| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Conditional probability with Poisson |
| Difficulty | Standard +0.3 Part (a) uses the standard result that sum of independent Poissons is Poisson with combined parameter, requiring routine probability calculation. Part (b) involves straightforward conditional probability using the definition P(A∩B)/P(B). Part (c) tests conceptual understanding that Poisson variables are non-negative. This is a standard S2 question with multiple accessible parts requiring known techniques rather than novel problem-solving. |
| 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!5.02k Calculate Poisson probabilities5.02n Sum of Poisson variables: is Poisson |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\lambda = 3\) | B1 | |
| \(e^{-3}(1+3)\) | M1 | |
| \(= 0.199\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(A_1 = 1 \text{ and } A_1 + A_2 < 2) = P(A_1 = 1) \times P(A_2 = 0)\) | M1 | |
| \(e^{-1.5} \times 1.5 \times e^{-1.5} = 0.0747\) | A1 | |
| \(P(A_1 = 1 \mid A_1 + A_2 < 2) = \frac{P(A_1 = 1 \text{ and } A_1 + A_2 < 2)}{P(A_1 + A_2 < 2)}\) | M1 | |
| \(= \frac{1.5 \times (e^{-1.5})^2}{4e^{-3}} = \frac{\text{'0.0747'}}{\text{'0.199'}}\) | ||
| \(\frac{3}{8}\) or \(0.375\) (3 sf) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Takes negative values | B1 |
## Question 4(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\lambda = 3$ | B1 | |
| $e^{-3}(1+3)$ | M1 | |
| $= 0.199$ (3 sf) | A1 | |
---
## Question 4(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(A_1 = 1 \text{ and } A_1 + A_2 < 2) = P(A_1 = 1) \times P(A_2 = 0)$ | M1 | |
| $e^{-1.5} \times 1.5 \times e^{-1.5} = 0.0747$ | A1 | |
| $P(A_1 = 1 \mid A_1 + A_2 < 2) = \frac{P(A_1 = 1 \text{ and } A_1 + A_2 < 2)}{P(A_1 + A_2 < 2)}$ | M1 | |
| $= \frac{1.5 \times (e^{-1.5})^2}{4e^{-3}} = \frac{\text{'0.0747'}}{\text{'0.199'}}$ | | |
| $\frac{3}{8}$ or $0.375$ (3 sf) | A1 | |
---
## Question 4(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| Takes negative values | B1 | |
---4 The random variable $A$ has the distribution $\operatorname { Po } ( 1.5 ) . A _ { 1 }$ and $A _ { 2 }$ are independent values of $A$.
\begin{enumerate}[label=(\alph*)]
\item Find $\mathrm { P } \left( A _ { 1 } + A _ { 2 } < 2 \right)$.
\item Given that $A _ { 1 } + A _ { 2 } < 2$, find $\mathrm { P } \left( A _ { 1 } = 1 \right)$.
\item Give a reason why $A _ { 1 } - A _ { 2 }$ cannot have a Poisson distribution.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2020 Q4 [8]}}