Pre-U Pre-U 9795/2 2016 Specimen — Question 4 10 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2016
SessionSpecimen
Marks10
TopicPoisson distribution
TypeConditional probability with Poisson
DifficultyChallenging +1.2 Part (i) is a standard PGF derivation requiring manipulation of the exponential series. Part (ii) is a textbook application of PGF properties for sums of independent variables. Part (iii) requires conditional probability with Poisson distributions, which is less routine but follows a standard technique (conditioning on the sum gives a binomial distribution). This is moderately above average difficulty due to the multi-step nature and the conditional probability calculation, but all techniques are standard for Further Maths statistics.
Spec5.02i Poisson distribution: random events model5.02n Sum of Poisson variables: is Poisson

4
  1. The random variable \(X\) has the distribution \(\operatorname { Po } ( \lambda )\). Prove that the probability generating function, \(\mathrm { G } _ { X } ( t )\), is given by $$\mathrm { G } _ { X } ( t ) = \mathrm { e } ^ { \lambda ( t - 1 ) }$$
  2. The independent random variables \(X\) and \(Y\) have distributions \(\operatorname { Po } ( \lambda )\) and \(\operatorname { Po } ( \mu )\) respectively. Use probability generating functions to show that the distribution of \(X + Y\) is \(\operatorname { Po } ( \lambda + \mu )\).
  3. Given that \(X \sim \operatorname { Po } ( 1.5 )\) and \(Y \sim \operatorname { Po } ( 2.5 )\), find \(\mathrm { P } ( X \leqslant 2 \mid X + Y = 4 )\).

(i) \(P(X = r) = e^{-\lambda}\frac{\lambda^r}{r!}\) (may be implied by next line) B1
\(G_X(t) = \sum_0^\infty p_r t^r = \sum_0^\infty e^{-\lambda}\frac{(\lambda t)^r}{r!} = e^{-\lambda}e^{\lambda t} = e^{\lambda(t-1)}\) AG M1 A1
(ii) \(G_{X+Y}(t) = e^{\lambda(t-1)} \cdot e^{\mu(t-1)} = e^{(\lambda+\mu)(t-1)} \Rightarrow (X+Y) \sim \text{Po}(\lambda + \mu)\) M1 A1
(iii) \(P([0,4]\) or \([1,3]\) or \([2,2]) = 0.2231 \times 0.1336 + 0.3347 \times 0.2138 + 0.2510 \times 0.2565\) B2,1,0
(B1 for any two correct)
\(P(X \leqslant 2 \mid X + Y = 4) = \frac{0.1657}{0.1954}\) M1 A1ft
\(= 0.848\) art A1
**(i)** $P(X = r) = e^{-\lambda}\frac{\lambda^r}{r!}$ (may be implied by next line) **B1**

$G_X(t) = \sum_0^\infty p_r t^r = \sum_0^\infty e^{-\lambda}\frac{(\lambda t)^r}{r!} = e^{-\lambda}e^{\lambda t} = e^{\lambda(t-1)}$ AG **M1 A1**

**(ii)** $G_{X+Y}(t) = e^{\lambda(t-1)} \cdot e^{\mu(t-1)} = e^{(\lambda+\mu)(t-1)} \Rightarrow (X+Y) \sim \text{Po}(\lambda + \mu)$ **M1 A1**

**(iii)** $P([0,4]$ or $[1,3]$ or $[2,2]) = 0.2231 \times 0.1336 + 0.3347 \times 0.2138 + 0.2510 \times 0.2565$ **B2,1,0**

(B1 for any two correct)

$P(X \leqslant 2 \mid X + Y = 4) = \frac{0.1657}{0.1954}$ **M1 A1ft**

$= 0.848$ art **A1**
4 (i) The random variable $X$ has the distribution $\operatorname { Po } ( \lambda )$. Prove that the probability generating function, $\mathrm { G } _ { X } ( t )$, is given by

$$\mathrm { G } _ { X } ( t ) = \mathrm { e } ^ { \lambda ( t - 1 ) }$$

(ii) The independent random variables $X$ and $Y$ have distributions $\operatorname { Po } ( \lambda )$ and $\operatorname { Po } ( \mu )$ respectively. Use probability generating functions to show that the distribution of $X + Y$ is $\operatorname { Po } ( \lambda + \mu )$.\\
(iii) Given that $X \sim \operatorname { Po } ( 1.5 )$ and $Y \sim \operatorname { Po } ( 2.5 )$, find $\mathrm { P } ( X \leqslant 2 \mid X + Y = 4 )$.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2016 Q4 [10]}}