FS1-7 Probability generating functions — revision question pack

3 specification points · notes, questions, answers and worked methods

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.

How this checking works

FS1-7.1 · Definitions, derivations and applications. Use of the probability generating function for the negative binomial, geometric, binomial and Poisson distributions.

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

1.
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)
2.
Write down $G_X(t)$ for $X\sim\operatorname{Po}(1.8)$ . Verify that $G_X(1)=1$ , and use the probability generating function to find $P(X\geq2)$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
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 $X$ is $\dfrac{0.4t}{1-0.6t}$ and hence find $P(X=5)$ .
(5)
(Total for Question 1 is 5 marks)
2.
$X$ is a discrete random variable with probability generating function $G_X(t)=\left(\dfrac{0.3t}{1-0.7t}\right)^4$ . State the name of its distribution, the values of its parameters and the values $X$ can take. Justify your answer from the standard PGF form, and find $P(X=6)$ .
(5)
(Total for Question 2 is 5 marks)
3.
For $X\sim B(7,0.45)$ , use the definition of a probability generating function to show that $G_X(t)=(0.55+0.45t)^7$ . Hence find $P(X\leq1)$ , giving your answer to $4$ decimal places.
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
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)
2.
$X$ is a discrete random variable with probability generating function $G_X(t)=e^{\lambda(t-1)}$ , where $\lambda>0$ . Its $t^4$ coefficient is $0.7$ times its $t^3$ coefficient. Determine $\lambda$ and hence find $P(X\leq1)$ .
(6)
(Total for Question 2 is 6 marks)
3.
The function $G_W(t)=(1-p+pt)^6$ is the probability generating function of $W$ , where $0<p<1$ . Given that $P(W=0)=64/729$ , state the distribution of $W$ in the form $W\sim B(n,p)$ and determine $p$ . By considering $G_W(-1)$ , find the exact probability that $W$ is even.
(7)
(Total for Question 3 is 7 marks)
4.
The discrete random variable $X$ has a geometric distribution with parameter $p$ , where $X$ counts the trial of the first success and $0<p<1$ . Given that $G_X(1/2)=1/3$ , determine $p$ , find the exact value of $P(X>4)$ and find the least integer $k$ for which $P(X>k)<0.01$ .
(7)
(Total for Question 4 is 7 marks)
5.
For a discrete random variable $X$ , $G_X(t)=a\left(\dfrac{1+t}{2}\right)^4+(1-a)\left(\dfrac{3+t}{4}\right)^3$ , where $0<a<1$ . Given that $P(X=1)=35/96$ , determine $a$ . Find the exact values of $P(X=2)$ and $P(X=4)$ .
(7)
(Total for Question 5 is 7 marks)

FS1-7.2 · Use to find the mean and variance.

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

1.
The PGF of $X$ is $G_X(t)=(0.8+0.2t)^6$ . Use derivatives of the PGF to find $E(X)$ and $\operatorname{Var}(X)$ .
(4)
(Total for Question 1 is 4 marks)
2.
$X$ is a discrete random variable with probability generating function $G_X(t)=e^{2.5(t-1)}$ . Use calculus to show that $E(X)=2.5$ and $\operatorname{Var}(X)=2.5$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
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)
2.
A geometric random variable $X$ counts the trial on which the first success occurs and has PGF $G_X(t)=\dfrac{pt}{1-(1-p)t}$ . By differentiating this PGF, prove the standard results for $E(X)$ and $\operatorname{Var}(X)$ . You may write $q=1-p$ .
(5)
(Total for Question 2 is 5 marks)
3.
The function $G_X(t)=e^{\frac32(t^2-1)}$ is the probability generating function of a discrete random variable $X$ . Verify that $G_X(t)$ is a valid probability generating function by evaluating $G_X(1)$ , then use derivatives to find $E(X)$ and $\operatorname{Var}(X)$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
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)
2.
A random variable has PGF $G_X(t)=\dfrac{1}{3}\dfrac{1+2t}{(2-t)^2}$ . Use calculus to find $\operatorname{Var}(X)$ . You must show all your working. Find the exact value of $P(X=0)$ .
(6)
(Total for Question 2 is 6 marks)
3.
Consider $G_X(t)=\dfrac{(1-a)e^{2(t-1)}}{1-at}$ for a discrete random variable $X$ , where $0<a<1$ . Verify that this is a valid probability generating function by evaluating $G_X(1)$ . Given that $\operatorname{Var}(X)=4$ , use derivatives to find $a$ and $E(X)$ .
(6)
(Total for Question 3 is 6 marks)
4.
For a discrete random variable $X$ , $G_X(t)=k(1+t+t^2+\cdots+t^m)$ is a probability generating function, where $m$ is a positive integer. Given that $E(X)=7/2$ , use derivatives to find the value of $m$ , the value of $k$ and $\operatorname{Var}(X)$ .
(7)
(Total for Question 4 is 7 marks)
5.
For a discrete random variable $X$ , $G_X(t)=1-c+ce^{\lambda(t-1)}$ is a probability generating function, where $0<c<1$ and $\lambda>0$ . Given that $E(X)=2$ and $\operatorname{Var}(X)=5$ , use derivatives to find $c$ and $\lambda$ . Hence find $P(X=0)$ , giving your answer to $4$ decimal places.
(7)
(Total for Question 5 is 7 marks)

FS1-7.3 · Probability generating function of the sum of independent random variables.

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

1.
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)
2.
Independent variables satisfy $P(X=1)=0.3$ , $P(X=0)=0.7$ and $Y\sim\operatorname{Po}(1.2)$ . Form the PGF of $S=X+Y$ and extract the coefficient of $t^2$ to find $P(S=2)$ .
(5)
(Total for Question 2 is 5 marks)

Tier 2 · Standard

1.
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)
2.
The independent random variables $X$ and $Y$ each have a negative binomial distribution with parameters $r=2$ and $p=0.25$ , and each counts the trial on which the second success occurs. Use probability generating functions to identify the distribution of $S=X+Y$ . Hence find $P(S=5)$ .
(6)
(Total for Question 2 is 6 marks)
3.
The independent random variables $X$ and $Y$ have sum $S=X+Y$ . Given that $X\sim B(3,0.4)$ and $G_S(t)=(0.6+0.4t)^3e^{1.7(t-1)}$ , use probability generating functions to determine the distribution of $Y$ . Hence find $P(Y\geq2)$ , giving your answer to $3$ significant figures.
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
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)
2.
The independent random variables $X$ , $Y$ and $Z$ have Poisson distributions with means $0.8$ , $1.6$ and $2.3$ respectively. Use probability generating functions to identify the distribution of $S=X+Y+Z$ . Hence find $P(S\geq7)$ , giving your answer to $4$ decimal places.
(6)
(Total for Question 2 is 6 marks)
3.
The independent random variables $X$ and $Y$ satisfy $X\sim B(3,1/2)$ and $Y\sim\operatorname{Geo}(3/8)$ , where $Y$ counts the trial of the first success. Let $S=X+Y$ . Form the probability generating function of $S$ . For the event $S=4$ , calculate the exact conditional probability $P(X=2\mid S=4)$ .
(7)
(Total for Question 3 is 7 marks)
4.
Let $S=X+Y$ , where $X\sim B(2,1/3)$ and $Y\sim B(4,1/2)$ are independent. Form the probability generating function of $S$ , explain why $S$ does not have a binomial distribution, and find the exact value of $P(S=3)$ .
(7)
(Total for Question 4 is 7 marks)
5.
Let $S=X+Y$ for independent random variables with probability generating functions $G_X(t)=(3+t^2)/4$ and $G_Y(t)=t(3+t^2)^2/16$ . Find $G_S(t)$ . State the set of possible values of $S$ and find the exact value of $P(S\geq5)$ .
(6)
(Total for Question 5 is 6 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

FS1-7.1 · Definitions, derivations and applications. Use of the probability generating function for the negative binomial, geometric, binomial and Poisson distributions.

Tier 1 · Easy

Mark scheme for FS1-7.1 Tier 1 · Easy
Question	Scheme	Marks
1	$G_X(t)=(0.7+0.3t)^4$ $P(X=2)=0.2646$	3
(3 marks)		3
Notes
The binomial PGF is $(1-p+pt)^n$ , giving $(0.7+0.3t)^4$ . Its $t^2$ coefficient is $\binom42(0.3)^2(0.7)^2=0.2646$ .
2	$G_X(t)=e^{1.8(t-1)}$ $G_X(1)=e^0=1$ $P(X\geq2)=1-e^{-1.8}(1+1.8)=0.5372$ to $4$ decimal places	3
(3 marks)		3
Notes
The Poisson PGF is $G_X(t)=e^{\lambda(t-1)}$ , so here $G_X(t)=e^{1.8(t-1)}$ and $G_X(1)=e^0=1$ . Expanding, $G_X(t)=e^{-1.8}e^{1.8t}$ , so the $t^0$ and $t^1$ coefficients are $e^{-1.8}$ and $1.8e^{-1.8}$ . Therefore $P(X\geq2)=1-P(X=0)-P(X=1)=1-e^{-1.8}(1+1.8)=0.5371631129\ldots$ , which rounds to $0.5372$ .

Tier 2 · Standard

Mark scheme for FS1-7.1 Tier 2 · Standard
Question	Scheme	Marks
1	$G_X(t)=\dfrac{0.4t}{1-0.6t}$ $P(X=5)=0.05184$	5
(5 marks)		5
Notes
$G_X(t)=\sum_{x=1}^{\infty}0.4(0.6)^{x-1}t^x=0.4t\sum_{j=0}^{\infty}(0.6t)^j=\dfrac{0.4t}{1-0.6t}$ . The coefficient of $t^5$ is $0.4(0.6)^4=0.05184$ .
2	$G(t)=[pt/(1-(1-p)t)]^r$ with $r=4$ and $p=0.3$ $X$ has the negative binomial distribution with $r=4$ , $p=0.3$ $X$ counts the trial on which the fourth success occurs, so $X=4,5,6,\ldots$ $P(X=6)=\binom{5}{3}(0.3)^4(0.7)^2$ $P(X=6)=0.03969$	5
(5 marks)		5
Notes
The standard PGF for the trial number of the $r$ th success is $[pt/(1-(1-p)t)]^r$ . Comparison gives $r=4$ and $p=0.3$ , so $X$ is negative binomial on $4,5,6,\ldots$ . For $X=6$ , the first five trials must contain three successes and the sixth must be a success. Hence $P(X=6)=\binom{5}{3}(0.3)^3(0.7)^2(0.3)=0.03969$ .
3	$G_X(t)=\displaystyle\sum_{x=0}^{7}\binom7x(0.45)^x(0.55)^{7-x}t^x$ $G_X(t)=\displaystyle\sum_{x=0}^{7}\binom7x(0.45t)^x(0.55)^{7-x}$ $G_X(t)=(0.55+0.45t)^7$ by the binomial theorem $P(X\leq1)=(0.55)^7+7(0.45)(0.55)^6$ $P(X\leq1)=0.1024183703\ldots=0.1024$	5
(5 marks)		5
Notes
By definition, $G_X(t)=E(t^X)=\sum_{x=0}^{7}P(X=x)t^x$ . Substituting the binomial probability gives $\sum_{x=0}^{7}\binom7x(0.45)^x(0.55)^{7-x}t^x=\sum_{x=0}^{7}\binom7x(0.45t)^x(0.55)^{7-x}=(0.55+0.45t)^7$ by the binomial theorem. Hence $P(X\leq1)$ is the sum of the constant and linear coefficients: $(0.55)^7+7(0.45)(0.55)^6=0.1024183703\ldots$ , which is $0.1024$ to $4$ decimal places.

Tier 3 · Hard

Mark scheme for FS1-7.1 Tier 3 · Hard
Question	Scheme	Marks
1	$G_X(t)=\left(\dfrac{0.25t}{1-0.75t}\right)^3$ $P(X=6)=0.06592$ to $4$ significant figures	7
(7 marks)		7
Notes
A waiting time to the third success is the sum of three independent geometric waiting times, so its PGF is the cube of the geometric PGF: $G_X(t)=[0.25t/(1-0.75t)]^3$ . Using $(1-z)^{-3}=\sum_{j=0}^{\infty}\binom{j+2}{2}z^j$ , the coefficient of $t^6$ is $\binom52(0.25)^3(0.75)^3=0.06591797\ldots$ .
2	$G_X(t)=e^{-\lambda}\sum_{x=0}^{\infty}\lambda^xt^x/x!$ $P(X=4)/P(X=3)=\lambda/4$ $\lambda/4=0.7$ $\lambda=2.8$ $P(X\leq1)=e^{-2.8}(1+2.8)$ $P(X\leq1)=0.2311$ to $4$ decimal places	6
(6 marks)		6
Notes
Expanding the PGF gives coefficient $P(X=x)=e^{-\lambda}\lambda^x/x!$ . Hence $P(X=4)/P(X=3)=\lambda/4=0.7$ , so $\lambda=2.8$ . Therefore $P(X\leq1)=e^{-2.8}[1+2.8]=0.2310782379\ldots$ , which rounds to $0.2311$ .
3	$W\sim B(6,p)$ $P(W=0)=(1-p)^6$ $(1-p)^6=64/729=(2/3)^6$ $p=1/3$ $G_W(-1)=P(W\text{ is even})-P(W\text{ is odd})$ $G_W(-1)=(1-2p)^6=1/729$ $P(W\text{ is even})=\tfrac12(1+1/729)=365/729$	7
(7 marks)		7
Notes
The standard binomial probability generating function identifies $W\sim B(6,p)$ . Its constant coefficient is $P(W=0)=(1-p)^6$ . Since $0<p<1$ , the equation $(1-p)^6=64/729=(2/3)^6$ gives $p=1/3$ . At $t=-1$ , the even-power coefficients enter with sign $+1$ and the odd-power coefficients with sign $-1$ , so $G_W(-1)=P(W\text{ is even})-P(W\text{ is odd})$ . Also $G_W(-1)=(1-2p)^6=(1/3)^6=1/729$ . Together with the total probability $1$ , this gives $P(W\text{ is even})=(1+1/729)/2=365/729$ .
4	$G_X(t)=\dfrac{pt}{1-(1-p)t}$ $G_X(1/2)=\dfrac{p}{1+p}=1/3$ $3p=1+p$ , so $p=1/2$ $P(X>4)=(1-p)^4$ $P(X>4)=1/16$ $P(X>k)=(1/2)^k$ , so $(1/2)^k<0.01$ Since $(1/2)^6=0.015625$ and $(1/2)^7=0.0078125$ , the least integer is $k=7$	7
(7 marks)		7
Notes
For a geometric variable that counts the trial of the first success, $G_X(t)=pt/[1-(1-p)t]$ . Substitution of $t=1/2$ gives $G_X(1/2)=p/(1+p)$ . Equating this to $1/3$ gives $3p=1+p$ and hence $p=1/2$ . The event $X>4$ means that the first four trials are failures, so $P(X>4)=(1-p)^4=(1/2)^4=1/16$ . More generally, $P(X>k)=(1/2)^k$ . Since $(1/2)^6=0.015625$ is not below $0.01$ but $(1/2)^7=0.0078125$ is, the least possible integer is $k=7$ .
5	$P(X=1)=a/4+27(1-a)/64$ $a/4+27(1-a)/64=35/96$ $a=1/3$ The first term contributes $(1/3)\binom42/2^4=1/8$ to $P(X=2)$ The second term contributes $(2/3)\binom32(1/4)^2(3/4)=3/32$ to $P(X=2)$ $P(X=2)=1/8+3/32=7/32$ $P(X=4)=(1/3)(1/2)^4=1/48$	7
(7 marks)		7
Notes
The coefficient of $t$ is $a\binom41/2^4+(1-a)\binom31(3/4)^2(1/4)=a/4+27(1-a)/64$ . Equating this to $35/96$ gives $a=1/3$ . The coefficient of $t^2$ is $(1/3)\binom42/2^4+(2/3)\binom32(3/4)(1/4)^2=1/8+3/32=7/32$ . Only the first component has a $t^4$ term, so $P(X=4)=(1/3)(1/2)^4=1/48$ .

FS1-7.2 · Use to find the mean and variance.

Tier 1 · Easy

Mark scheme for FS1-7.2 Tier 1 · Easy
Question	Scheme	Marks
1	$E(X)=1.2$ $\operatorname{Var}(X)=0.96$	4
(4 marks)		4
Notes
$G'_X(t)=1.2(0.8+0.2t)^5$ , so $G'_X(1)=1.2$ . Also $G''_X(1)=6(5)(0.2)^2=1.2$ . Hence $\operatorname{Var}(X)=1.2+1.2-1.2^2=0.96$ .
2	$G'_X(t)=2.5e^{2.5(t-1)}$ , so $G'_X(1)=2.5$ $G''_X(t)=6.25e^{2.5(t-1)}$ , so $G''_X(1)=6.25$ $E(X)=G'_X(1)=2.5$ $\operatorname{Var}(X)=G''_X(1)+G'_X(1)-[G'_X(1)]^2=6.25+2.5-2.5^2=2.5$	4
(4 marks)		4
Notes
Differentiation gives $G'_X(t)=2.5e^{2.5(t-1)}$ and $G''_X(t)=6.25e^{2.5(t-1)}$ . Hence $G'_X(1)=2.5$ and $G''_X(1)=6.25$ . Therefore $E(X)=G'_X(1)=2.5$ and $\operatorname{Var}(X)=G''_X(1)+G'_X(1)-[G'_X(1)]^2=6.25+2.5-6.25=2.5$ .

Tier 2 · Standard

Mark scheme for FS1-7.2 Tier 2 · Standard
Question	Scheme	Marks
1	$E(X)=\frac32$ $\operatorname{Var}(X)=\frac34$	5
(5 marks)		5
Notes
$G'_X(t)=6/(3-t)^2$ and $G''_X(t)=12/(3-t)^3$ . Thus $G'_X(1)=3/2$ and $G''_X(1)=3/2$ . Therefore $\operatorname{Var}(X)=3/2+3/2-(3/2)^2=3/4$ .
2	With $q=1-p$ , $G_X(t)=pt/(1-qt)$ $G'_X(t)=p/(1-qt)^2$ $G''_X(t)=2pq/(1-qt)^3$ $E(X)=G'_X(1)=1/p$ $\operatorname{Var}(X)=G''_X(1)+G'_X(1)-[G'_X(1)]^2=(1-p)/p^2$	5
(5 marks)		5
Notes
Put $q=1-p$ , so $G_X(t)=pt(1-qt)^{-1}$ . Differentiation and simplification give $G'_X(t)=p(1-qt)^{-2}$ and $G''_X(t)=2pq(1-qt)^{-3}$ . Since $1-q=p$ , $G'_X(1)=1/p$ and $G''_X(1)=2q/p^2$ . Hence $E(X)=1/p$ and $\operatorname{Var}(X)=2q/p^2+1/p-1/p^2=q/p^2=(1-p)/p^2$ .
3	$G_X(1)=1$ $G'_X(t)=3tG_X(t)$ $E(X)=G'_X(1)=3$ $G''_X(t)=(3+9t^2)G_X(t)$ , so $G''_X(1)=12$ $\operatorname{Var}(X)=12+3-3^2=6$	5
(5 marks)		5
Notes
Substitution gives $G_X(1)=e^0=1$ . Differentiating the exponential gives $G'_X(t)=3tG_X(t)$ , so $E(X)=G'_X(1)=3$ . Differentiating again gives $G''_X(t)=3G_X(t)+9t^2G_X(t)=(3+9t^2)G_X(t)$ , hence $G''_X(1)=12$ . Therefore $\operatorname{Var}(X)=G''_X(1)+G'_X(1)-[G'_X(1)]^2=12+3-9=6$ .

Tier 3 · Hard

Mark scheme for FS1-7.2 Tier 3 · Hard
Question	Scheme	Marks
1	$k=0.1$ $E(X)=2$ $\operatorname{Var}(X)=1$	6
(6 marks)		6
Notes
$G_X(1)=10k=1$ , so $k=0.1$ . Then $G'_X(t)=0.1(2+6t+12t^2)$ , giving $G'_X(1)=2$ , and $G''_X(t)=0.1(6+24t)$ , giving $G''_X(1)=3$ . Hence $\operatorname{Var}(X)=3+2-2^2=1$ .
2	$G'_X(t)=\dfrac{2(3+t)}{3(2-t)^3}$ $E(X)=G'_X(1)=8/3$ $G''_X(t)=\dfrac{2(11+2t)}{3(2-t)^4}$ $G''_X(1)=26/3$ $\operatorname{Var}(X)=26/3+8/3-(8/3)^2=38/9$ $P(X=0)=G_X(0)=1/12$	6
(6 marks)		6
Notes
Direct differentiation gives $G'_X(t)=2(3+t)/[3(2-t)^3]$ and $G''_X(t)=2(11+2t)/[3(2-t)^4]$ . Therefore $G'_X(1)=8/3$ and $G''_X(1)=26/3$ . It follows that $E(X)=8/3$ and $\operatorname{Var}(X)=G''_X(1)+G'_X(1)-[G'_X(1)]^2=26/3+8/3-64/9=38/9$ . The constant coefficient of a PGF is $P(X=0)$ , so $P(X=0)=G_X(0)=1/12$ .
3	$G_X(1)=1$ $G'_X(t)=G_X(t)\{2+a/(1-at)\}$ $G''_X(t)=G'_X(t)\{2+a/(1-at)\}+G_X(t)a^2/(1-at)^2$ $\operatorname{Var}(X)=2+a/(1-a)^2$ $4=2+a/(1-a)^2$ , so $2a^2-5a+2=0$ and $a=1/2$ or $2$ $0<a<1$ gives $a=1/2$ , and $E(X)=2+a/(1-a)=3$	6
(6 marks)		6
Notes
The function satisfies $G_X(1)=(1-a)/(1-a)=1$ . Differentiation gives $G'_X(t)=G_X(t)(2+a/(1-at))$ . Applying the product rule to this expression gives $G''_X(t)=G'_X(t)(2+a/(1-at))+G_X(t)a^2/(1-at)^2$ . Since $G_X(1)=1$ , $E(X)=G'_X(1)=2+a/(1-a)$ . Substitution into $\operatorname{Var}(X)=G''_X(1)+G'_X(1)-[G'_X(1)]^2$ simplifies to $\operatorname{Var}(X)=2+a/(1-a)^2$ . Hence $4=2+a/(1-a)^2$ , giving $2a^2-5a+2=0$ and therefore $a=1/2$ or $a=2$ . The condition $0<a<1$ gives $a=1/2$ , and then $E(X)=2+(1/2)/(1/2)=3$ .
4	$G_X(1)=k(m+1)=1$ , so $k=1/(m+1)$ $G'_X(1)=k\displaystyle\sum_{x=0}^{m}x=k\,m(m+1)/2$ $E(X)=m/2$ $G''_X(1)=k\displaystyle\sum_{x=0}^{m}x(x-1)=m(m-1)/3$ $\operatorname{Var}(X)=m(m-1)/3+m/2-m^2/4=m(m+2)/12$ $m/2=7/2$ , so $m=7$ $k=1/8$ and $\operatorname{Var}(X)=21/4$	7
(7 marks)		7
Notes
Normalisation gives $k=1/(m+1)$ . Using $\sum x=m(m+1)/2$ and $\sum x(x-1)=m(m+1)(m-1)/3$ , the PGF derivatives give $G'_X(1)=m/2$ and $G''_X(1)=m(m-1)/3$ . The given mean therefore gives $m/2=7/2$ , so $m=7$ and $k=1/8$ . Finally $\operatorname{Var}(X)=G''_X(1)+G'_X(1)-[G'_X(1)]^2=m(m+2)/12=21/4$ .
5	$G'_X(t)=c\lambda e^{\lambda(t-1)}$ , so $c\lambda=2$ $G''_X(t)=c\lambda^2e^{\lambda(t-1)}$ , so $G''_X(1)=c\lambda^2$ $5=c\lambda^2+2-2^2$ , so $c\lambda^2=7$ Dividing $c\lambda^2=7$ by $c\lambda=2$ gives $\lambda=7/2$ $c=2/\lambda=4/7$ $P(X=0)=G_X(0)=1-c+ce^{-\lambda}=3/7+(4/7)e^{-7/2}$ $P(X=0)=0.4458$ to $4$ decimal places	7
(7 marks)		7
Notes
Differentiation gives $G'_X(1)=c\lambda$ and $G''_X(1)=c\lambda^2$ . The mean condition gives $c\lambda=2$ , while the variance identity gives $5=c\lambda^2+2-4$ , so $c\lambda^2=7$ . Division yields $\lambda=7/2$ and then $c=4/7$ . The constant coefficient is $G_X(0)=1-c+ce^{-\lambda}=3/7+(4/7)e^{-7/2}=0.4458270762\ldots$ , which is $0.4458$ to $4$ decimal places.

FS1-7.3 · Probability generating function of the sum of independent random variables.

Tier 1 · Easy

Mark scheme for FS1-7.3 Tier 1 · Easy
Question	Scheme	Marks
1	$X+Y\sim\operatorname{Po}(4)$ $P(X+Y=0)=e^{-4}=0.01832$	4
(4 marks)		4
Notes
$G_X(t)G_Y(t)=e^{1.4(t-1)}e^{2.6(t-1)}=e^{4(t-1)}$ , which is the PGF of $\operatorname{Po}(4)$ . Its constant coefficient is $P(X+Y=0)=e^{-4}=0.0183156\ldots$ .
2	$G_X(t)=0.7+0.3t$ $G_Y(t)=e^{1.2(t-1)}$ $G_S(t)=(0.7+0.3t)e^{1.2(t-1)}$ $P(S=2)=e^{-1.2}[0.7(1.2)^2/2!+0.3(1.2)]$ $P(S=2)=0.2602$ to $4$ decimal places	5
(5 marks)		5
Notes
Independence gives $G_S(t)=G_X(t)G_Y(t)=(0.7+0.3t)e^{1.2(t-1)}$ . The coefficient of $t^2$ is $e^{-1.2}[0.7(1.2)^2/2!+0.3(1.2)]=0.2602317991\ldots$ , which rounds to $0.2602$ .

Tier 2 · Standard

Mark scheme for FS1-7.3 Tier 2 · Standard
Question	Scheme	Marks
1	$S\sim\operatorname{Bin}(12,0.4)$ $P(S=3)=0.1419$ to $4$ significant figures	5
(5 marks)		5
Notes
$G_S(t)=(0.6+0.4t)^5(0.6+0.4t)^7=(0.6+0.4t)^{12}$ , so $S\sim\operatorname{Bin}(12,0.4)$ . The coefficient of $t^3$ is $\binom{12}{3}(0.4)^3(0.6)^9=0.14189396\ldots$ .
2	$G_X(t)=G_Y(t)=\left(\dfrac{0.25t}{1-0.75t}\right)^2$ By independence, $G_S(t)=G_X(t)G_Y(t)=\left(\dfrac{0.25t}{1-0.75t}\right)^4$ $S$ is negative binomial with parameters $r=4$ and $p=0.25$ $P(S=s)=\binom{s-1}{3}(0.25)^4(0.75)^{s-4}$ for $s=4,5,6,\ldots$ $P(S=5)=\binom{4}{3}(0.25)^4(0.75)$ $P(S=5)=3/256$	6
(6 marks)		6
Notes
Each variable counts the trial of the second success, so each has PGF $[0.25t/(1-0.75t)]^2$ . Independence gives $G_S(t)=[0.25t/(1-0.75t)]^4$ , the PGF of the trial count for the fourth success. Thus $P(S=s)=\binom{s-1}{3}(0.25)^4(0.75)^{s-4}$ for $s=4,5,6,\ldots$ . In particular, $P(S=5)=\binom43(0.25)^4(0.75)=4(1/256)(3/4)=3/256$ .
3	$G_X(t)=(0.6+0.4t)^3$ $G_S(t)=G_X(t)G_Y(t)$ by independence $G_Y(t)=G_S(t)/G_X(t)=e^{1.7(t-1)}$ $Y\sim\operatorname{Po}(1.7)$ $P(Y\geq2)=1-e^{-1.7}(1+1.7)$ $P(Y\geq2)=0.507$ to $3$ significant figures	6
(6 marks)		6
Notes
For $X\sim B(3,0.4)$ , $G_X(t)=(0.6+0.4t)^3$ . Independence gives $G_S(t)=G_X(t)G_Y(t)$ , so division by $G_X(t)$ yields $G_Y(t)=e^{1.7(t-1)}$ . This is the probability generating function of $\operatorname{Po}(1.7)$ . Therefore $P(Y\geq2)=1-P(Y=0)-P(Y=1)=1-e^{-1.7}(1+1.7)=0.5067544851\ldots$ , which is $0.507$ to $3$ significant figures.

Tier 3 · Hard

Mark scheme for FS1-7.3 Tier 3 · Hard
Question	Scheme	Marks
1	$S$ is negative binomial with $r=2$ , $p=0.4$ $P(S>5)=0.33696$	7
(7 marks)		7
Notes
$G_S(t)=\left(\dfrac{0.4t}{1-0.6t}\right)^2$ , the PGF of the trial of the second success. Hence $P(S>5)=1-\sum_{s=2}^{5}\binom{s-1}{1}(0.4)^2(0.6)^{s-2}=0.33696$ .
2	$G_X(t)=e^{0.8(t-1)}$ , $G_Y(t)=e^{1.6(t-1)}$ and $G_Z(t)=e^{2.3(t-1)}$ By independence, $G_S(t)=G_X(t)G_Y(t)G_Z(t)$ $G_S(t)=e^{(0.8+1.6+2.3)(t-1)}=e^{4.7(t-1)}$ $S\sim\operatorname{Po}(4.7)$ $P(S\geq7)=1-e^{-4.7}\displaystyle\sum_{s=0}^{6}\dfrac{4.7^s}{s!}$ $P(S\geq7)=0.1954$ to $4$ decimal places	6
(6 marks)		6
Notes
The three PGFs are $e^{0.8(t-1)}$ , $e^{1.6(t-1)}$ and $e^{2.3(t-1)}$ . Independence gives $G_S(t)=e^{(0.8+1.6+2.3)(t-1)}=e^{4.7(t-1)}$ , the PGF of $\operatorname{Po}(4.7)$ . Therefore $P(S\geq7)=1-P(S\leq6)=1-e^{-4.7}\sum_{s=0}^{6}4.7^s/s!=0.1953949169\ldots$ , which rounds to $0.1954$ .
3	$G_X(t)=(1/2+t/2)^3$ $G_Y(t)=3t/(8-5t)$ $G_S(t)=(1/2+t/2)^3\,3t/(8-5t)$ by independence $P(S=4)=375/32768+225/4096+45/512+3/64$ $P(S=4)=6591/32768$ $P(X=2,S=4)=P(X=2,Y=2)=45/512$ $P(X=2\mid S=4)=(45/512)/(6591/32768)=960/2197$	7
(7 marks)		7
Notes
The two probability generating functions are $(1/2+t/2)^3$ and $3t/(8-5t)$ ; the latter is normalised because $G_Y(1)=3/(8-5)=1$ . Independence gives $G_S(t)=(1/2+t/2)^3\,3t/(8-5t)$ . The coefficient of $t^4$ is the convolution over $(X,Y)=(0,4),(1,3),(2,2),(3,1)$ : $375/32768+225/4096+45/512+3/64=6591/32768$ . The event $X=2$ together with $S=4$ requires $Y=2$ , and its probability is $[\binom32/2^3][(3/8)(5/8)]=45/512$ . Therefore the conditional probability is $(45/512)/(6591/32768)=960/2197$ .
4	$G_X(t)=(2/3+t/3)^2$ $G_Y(t)=(1/2+t/2)^4$ $G_S(t)=(2/3+t/3)^2(1/2+t/2)^4$ by independence The two distinct linear factors have roots $-2$ and $-1$ , whereas a non-degenerate binomial PGF has only one repeated root $P(S=3)=P(X=0,Y=3)+P(X=1,Y=2)+P(X=2,Y=1)$ $P(S=3)=1/9+1/6+1/36$ $P(S=3)=11/36$	7
(7 marks)		7
Notes
Independence gives the product $G_S(t)=(2/3+t/3)^2(1/2+t/2)^4$ . A binomial PGF $(1-p+pt)^n$ with $0<p<1$ is a power of one linear factor, but this product has distinct roots $-2$ and $-1$ , so it is not binomial. The $t^3$ coefficient is the convolution over $(X,Y)=(0,3),(1,2),(2,1)$ , giving $(4/9)(1/4)+(4/9)(3/8)+(1/9)(1/4)=1/9+1/6+1/36=11/36$ .
5	$G_S(t)=G_X(t)G_Y(t)$ by independence $G_S(t)=t(3+t^2)^3/64$ $G_S(t)=(27t+27t^3+9t^5+t^7)/64$ The set of possible values of $S$ is $\{1,3,5,7\}$ $P(S\geq5)=(9+1)/64$ $P(S\geq5)=5/32$	6
(6 marks)		6
Notes
Independence gives $G_S(t)=G_X(t)G_Y(t)=t(3+t^2)^3/64$ . Direct expansion gives $G_S(t)=(27t+27t^3+9t^5+t^7)/64$ , so the non-zero coefficients show that the support is $\{1,3,5,7\}$ . Adding the coefficients of $t^5$ and $t^7$ gives $P(S\geq5)=(9+1)/64=5/32$ .