FS1-1 Discrete probability distributions — revision question pack

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

Checked against Edexcel 9FM0 section FS1-1. 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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FS1-1.1 · Calculation of the mean and variance of discrete probability distributions. Extension of expected value function to include E(g(X)).

Explanation

For a discrete random variable, $E(X)=\sum xP(X=x)$ and $\operatorname{Var}(X)=E(X^2)-[E(X)]^2$ . The probabilities over the full support must sum to $1$ .
For any function $g$ , calculate $E[g(X)]=\sum g(x)P(X=x)$ by applying $g$ to each support value before weighting. Algebra can reduce repeated work: $E[(X-a)^2]=E(X^2)-2aE(X)+a^2$ .
In general $E[g(X)]\ne g(E(X))$.
Questions may ask whether a model is suitable, so the calculated mean, spread or transformed expectation must be interpreted against the context.
Examiners expect a complete probability table and clear separation of $E(X^2)$ from $[E(X)]^2$ .

Worked example

$X$ takes values $-1,2,4$ with probabilities $0.3,0.5,0.2$ . Find $E(X)$ , $\operatorname{Var}(X)$ and $E[(X-1)^2]$ .

1. $E(X)=-1(0.3)+2(0.5)+4(0.2)=1.5$ .
2. $E(X^2)=1(0.3)+4(0.5)+16(0.2)=5.5$ , so $\operatorname{Var}(X)=5.5-1.5^2=3.25$ .
3. $E[(X-1)^2]=E(X^2)-2E(X)+1=5.5-3+1=3.5$ .

Answer: $E(X)=1.5$ , $\operatorname{Var}(X)=3.25$ , and $E[(X-1)^2]=3.5$ .

Common mistakes

Don't subtract $E(X)$ rather than $[E(X)]^2$ when calculating the variance.
Don't replace $E[(X-1)^2]$ by $(E(X)-1)^2$ .
Don't omit a support value whose probability is non-zero.

Exam tip

Add columns for $xP(X=x)$ and $x^2P(X=x)$ so the two expectation totals remain visible.

Tier 1 · Easy

1.
The random variable $X$ takes values $0,1,2,4$ with probabilities $0.15,0.25,0.35,0.25$ respectively. Find $E(X)$ and $\operatorname{Var}(X)$ .
(4)
(Total for Question 1 is 4 marks)
2.
The random variable $X$ takes values $0,1,4,9$ with probabilities $0.1,0.3,0.4,0.2$ respectively. A service cost is $g(x)=5+4\sqrt{x}$ , giving the cost table $g(0)=5$ , $g(1)=9$ , $g(4)=13$ , $g(9)=17$ . Find $E[g(X)]$ and compare it with $g(E[X])$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
The random variable $X$ takes values $-2,1,3$ with probabilities $0.2,0.5,0.3$ respectively. Calculate $E[(X+2)^2]$ .
(4)
(Total for Question 1 is 4 marks)
2.
A proposed distribution for $X$ assigns probabilities $a,b,0.25,0.15$ to the values $-1,0,2,4$ respectively. Given that $E(X)=0.8$ , find $a$ and $b$ . An analyst wants to use $X$ for the number of failed modules in a device. Comment on the suitability of this model.
(4)
(Total for Question 2 is 4 marks)
3.
The random variable $X$ takes values $-2,0,1,4$ with probabilities $0.2,a,b,0.3$ respectively. Given that $P(X=0\mid X\leq1)=2/7$ , find $a$ , $b$ , $E(X)$ and $\operatorname{Var}(X)$ .
(6)
(Total for Question 3 is 6 marks)

Tier 3 · Hard

1.
The random variable $X$ takes values $0,1,3,5$ with probabilities $k,2k,3k,4k$ respectively. Find $k$ , $E(X)$ , $\operatorname{Var}(X)$ and $E[(X-2)^2]$ .
(7)
(Total for Question 1 is 7 marks)
2.
The random variable $X$ takes values $0,1,2,4$ with probabilities $a,b,c,0.1$ respectively. It is known that $E(X)=1.4$ and $E[X(X-1)]=1.9$ . Determine $a,b,c$ and $\operatorname{Var}(X)$ . The same distribution is proposed for repair counts on shifts lasting between $6$ and $10$ hours; comment on this proposal.
(7)
(Total for Question 2 is 7 marks)
3.
The random variable $X$ takes values $0,2,3,7$ with probabilities $a,b,c,0.2$ respectively. It is known that $P(X\leq2\mid X<7)=3/8$ and $P(X=3\mid X>0)=5/9$ . Determine $a$ , $b$ and $c$ , then find $E[(X-1)(X-4)]$ and the standard deviation of $X$ .
(8)
(Total for Question 3 is 8 marks)
4.
The random variable $X$ takes values $0,2,5$ with probabilities $p,q,1-p-q$ respectively, where probabilities may be zero. Given that $E(X)=3$ , express $q$ and $1-p-q$ in terms of $p$ , and find the set of possible values of $p$ . Determine the greatest possible value of $\operatorname{Var}(X)$ and the corresponding probabilities. For the distribution of greatest variance, find $E\!\left(\dfrac{1}{X+1}\right)$ exactly.
(8)
(Total for Question 4 is 8 marks)
5.
The random variable $X$ takes values $0,1,3,4$ with probabilities $0.1,0.4,0.3,0.2$ respectively. Two daily-cost models, in pounds, are $A=50+8X$ and $B=20+6X^2+cI$ , where $I=1$ when $X\geq3$ and $I=0$ otherwise. Find the value of $c$ for which $E(A)=E(B)$ , then find $\operatorname{Var}(B)$ for this value of $c$ .
(8)
(Total for Question 5 is 8 marks)

Answer key

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

FS1-1.1 · Calculation of the mean and variance of discrete probability distributions. Extension of expected value function to include E(g(X)).

Tier 1 · Easy

Mark scheme for FS1-1.1 Tier 1 · Easy
Question	Scheme	Marks
1	$E(X)=1.95$ $\operatorname{Var}(X)=1.8475$	4
(4 marks)		4
Notes
$E(X)=0(0.15)+1(0.25)+2(0.35)+4(0.25)=1.95$ . Also $E(X^2)=0+1(0.25)+4(0.35)+16(0.25)=5.65$ . Hence $\operatorname{Var}(X)=5.65-1.95^2=1.8475$ .
2	$E[g(X)]=5(0.1)+9(0.3)+13(0.4)+17(0.2)=11.8$ $E(X)=0(0.1)+1(0.3)+4(0.4)+9(0.2)=3.7$ $g(E[X])=5+4\sqrt{3.7}=12.694\ldots$ $E[g(X)]\ne g(E[X])$ for this non-linear cost function	4
(4 marks)		4
Notes
Apply the non-polynomial cost function before averaging: $E[g(X)]=0.5+2.7+5.2+3.4=11.8$ . In contrast, $E(X)=0.3+1.6+1.8=3.7$ , so $g(E[X])=5+4\sqrt{3.7}=12.694153\ldots$ . The two operations give different values.

Tier 2 · Standard

Mark scheme for FS1-1.1 Tier 2 · Standard
Question	Scheme	Marks
1	$E[(X+2)^2]=12$	4
(4 marks)		4
Notes
Evaluate the function at each support value: $(X+2)^2$ is $0,9,25$ . Therefore $E[(X+2)^2]=0(0.2)+9(0.5)+25(0.3)=12$ .
2	$a+b=0.6$ $a=0.3$ $b=0.3$ The model is unsuitable because it gives the impossible count $-1$ positive probability	4
(4 marks)		4
Notes
Total probability gives $a+b+0.25+0.15=1$ , so $a+b=0.6$ . Also $E(X)=-a+2(0.25)+4(0.15)=1.1-a=0.8$ , hence $a=0.3$ and $b=0.3$ . Although these are valid probabilities, a count of failed modules cannot be negative, while the model assigns $P(X=-1)=0.3$ .
3	$a+b=0.5$ $P(X\leq1)=0.7$ , so $\dfrac{a}{0.7}=\dfrac27$ $a=0.2$ and $b=0.3$ $E(X)=1.1$ $E(X^2)=5.9$ $\operatorname{Var}(X)=4.69$	6
(6 marks)		6
Notes
Total probability gives $a+b=1-0.2-0.3=0.5$ . Therefore $P(X\leq1)=0.2+a+b=0.7$ , so the conditional probability gives $a/0.7=2/7$ and hence $a=0.2$ and $b=0.3$ . Then $E(X)=-2(0.2)+1(0.3)+4(0.3)=1.1$ and $E(X^2)=4(0.2)+1(0.3)+16(0.3)=5.9$ . Thus $\operatorname{Var}(X)=5.9-1.1^2=4.69$ .

Tier 3 · Hard

Mark scheme for FS1-1.1 Tier 3 · Hard
Question	Scheme	Marks
1	$k=0.1$ $E(X)=3.1$ $\operatorname{Var}(X)=3.29$ $E[(X-2)^2]=4.5$	7
(7 marks)		7
Notes
$10k=1$ , so $k=0.1$ . Then $E(X)=0+1(0.2)+3(0.3)+5(0.4)=3.1$ and $E(X^2)=1(0.2)+9(0.3)+25(0.4)=12.9$ . Thus $\operatorname{Var}(X)=12.9-3.1^2=3.29$ . Finally, $E[(X-2)^2]=E(X^2)-4E(X)+4=12.9-12.4+4=4.5$ .
2	$a+b+c=0.9$ $b+2c=1$ $c=0.35$ $b=0.30$ $a=0.25$ $\operatorname{Var}(X)=1.34$ A single distribution is doubtful when the exposure time varies unless duration has no effect or is adjusted for	7
(7 marks)		7
Notes
The probability total is $a+b+c=0.9$ . The mean condition gives $b+2c+4(0.1)=1.4$ , so $b+2c=1$ . Since $x(x-1)$ is $0,0,2,12$ on the support, $2c+12(0.1)=1.9$ , giving $c=0.35$ , $b=0.30$ and $a=0.25$ . Also $E(X^2)=b+4c+16(0.1)=0.30+1.40+1.60=3.30$ , so $\operatorname{Var}(X)=3.30-1.4^2=1.34$ . Repair opportunities normally increase with shift length, so applying one unchanged count distribution to unequal exposures needs justification or a duration adjustment.
3	$a+b+c=0.8$ $\dfrac{a+b}{0.8}=\dfrac38$ , so $a+b=0.3$ $c=0.5$ $\dfrac{c}{b+c+0.2}=\dfrac59$ $b=0.2$ $a=0.1$ $E[(X-1)(X-4)]=2.6$ $\operatorname{sd}(X)=\sqrt{4.21}=2.05$ to $3$ significant figures	8
(8 marks)		8
Notes
The probability total is $a+b+c=0.8$ . Since $P(X<7)=0.8$ , the first conditional probability gives $a+b=0.3$ , so $c=0.5$ . The second condition is $c/(b+c+0.2)=5/9$ , giving $0.5/(b+0.7)=5/9$ and hence $b=0.2$ and $a=0.1$ . The values of $(X-1)(X-4)$ are $4,-2,-2,18$ , so the required expectation is $4(0.1)-2(0.2)-2(0.5)+18(0.2)=2.6$ . Also $E(X)=3.3$ and $E(X^2)=15.1$ , so $\operatorname{Var}(X)=15.1-3.3^2=4.21$ and $\operatorname{sd}(X)=\sqrt{4.21}=2.0518\ldots$ .
4	$2q+5(1-p-q)=3$ $q=(2-5p)/3$ $1-p-q=(1+2p)/3$ $0\leq p\leq2/5$ $E(X^2)=11+10p$ $\operatorname{Var}(X)=2+10p$ The greatest variance is $6$ , attained when $(p,q,1-p-q)=(2/5,0,3/5)$ $E\!\left(\dfrac{1}{X+1}\right)=\dfrac12$	8
(8 marks)		8
Notes
The mean condition is $2q+5(1-p-q)=3$ , so $q=(2-5p)/3$ and the remaining probability is $(1+2p)/3$ . Non-negative probabilities require $0\leq p\leq2/5$ . Also $E(X^2)=4q+25(1-p-q)=11+10p$ , hence $\operatorname{Var}(X)=11+10p-3^2=2+10p$ . This increases with $p$ , so its greatest value is $6$ at $p=2/5$ , where the probabilities are $(2/5,0,3/5)$ . The required expectation is then $(2/5)/1+(3/5)/6=1/2$ .
5	$E(X)=2.1$ $E(X^2)=6.3$ $P(X\geq3)=0.5$ $E(A)=66.8$ $E(B)=57.8+0.5c$ $c=18$ $E(B^2)=6440.8$ $\operatorname{Var}(B)=1978.56$	8
(8 marks)		8
Notes
The distribution gives $E(X)=2.1$ , $E(X^2)=6.3$ and $E(I)=P(X\geq3)=0.5$ . Thus $E(A)=50+8(2.1)=66.8$ and $E(B)=20+6(6.3)+0.5c=57.8+0.5c$ , so equality requires $c=18$ . The corresponding values of $B$ are $20,26,92,134$ . Hence $E(B^2)=0.1(20^2)+0.4(26^2)+0.3(92^2)+0.2(134^2)=6440.8$ , and $\operatorname{Var}(B)=6440.8-66.8^2=1978.56$ .