CAIE Further Paper 4 2021 June — Question 4 5 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2021
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProbability Generating Functions
TypeUse PGF to find mean and variance
DifficultyStandard +0.8 This is a Further Maths probability generating function question requiring differentiation of a rational function and application of standard PGF formulas for mean and variance. While the algebra is moderately involved (chain rule, quotient rule), the method is direct and well-practiced by Further Maths students. The unusual feature of even values only (requiring t² terms) adds slight conceptual complexity but follows standard PGF theory.
Spec5.02b Expectation and variance: discrete random variables
\(X\) is a discrete random variable which takes the values 0, 2, 4, ... . The probability generating function of \(X\) is given by $$G_X(t) = \frac{1}{3 - 2t^2}.$$
  1. Find E\((X)\) and Var\((X)\). [5]
Question 4:
AnswerMarks
4Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
AnswerMarks
4(a)4t
G’(t) =
( )2
AnswerMarks Guidance
3−2t2M1 kt
Differentiate to obtain OE.
( )2
3−2t2
AnswerMarks Guidance
so E(X) = G’(1) = 4A1 CAO WWW
12+24t2 ( )−2 ( )−3
G’’(t) = or 4 3−2t2 +32t2 3−2t2
( )3
AnswerMarks Guidance
3−2t2M1 OE. Differentiate, allow only numerical or sign slips.
( ( ))2
AnswerMarks Guidance
Var(X) = G' '(1) + G'(1) – G' 1 = 36 + 4 – 16M1 Substitute their values into correct formula, dependent on
attempt at G' '(t).
AnswerMarks Guidance
[Var(X)] = 24A1 CAO WWW
5
AnswerMarks
4(b)G ( t )= 1 = ( 3−2t2 )−1
X 3−2t2
1 2 4 
= 1+ t2 + t4 +….
AnswerMarks Guidance
3 3 9 M1 Expand given expression or give expression for the term in t4 .
P(X = 4) = their coefficient of t4M1 Found from legitimate method.
4
[P(X = 4) =]
AnswerMarks Guidance
27A1 WWW
3
AnswerMarks Guidance
QuestionAnswer Marks 
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | 4t
G’(t) =
( )2
3−2t2 | M1 | kt
Differentiate to obtain OE.
( )2
3−2t2
so E(X) = G’(1) = 4 | A1 | CAO WWW
12+24t2 ( )−2 ( )−3
G’’(t) = or 4 3−2t2 +32t2 3−2t2
( )3
3−2t2 | M1 | OE. Differentiate, allow only numerical or sign slips.
( ( ))2
Var(X) = G' '(1) + G'(1) – G' 1 = 36 + 4 – 16 | M1 | Substitute their values into correct formula, dependent on
attempt at G' '(t).
[Var(X)] = 24 | A1 | CAO WWW
5
--- 4(b) ---
4(b) | G ( t )= 1 = ( 3−2t2 )−1
X 3−2t2
1 2 4 
= 1+ t2 + t4 +….
3 3 9  | M1 | Expand given expression or give expression for the term in t4 .
P(X = 4) = their coefficient of t4 | M1 | Found from legitimate method.
4
[P(X = 4) =]
27 | A1 | WWW
3
Question | Answer | Marks | Guidance
$X$ is a discrete random variable which takes the values 0, 2, 4, ... . The probability generating function of $X$ is given by

$$G_X(t) = \frac{1}{3 - 2t^2}.$$

\begin{enumerate}[label=(\alph*)]
\item Find E$(X)$ and Var$(X)$. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 4 2021 Q4 [5]}}
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