| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2013 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Conditional probability with Poisson |
| Difficulty | Standard +0.8 Part (i) requires understanding conditional probability with independent Poisson variables, computing P(X=1, Y=4 | X+Y=5) which involves recognizing that the conditional distribution follows a binomial pattern. Part (ii) involves algebraic manipulation of Poisson probability formulas and verification. This goes beyond routine Poisson calculations, requiring conceptual understanding of conditional distributions and multi-step reasoning, but remains within standard S2 scope. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \(e^{-2} \times 2\ (\times)\ e^{-3} \times \frac{3^4}{4!}\); \(e^{-5} \times \frac{5^4}{5!}\); \(\div\); \(\frac{162}{625}\) or \(0.259\) (3 sf) | M1, B1, M1, A1 [4] |
| (ii) | \(\left(e^{-2} \times \frac{2^r}{r!} = \frac{2}{3}e^{-2} \Rightarrow\right)\); \(3 \times 2^r = 2 \times r!\) OR \(2^{r-1} = \frac{1}{3} \times r!\); \(\left(\Rightarrow 3 \times 2^{r-1} = r!\right)\); \(3 \times 2^3 = 24\) OR \(3! = 24\) seen | B1, B1 [2] |
## Question 4:
**(i)** | $e^{-2} \times 2\ (\times)\ e^{-3} \times \frac{3^4}{4!}$; $e^{-5} \times \frac{5^4}{5!}$; $\div$; $\frac{162}{625}$ or $0.259$ (3 sf) | M1, B1, M1, A1 [4] | Correct exp'n for $P(1)$ with $\lambda=2$ OR $P(4)$ with $\lambda=3$; correct exp'n; dep M1B1 |
**(ii)** | $\left(e^{-2} \times \frac{2^r}{r!} = \frac{2}{3}e^{-2} \Rightarrow\right)$; $3 \times 2^r = 2 \times r!$ OR $2^{r-1} = \frac{1}{3} \times r!$; $\left(\Rightarrow 3 \times 2^{r-1} = r!\right)$; $3 \times 2^3 = 24$ OR $3! = 24$ seen | B1, B1 [2] | Legitimately shown; legitimately shown on either equation |
---4 The independent random variables $X$ and $Y$ have the distributions $\operatorname { Po } ( 2 )$ and $\operatorname { Po } ( 3 )$ respectively.\\
(i) Given that $X + Y = 5$, find the probability that $X = 1$ and $Y = 4$.\\
(ii) Given that $\mathrm { P } ( X = r ) = \frac { 2 } { 3 } \mathrm { P } ( X = 0 )$, show that $3 \times 2 ^ { r - 1 } = r$ ! and verify that $r = 4$ satisfies this equation.
\hfill \mbox{\textit{CAIE S2 2013 Q4 [6]}}