| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9795/2 (Pre-U Further Mathematics Paper 2) |
| Session | Specimen |
| Marks | 8 |
| Topic | Probability Generating Functions |
| Type | Find PGF from probability distribution |
| Difficulty | Standard +0.8 This is a multi-part PGF question requiring: (i) writing the PGF with a negative power of t (non-standard), (ii) differentiation for mean/variance, (iii) recognizing that the PGF of a sum equals the product of individual PGFs and algebraic manipulation to the given form, and (iv) extracting specific probabilities from the coefficient of the resulting expression. The negative value in the distribution and the algebraic manipulation in part (iii) elevate this above a routine PGF question, but it remains a standard Further Maths exercise testing core PGF techniques. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | - 1 | 1 | 3 |
| \(\mathrm { P } ( X = x )\) | \(\frac { 16 } { 25 }\) | \(\frac { 8 } { 25 }\) | \(\frac { 1 } { 25 }\) |
**Question 12**
**(i)** $G(t) = \Sigma p_r t^r = \frac{1}{25}t^{-1} + \frac{8}{25}t + \frac{1}{25}t^3 = \frac{1}{25}(t^3 + 8t + 16t^{-1})$ B1 **[1 mark]**
**(ii)** $G'(t) = -\dfrac{16}{25}t^{-2} + \dfrac{8}{25} + \dfrac{3}{25}t^2$ M1
$\mathrm{E}(X) = G'(1) = -\dfrac{16}{25} + \dfrac{8}{25} + \dfrac{3}{25} = -\dfrac{5}{25} = -\dfrac{1}{5}$ A1
$G''(t) = \dfrac{32}{25}t^{-3} + \dfrac{6}{25}t$ M1
$G''(1) = \dfrac{32}{25} + \dfrac{6}{25}$ A1
$\sigma^2 = \dfrac{32}{25} + \dfrac{6}{25} - \dfrac{6}{25} = \dfrac{32}{25} = 1.28$ A1 **[5 marks]**
**(iii)** $G_Y(t) = \left(\dfrac{1}{25}\right)^4\left(\dfrac{t^4 + 8t^2 + 16}{t}\right)^4$ M1A1
$$= \left(\dfrac{1}{25}\right)^4\dfrac{(t^2+4)^8}{t^4}$$ A1 **[3 marks]**
**(iv)** $G_Y(t) = \left(\dfrac{1}{25}\right)^4\left(\dfrac{4^8}{t^4} + \dbinom{8}{1}4^7 t^{-2} + \cdots\right)$ M1A1
$$\mathrm{P}(Y < 0) = \frac{4^8 + 8 \times 4^7}{25^4} = 0.503$$ A1 **[3 marks]**12 A game is played in which the number of points scored, $X$, has the probability distribution given in the following table.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
$x$ & - 1 & 1 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & $\frac { 16 } { 25 }$ & $\frac { 8 } { 25 }$ & $\frac { 1 } { 25 }$ \\
\hline
\end{tabular}
\end{center}
(i) Write down the probability generating function of $X$.\\
(ii) Use this generating function to find the mean and variance of $X$.\\
(iii) The game is played 4 times (independently) and the total number of points scored is denoted by $Y$. Show that the probability generating function of $Y$ can be written in the form
$$\frac { \left( a + t ^ { 2 } \right) ^ { 8 } } { b t ^ { 4 } }$$
where $a$ and $b$ are constants.\\
(iv) Hence find $\mathrm { P } ( Y < 0 )$.
\hfill \mbox{\textit{Pre-U Pre-U 9795/2 Q12 [8]}}