| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 4 |
| Topic | Poisson distribution |
| Type | Conditional probability with Poisson |
| Difficulty | Standard +0.3 This is a straightforward application of Poisson distribution properties with clear structure. Part (i) involves routine probability calculations and verification of the well-known result that sum of independent Poissons is Poisson. Part (ii) applies these formulas directly with given parameters, and part (c) is a standard conditional probability calculation. While it requires careful bookkeeping across multiple parts, no novel insight is needed—just methodical application of standard techniques. |
| 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 |
**Question 10**
**(i)**
$\mathrm{P}(X=1 \text{ and } Y=1) = \lambda\mathrm{e}^{-\lambda} \times \mu\mathrm{e}^{-\mu} = \lambda\mu\mathrm{e}^{-(\lambda+\mu)}$ acf B1
$\mathrm{P}(X=0 \text{ and } Y=2) = \mathrm{e}^{-\lambda} \times \mathrm{e}^{-\mu}\dfrac{\mu^2}{2!} = \mathrm{e}^{-(\lambda+\mu)}\dfrac{\mu^2}{2!}$ acf B1
$\mathrm{P}(X=2 \text{ and } Y=0) = \mathrm{e}^{-\lambda}\dfrac{\lambda^2}{2!} \times \mathrm{e}^{-\mu} = \mathrm{e}^{-(\lambda+\mu)}\dfrac{\lambda^2}{2!}$ acf B1
Adding gives $\mathrm{e}^{-(\lambda+\mu)}\left(\dfrac{\lambda^2 + 2\lambda\mu + \mu^2}{2!}\right) = \mathrm{e}^{-(\lambda+\mu)}\dfrac{(\lambda+\mu)^2}{2!} = \mathrm{P}(Z=2)$ B1 **[4 marks]**
**(ii)(a)** $\mathrm{e}^{-4.5}\dfrac{4.5^3}{3!} \times \mathrm{e}^{-4.1}\dfrac{4.1^3}{3!} = 0.168\,71 \times 0.190\,36 = 0.0321$ M1A1A1 **[3 marks]**
**(b)** $\mathrm{e}^{-8.6}\dfrac{8.6^6}{6!} = 0.103$ M1A1 **[2 marks]**
**(c)** $\dfrac{0.032\,11...}{0.103\,44...} = 0.310$ (3sf) (allow 0.311) M1A1 **[2 marks]**10 (i) $X , Y$ and $Z$ are independent random variables having Poisson distributions with means $\lambda , \mu$ and $\lambda + \mu$ respectively. Find $\mathrm { P } ( X = 0$ and $Y = 2 ) , \mathrm { P } ( X = 1$ and $Y = 1 )$ and $\mathrm { P } ( X = 2$ and $Y = 0 )$. Hence verify that $\mathrm { P } ( X + Y = 2 ) = \mathrm { P } ( Z = 2 )$.\\
(ii) In an office the male absence rate, i.e. the number of working days lost each month due to the absence of male employees, has a Poisson distribution with mean 4.5. In the same office the female absence rate has an independent Poisson distribution with mean 4.1. Calculate the probability that
\begin{enumerate}[label=(\alph*)]
\item during a particular month both the male absence rate and the female absence rate are equal to 3,
\item during a particular month the total of the male and female absence rates is equal to 6,
\item during a particular month the male and female absence rates were each equal to 3 , given that the total of the male and female absence rates was equal to 6 .
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q10 [4]}}