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Edexcel A-level Further Maths revision notes

Discrete probability distributions

Section 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 1

Checked against Edexcel 9FM0 section 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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Year 1
Year 1: this is the AS subject content the exam board publishes, which is what most schools teach in Year 12.

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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