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Edexcel A-level Further Maths revision notes

Discrete probability distributions

Section FS1-1
Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.
1 specification point

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

Checked against Edexcel 9FM0 section FS1-1

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)).

Notes
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Explanation

  • For a discrete random variable, $E(X)=\sum xP(X=x)$ and Var⁡(X)=E(X2)−[E(X)]2\operatorname{Var}(X)=E(X^2)-[E(X)]^2. The probabilities over the full support must sum to 11.
  • For any function gg, calculate $E[g(X)]=\sum g(x)P(X=x)$ by applying gg to each support value before weighting. Algebra can reduce repeated work: E[(X−a)2]=E(X2)−2aE(X)+a2E[(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(X2)E(X^2) from [E(X)]2[E(X)]^2.
Worked example

XX takes values −1,2,4-1,2,4 with probabilities 0.3,0.5,0.20.3,0.5,0.2. Find E(X)E(X), Var⁡(X)\operatorname{Var}(X) and E[(X−1)2]E[(X-1)^2].

  1. 1.E(X)=−1(0.3)+2(0.5)+4(0.2)=1.5E(X)=-1(0.3)+2(0.5)+4(0.2)=1.5.
  2. 2.E(X2)=1(0.3)+4(0.5)+16(0.2)=5.5E(X^2)=1(0.3)+4(0.5)+16(0.2)=5.5, so Var⁡(X)=5.5−1.52=3.25\operatorname{Var}(X)=5.5-1.5^2=3.25.
  3. 3.E[(X−1)2]=E(X2)−2E(X)+1=5.5−3+1=3.5E[(X-1)^2]=E(X^2)-2E(X)+1=5.5-3+1=3.5.

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

Common mistakes

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

Exam tip

Add columns for xP(X=x)xP(X=x) and x2P(X=x)x^2P(X=x) so the two expectation totals remain visible.

Tier 1 · Easy

ORIGINAL

1.

The random variable XX takes values 0,1,2,40,1,2,4 with probabilities 0.15,0.25,0.35,0.250.15,0.25,0.35,0.25 respectively. Find E(X)E(X) and Var⁡(X)\operatorname{Var}(X).

(4)

(Total for Question 1 is 4 marks)

Tier 2 · Standard

ORIGINAL

1.

The random variable XX takes values −2,1,3-2,1,3 with probabilities 0.2,0.5,0.30.2,0.5,0.3 respectively. Calculate E[(X+2)2]E[(X+2)^2].

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1.

The random variable XX takes values 0,1,3,50,1,3,5 with probabilities k,2k,3k,4kk,2k,3k,4k respectively. Find kk, E(X)E(X), Var⁡(X)\operatorname{Var}(X) and E[(X−2)2]E[(X-2)^2].

(7)

(Total for Question 1 is 7 marks)

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