Question 6 - A-Level Maths

Edexcel FS1 2020 June — Question 6 13 marks

Exam Board	Edexcel
Module	FS1 (Further Statistics 1)
Year	2020
Session	June
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Probability Generating Functions
Type	Solve for parameters using PGF coefficients
Difficulty	Standard +0.8 This Further Maths Statistics question requires understanding PGF coefficient extraction, solving simultaneous equations from probability constraints, using PGF properties for expectation, and applying linear transformations. While systematic, it demands multiple technical steps and conceptual understanding beyond standard A-level, placing it moderately above average difficulty.
Spec	5.02a Discrete probability distributions: general 5.02b Expectation and variance: discrete random variables

A discrete random variable $X$ has probability generating function given by

$$\mathrm { G } _ { X } ( t ) = \frac { 1 } { 64 } \left( a + b t ^ { 2 } \right) ^ { 2 }$$ where $a$ and $b$ are positive constants.

Write down the value of $\mathrm { P } ( X = 3 )$ Given that $\mathrm { P } ( X = 4 ) = \frac { 25 } { 64 }$
1. find $\mathrm { P } ( X = 2 )$
2. find $\mathrm { E } ( X )$ The random variable $Y = 3 X + 2$
Find the probability generating function of $Y$

Show mark scheme Show mark scheme source

Question 6:

Part (a):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$P(X=3) = \mathbf{0}$	B1	Since there is no term in $t^3$

Part (b)(i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Coefficient of $t^4 = \frac{1}{64}b^2$	M1	Realising $\frac{1}{64}b^2$ is the coefficient of $t^4$
$\frac{1}{64}b^2 = \frac{25}{64}$	M1	Equating coefficient of $t^4$ to $\frac{25}{64}$
$b = 5$ (reject $b=-5$ since $b>0$)	A1	$b=5$ only
$G_X(1) = 1$; $\frac{1}{64}(a + \text{"5"})^2 = 1$	M1	Realising $G_X(1)=1$ is required
$a = 3$ (reject $a=-13$ since $a>0$)	A1	$a=3$ only
$P(X=2) = $ coefficient of $t^2 = \frac{1}{64}(2ab)$	M1	Finding coefficient of $t^2$ with $a>0$ and $b>0$
$= \frac{15}{32}$	A1	Condone awrt 0.469

Part (b)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$\text{E}(X) = G'_X(1)$	M1	Realising $G'_X(1)$ is needed
$G'_X(t) = \frac{2}{64}(\text{"3"}+\text{"5"}t^2)\times\text{"10"}t$ or $G'_X(t) = \frac{1}{64}(\text{"60"}t + \text{"100"}t^3)$	M1	Attempt to differentiate $G_X(t)$ with their values of $a$ and $b$
$G'_X(1) = 2.5$	A1ft	2.5 (ft their $a,b$, $a>0$, $b>0$); $\text{E}(X) = \frac{ab+b^2}{16}$

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$G_Y(t) = t^2 G_X(t^3) \left[= \frac{t^2}{64}(a+b(t^3)^2)^2\right]$	M1	Either $G_X(t^3)$ or $\times t^2$ or using $Y=2,8,14$
$G_Y(t) = \frac{t^2}{64}(\text{"3"}+\text{"5"}t^6)^2$	A1ft	ft their values of $a$ and $b$, $a>0$ and $b>0$

# Question 6:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X=3) = \mathbf{0}$ | B1 | Since there is no term in $t^3$ |

## Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Coefficient of $t^4 = \frac{1}{64}b^2$ | M1 | Realising $\frac{1}{64}b^2$ is the coefficient of $t^4$ |
| $\frac{1}{64}b^2 = \frac{25}{64}$ | M1 | Equating coefficient of $t^4$ to $\frac{25}{64}$ |
| $b = 5$ (reject $b=-5$ since $b>0$) | A1 | $b=5$ only |
| $G_X(1) = 1$; $\frac{1}{64}(a + \text{"5"})^2 = 1$ | M1 | Realising $G_X(1)=1$ is required |
| $a = 3$ (reject $a=-13$ since $a>0$) | A1 | $a=3$ only |
| $P(X=2) = $ coefficient of $t^2 = \frac{1}{64}(2ab)$ | M1 | Finding coefficient of $t^2$ with $a>0$ and $b>0$ |
| $= \frac{15}{32}$ | A1 | Condone awrt 0.469 |

## Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{E}(X) = G'_X(1)$ | M1 | Realising $G'_X(1)$ is needed |
| $G'_X(t) = \frac{2}{64}(\text{"3"}+\text{"5"}t^2)\times\text{"10"}t$ or $G'_X(t) = \frac{1}{64}(\text{"60"}t + \text{"100"}t^3)$ | M1 | Attempt to differentiate $G_X(t)$ with their values of $a$ and $b$ |
| $G'_X(1) = 2.5$ | A1ft | 2.5 (ft their $a,b$, $a>0$, $b>0$); $\text{E}(X) = \frac{ab+b^2}{16}$ |

## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $G_Y(t) = t^2 G_X(t^3) \left[= \frac{t^2}{64}(a+b(t^3)^2)^2\right]$ | M1 | Either $G_X(t^3)$ or $\times t^2$ or using $Y=2,8,14$ |
| $G_Y(t) = \frac{t^2}{64}(\text{"3"}+\text{"5"}t^6)^2$ | A1ft | ft their values of $a$ and $b$, $a>0$ and $b>0$ |

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Show LaTeX source

\begin{enumerate}
  \item A discrete random variable $X$ has probability generating function given by
\end{enumerate}

$$\mathrm { G } _ { X } ( t ) = \frac { 1 } { 64 } \left( a + b t ^ { 2 } \right) ^ { 2 }$$

where $a$ and $b$ are positive constants.\\
(a) Write down the value of $\mathrm { P } ( X = 3 )$

Given that $\mathrm { P } ( X = 4 ) = \frac { 25 } { 64 }$\\
(b) (i) find $\mathrm { P } ( X = 2 )$\\
(ii) find $\mathrm { E } ( X )$

The random variable $Y = 3 X + 2$\\
(c) Find the probability generating function of $Y$

\hfill \mbox{\textit{Edexcel FS1 2020 Q6 [13]}}

This paper (7 questions)

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Q1 13 Q2 4 Q3 9 Q4 8 Q5 13 Q6 13 Q7 15

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Coefficient of \(t^4 = \frac{1}{64}b^2\)	M1	Realising \(\frac{1}{64}b^2\) is the coefficient of \(t^4\)
\(\frac{1}{64}b^2 = \frac{25}{64}\)	M1	Equating coefficient of \(t^4\) to \(\frac{25}{64}\)
\(b = 5\) (reject \(b=-5\) since \(b>0\))	A1	\(b=5\) only
\(G_X(1) = 1\); \(\frac{1}{64}(a + \text{"5"})^2 = 1\)	M1	Realising \(G_X(1)=1\) is required
\(a = 3\) (reject \(a=-13\) since \(a>0\))	A1	\(a=3\) only
\(P(X=2) = \) coefficient of \(t^2 = \frac{1}{64}(2ab)\)	M1	Finding coefficient of \(t^2\) with \(a>0\) and \(b>0\)
\(= \frac{15}{32}\)	A1	Condone awrt 0.469