Question 6 - A-Level Maths

Edexcel FS1 Specimen — Question 6 14 marks

Exam Board	Edexcel
Module	FS1 (Further Statistics 1)
Session	Specimen
Marks	14
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Probability Generating Functions
Type	Given PGF manipulation and properties
Difficulty	Standard +0.8 This is a Further Maths Statistics question requiring multiple PGF properties: normalization condition, coefficient extraction, variance formula (E[X(X-1)] + E[X] - E[X]²), and transformation of random variables. While the techniques are standard for FS1, the multi-part nature, algebraic manipulation of the squared binomial, and the transformation in part (d) elevate this above typical A-level questions. It's moderately challenging for Further Maths but not requiring deep insight.
Spec	5.02a Discrete probability distributions: general 5.02b Expectation and variance: discrete random variables 5.02c Linear coding: effects on mean and variance

The probability generating function of the discrete random variable $X$ is given by

$$G _ { x } ( t ) = k \left( 3 + t + 2 t ^ { 2 } \right) ^ { 2 }$$

Show that $\mathrm { k } = \frac { 1 } { 36 }$
Find $\mathrm { P } ( \mathrm { X } = 3 )$
Show that $\operatorname { Var } ( \mathrm { X } ) = \frac { 29 } { 18 }$
Find the probability generating function of $2 \mathrm { X } + 1$ \section*{Q uestion 6 continued} \section*{Q uestion 6 continued} \section*{Q uestion 6 continued}

Show mark scheme Show mark scheme source

Question 6:

Part (a):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$G_X(1) = 1$ gives	M1	2.1 - Stating $G_X(1)=1$
$k \times 6^2 = 1$ so $k = \frac{1}{36}$	A1*cso	1.1b - Fully correct proof with no errors cso

Part (b):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$P(X=3)$ = coefficient of $t^3$ so $G_X(t) = k(\ldots + 4t^3 \ldots)$	M1	1.1b - Attempting to find coefficient of $t^3$; may be implied by obtaining $\frac{1}{9}$ or awrt 0.11
$P(X=3) = \frac{1}{9}$	A1	1.1b - allow awrt 0.111

Part (c):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$G'_x(t) = 2k(3+t+2t^2)\times(1+4t)$	M1	2.1 - Attempting to find $G'_x(t)$; allow Chain rule or multiplying out brackets and differentiating
$E(X) = G'_x(1) = 2k(3+1+2)\times(1+4)$	M1	1.1b - Substituting $t=1$ into $G'_x(t)$
$= \frac{5}{3}$	A1	1.1b - allow awrt 1.67
$G''_x(t) = 2k\Big[(3+t+2t^2)\times 4 + (1+4t)^2\Big]$	M1, A1	2.1, 1.1b - Attempting to find $G''_x(t)$; accept $2k\big[(3+t+2t^2)\times 4+(1+4t)^2\big]$ or $k(48t^2+24t+26)$ o.e.
$G''_x(1) = 2k[6\times 4 + 5^2]$ giving $\left\{=\frac{49}{18}\right\}$	M1	1.1b - $2k[6\times 4 + 5^2]$ o.e.
$\text{Var}(X) = G''_x(1) + G'_x(1) - [G'_x(1)]^2 = \frac{49}{18}+\frac{5}{3}-\frac{25}{9}$	M1	2.1 - Using $G''_x(1)+G'_x(1)-[G'_x(1)]^2$ to find Variance
$= \frac{29}{18}$	A1*cso	1.1b

Part (d):

Answer	Marks	Guidance
Working/Answer	Mark	Guidance
$G_{2X+1}(t) = \frac{t}{36}\left(3+t^2+2(t^2)^2\right)^2$ $[\times t$ or sub $t^2$ for $t]$	M1	3.1a - Realising the need to $\times t$ or sub $t^2$ for $t$
$= G_{2X+1}(t) = \frac{t}{36}(3+t^2+2t^4)^2$	A1	1.1b - $\frac{t}{36}(3+t^2+2t^4)^2$ or $\frac{t}{36}(9+6t^2+13t^4+4t^6+4t^8)$ o.e.

# Question 6:

## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $G_X(1) = 1$ gives | M1 | 2.1 - Stating $G_X(1)=1$ |
| $k \times 6^2 = 1$ so $k = \frac{1}{36}$ | A1*cso | 1.1b - Fully correct proof with no errors cso |

## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P(X=3)$ = coefficient of $t^3$ so $G_X(t) = k(\ldots + 4t^3 \ldots)$ | M1 | 1.1b - Attempting to find coefficient of $t^3$; may be implied by obtaining $\frac{1}{9}$ or awrt 0.11 |
| $P(X=3) = \frac{1}{9}$ | A1 | 1.1b - allow awrt 0.111 |

## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $G'_x(t) = 2k(3+t+2t^2)\times(1+4t)$ | M1 | 2.1 - Attempting to find $G'_x(t)$; allow Chain rule or multiplying out brackets and differentiating |
| $E(X) = G'_x(1) = 2k(3+1+2)\times(1+4)$ | M1 | 1.1b - Substituting $t=1$ into $G'_x(t)$ |
| $= \frac{5}{3}$ | A1 | 1.1b - allow awrt 1.67 |
| $G''_x(t) = 2k\Big[(3+t+2t^2)\times 4 + (1+4t)^2\Big]$ | M1, A1 | 2.1, 1.1b - Attempting to find $G''_x(t)$; accept $2k\big[(3+t+2t^2)\times 4+(1+4t)^2\big]$ or $k(48t^2+24t+26)$ o.e. |
| $G''_x(1) = 2k[6\times 4 + 5^2]$ giving $\left\{=\frac{49}{18}\right\}$ | M1 | 1.1b - $2k[6\times 4 + 5^2]$ o.e. |
| $\text{Var}(X) = G''_x(1) + G'_x(1) - [G'_x(1)]^2 = \frac{49}{18}+\frac{5}{3}-\frac{25}{9}$ | M1 | 2.1 - Using $G''_x(1)+G'_x(1)-[G'_x(1)]^2$ to find Variance |
| $= \frac{29}{18}$ | A1*cso | 1.1b |

## Part (d):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $G_{2X+1}(t) = \frac{t}{36}\left(3+t^2+2(t^2)^2\right)^2$ $[\times t$ or sub $t^2$ for $t]$ | M1 | 3.1a - Realising the need to $\times t$ or sub $t^2$ for $t$ |
| $= G_{2X+1}(t) = \frac{t}{36}(3+t^2+2t^4)^2$ | A1 | 1.1b - $\frac{t}{36}(3+t^2+2t^4)^2$ or $\frac{t}{36}(9+6t^2+13t^4+4t^6+4t^8)$ o.e. |

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Show LaTeX source

\begin{enumerate}
  \item The probability generating function of the discrete random variable $X$ is given by
\end{enumerate}

$$G _ { x } ( t ) = k \left( 3 + t + 2 t ^ { 2 } \right) ^ { 2 }$$

(a) Show that $\mathrm { k } = \frac { 1 } { 36 }$\\
(b) Find $\mathrm { P } ( \mathrm { X } = 3 )$\\
(c) Show that $\operatorname { Var } ( \mathrm { X } ) = \frac { 29 } { 18 }$\\
(d) Find the probability generating function of $2 \mathrm { X } + 1$\\

\section*{Q uestion 6 continued}

\section*{Q uestion 6 continued}

\section*{Q uestion 6 continued}

\hfill \mbox{\textit{Edexcel FS1  Q6 [14]}}

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Q1 5 Q2 12 Q3 14 Q4 4 Q5 8 Q6 14 Q7 18