Edexcel A-level Further Maths revision notes

Probability generating functions

Section FS1-7

Year 2

3 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9FM0 section FS1-7

Checked against Edexcel 9FM0 section FS1-7. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

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FS1-7.1

Definitions, derivations and applications. Use of the probability generating function for the negative binomial, geometric, binomial and Poisson distributions.

Notes

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Explanation

For a non-negative integer-valued random variable, the probability generating function is $G_X(t)=E(t^X)=\sum_xP(X=x)t^x$ .
The coefficient of $t^x$ is $P(X=x)$ , so the PGF encodes the distribution and must satisfy $G_X(1)=1$ .
Standard forms are $(1-p+pt)^n$ for binomial, $e^{\lambda(t-1)}$ for Poisson, $\dfrac{pt}{1-(1-p)t}$ for geometric, and its $r$ th power for negative binomial.
Derivations use binomial, exponential or geometric-series expansions.
Examiners may require standard results to be proved, so the support and series index must be shown; the geometric PGF includes $t$ because FS₁ counts trials from $1$ .

Worked example

Show that the probability generating function of a geometric variable with success probability $p$ , counting the trial of first success, is $\dfrac{pt}{1-(1-p)t}$ .

1. $G_X(t)=\sum_{x=1}^{\infty}p(1-p)^{x-1}t^x$ .
2.Factor $pt$ : $G_X(t)=pt\sum_{j=0}^{\infty}[(1-p)t]^j$ .
3.Use the geometric series to obtain $G_X(t)=\dfrac{pt}{1-(1-p)t}$ .

Answer: $G_X(t)=\dfrac{pt}{1-(1-p)t}$ , for values of $t$ where the series converges.

Common mistakes

Don't omit the factor $t$ and derive the PGF for failures before success instead.
Don't read the coefficient of $t^x$ as $E(X=x)$ rather than $P(X=x)$ .
Don't use a PGF for a variable whose support includes negative integers.

Exam tip

For a derivation, write the probability sum before quoting the matching power-series identity.

Tier 1 · Easy

ORIGINAL

Given

X\sim\operatorname{Bin}(4,0.3)

, write down

G_X(t)

and use its coefficient of

t^2

to find

P(X=2)

(3)

(Total for Question 1 is 3 marks)

Tier 2 · Standard

ORIGINAL

A geometric random variable

X

has parameter

p=0.4

and counts the trial of the first success. Show that the probability generating function of

X

\dfrac{0.4t}{1-0.6t}

and hence find

P(X=5)

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

The random variable

X

counts the trial on which the third success occurs in independent trials with success probability

0.25

. Show that

G_X(t)=\left(\dfrac{0.25t}{1-0.75t}\right)^3

and use it to find

P(X=6)

(7)

(Total for Question 1 is 7 marks)

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FS1-7.2

Use to find the mean and variance.

Notes

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Explanation

For a PGF $G_X$ , first check $G_X(1)=1$ . Differentiating term by term gives $G'_X(1)=E(X)$ and $G''_X(1)=E[X(X-1)]$ , the second factorial moment.
Since $X^2=X(X-1)+X$ , $E(X^2)=G''_X(1)+G'_X(1)$ . Therefore $\operatorname{Var}(X)=G''_X(1)+G'_X(1)-[G'_X(1)]^2$ .
Derivatives can establish standard means and variances or find moments from a given polynomial or rational PGF.
Higher derivatives similarly produce higher factorial moments and may be used in proofs of standard distribution results.
Examiners expect evaluation at $t=1$ , accurate differentiation and the extra first-derivative term in the variance; $G''_X(1)$ alone is not the second raw moment.

Worked example

A random variable has PGF $G(t)=(0.7+0.3t)^8$ . Use derivatives to find its mean and variance.

1. $G'(t)=2.4(0.7+0.3t)^7$ , so $G'(1)=2.4$ .
2. $G''(t)=5.04(0.7+0.3t)^6$ , so $G''(1)=5.04$ .
3. $\operatorname{Var}(X)=5.04+2.4-2.4^2=1.68$ .

Answer: $E(X)=2.4$ and $\operatorname{Var}(X)=1.68$ .

Common mistakes

Don't substitute $t=0$ rather than $t=1$ when finding moments.
Don't treat $G''(1)$ as $E(X^2)$ and omit $G'(1)$ .
Don't square $G'(t)$ before evaluating rather than squaring the scalar $G'(1)$ .

Exam tip

Write the factorial-moment identity beside $G''(1)$ before converting it to variance.

Tier 1 · Easy

ORIGINAL

The PGF of

X

G_X(t)=(0.8+0.2t)^6

. Use derivatives of the PGF to find

E(X)

and

\operatorname{Var}(X)

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

A random variable has PGF

G_X(t)=\dfrac{2t}{3-t}

. Use the PGF to find its mean and variance.

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

A random variable has PGF

G_X(t)=k(1+2t+3t^2+4t^3)

. Find

k

, then use derivatives to calculate

E(X)

and

\operatorname{Var}(X)

(6)

(Total for Question 1 is 6 marks)

FS1-7.3

Probability generating function of the sum of independent random variables.

Notes

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Your confidence:

Explanation

If $X$ and $Y$ are independent, then $G_{X+Y}(t)=G_X(t)G_Y(t)$ . Independence allows $E(t^{X+Y})=E(t^X)E(t^Y)$ .
Multiply and simplify before identifying a standard PGF or extracting a coefficient. This proves, for example, that independent Poisson parameters add and that independent binomial variables with the same $p$ combine by adding their trial counts.
A sum of independent geometric waiting times with common $p$ is negative binomial.
The product rule does not hold automatically for dependent variables.
Examiners expect independence to be stated, the product simplified into a recognisable form, and all distribution parameters and the support of the sum stated.

Worked example

Independent variables satisfy $X\sim\operatorname{Bin}(4,p)$ and $Y\sim\operatorname{Bin}(7,p)$ . Use PGFs to identify $X+Y$ .

1. $G_X(t)=(1-p+pt)^4$ and $G_Y(t)=(1-p+pt)^7$ .
2.Independence gives $G_{X+Y}(t)=G_X(t)G_Y(t)$ .
3.Thus $G_{X+Y}(t)=(1-p+pt)^{11}$ .

Answer: $X+Y\sim\operatorname{Bin}(11,p)$ .

Common mistakes

Don't multiply PGFs without stating or establishing independence.
Don't add binomial probabilities $p$ as well as adding the trial counts.
Don't identify a negative-binomial sum using the failures-before-success convention instead of trial number.

Exam tip

After multiplying PGFs, quote the exact standard form that identifies the sum's distribution.

Tier 1 · Easy

ORIGINAL

Independent variables have distributions

X\sim\operatorname{Po}(1.4)

and

Y\sim\operatorname{Po}(2.6)

. Use PGFs to identify the distribution of

X+Y

and find

P(X+Y=0)

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

Independent variables satisfy

X\sim\operatorname{Bin}(5,0.4)

and

Y\sim\operatorname{Bin}(7,0.4)

. Use PGFs to identify

S=X+Y

and calculate

P(S=3)

(5)

(Total for Question 1 is 5 marks)

Tier 3 · Hard

ORIGINAL

Independent geometric variables

X

and

Y

each have parameter

0.4

and count trials to first success. Use PGFs to identify the distribution of

S=X+Y

and find

P(S>5)

(7)

(Total for Question 1 is 7 marks)

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