| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 10 |
| Topic | Poisson distribution |
| Type | Conditional probability with Poisson |
| Difficulty | Standard +0.3 This is a straightforward Poisson distribution question requiring standard techniques: stating assumptions, direct probability calculations, conditional probability, and solving an inequality. All parts follow textbook methods with no novel insight required. Part (iii) uses P(A|B) = P(A∩B)/P(B) which is routine for Further Maths students. The most challenging aspect is part (iv) requiring logarithms to solve for the area, but this is still a standard application rather than 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 probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Fossils occur randomly (at a constant mean rate) | B1 | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(X \geq 7) = 1 - P(X \leq 6) = 1 - 0.7622 = 0.2378 = 0.238\) | M1, A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(Y > 6 | Y \geq 4) = \frac{1-P(Y \leq 6)}{1-P(Y \leq 3)} = \frac{1-0.9858}{1-0.7576} = 0.0586\) | M1, M1, A1 |
| Answer | Marks |
|---|---|
| \(\lambda = \frac{A}{2}\) | B1 |
| \(\Rightarrow 1 - e^{-\frac{A}{2}} > 0.999\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\therefore \frac{A}{2} > \ln 1000 \Rightarrow A > 13.8 \Rightarrow A = 14\) | M1, A1 | 4 |
### (i)
Fossils occur randomly (at a constant mean rate) | B1 | **1**
### (ii)
$X \sim \text{Po}(5)$
$P(X \geq 7) = 1 - P(X \leq 6) = 1 - 0.7622 = 0.2378 = 0.238$ | M1, A1 | **2**
### (iii)
$Y \sim \text{Po}(2.5)$
$P(Y > 6 | Y \geq 4) = \frac{1-P(Y \leq 6)}{1-P(Y \leq 3)} = \frac{1-0.9858}{1-0.7576} = 0.0586$ | M1, M1, A1 | **3**
### (iv)
$\lambda = \frac{A}{2}$ | B1 |
$\Rightarrow 1 - e^{-\frac{A}{2}} > 0.999$ | M1 |
$e^{-\frac{A}{2}} < 0.001 \Rightarrow e^{\frac{A}{2}} > 1000$
$\therefore \frac{A}{2} > \ln 1000 \Rightarrow A > 13.8 \Rightarrow A = 14$ | M1, A1 | **4**A certain type of fossil occurs at a mean rate of $0.5$ per square metre at a particular location.
\begin{enumerate}[label=(\roman*)]
\item State an assumption that must be made so that the above situation can be modelled by a Poisson distribution. [1]
\item Find the probability of at least 7 of these fossils occurring in an area of $10 \text{ m}^2$. [2]
\item Given that at least 4 such fossils have occurred in an area of $5 \text{ m}^2$, find the probability that there will be more than 6 found in this area of $5 \text{ m}^2$. [3]
\item Find the least area that must be searched in order that the probability of finding at least one fossil of this type is greater than $0.999$. Give your answer to the nearest square metre. [4]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q9 [10]}}