1 + 5.5 + \frac{5.5^2}{2!} + \frac{5.5^3}{3!} + \frac{5.5^4}{4!})$ | M1 | Condone one end error. Any $\lambda$. Accept fully correct $\Sigma$ notation. Expression must be seen. |
| $= 0.642$ or $0.643$ (3sf) | A1 | Correct answer with no working scores SC B1 B1 |
| | **3** | |
## Question 5(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X=3) \times P(Y=2) = e^{-3.1} \times \frac{3.1^3}{3!} \times e^{-2.4} \times \frac{2.4^2}{2!}$ or $0.223676 \times 0.261267 [= 0.05844]$ | M1 | Find P(3 in first half AND 2 in second half). Must see expression. |
| $P(\text{total } 5) = e^{-5.5} \times \frac{5.5^5}{5!}$ or $0.17140$ | M1 | Use of 5.5 to find P(5). |
| $P(\text{exactly 3 in 1st half} \mid \text{total } 5) = \frac{P(\text{exactly 3 in 1st half and total 5})}{P(\text{total 5})}$ | M1 | Attempt at conditional probability; numerator = *their* 0.05844 and denominator = P(total 5). Note: $(\frac{3.1^3}{3!} \times \frac{2.4^2}{2!}) \div (\frac{5.5^5}{5!})$ scores M1 M1 M1. |
| $[= \frac{0.05844}{0.17140}] = 0.341$ (3sf) | A1 | |
| | **4** | |5 The number of goals scored by a sports team in the first half of any match has the distribution $X \sim \mathrm { Po }$ (3.1). The number of goals scored by the same team in the second half of any match has the distribution $Y \sim \operatorname { Po } ( 2.4 )$. You may assume that the distributions of $X$ and $Y$ are independent.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } ( X < 4 )$.
\item Find the probability that, in a randomly chosen match, the team scores at least 5 goals.
\item Given that the team scores a total of 5 goals in a randomly chosen match, find the probability that they score exactly 3 goals in the first half.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q5 [9]}}