Discrete probability distributions

CIE A-Level Further Mathematics· Unit 4: Further Probability & Statistics· 15 min read

1. Core Definitions & Properties★★☆☆☆⏱ 5 min

📘 Definition

Discrete Probability Distribution

Described by a probability mass function (PMF) $p (x) = P (X = x)$

A distribution that describes all possible probabilities for a discrete random variable, which can only take a countable set of distinct values.

All probabilities satisfy $0 \leq P (X = x_{i}) \leq 1$ for all outcomes $x_{i}$
The sum of all probabilities equals 1: $\sum_{i} P (X = x_{i}) = 1$
The cumulative distribution function (CDF) $F (x) = P (X \leq x)$ is non-decreasing

📐 Worked Example

A discrete random variable $X$ has PMF $P (X = x) = k x$ for $x = 1, 2, 3$ , and 0 otherwise. Find the value of $k$ .

1
Use the core property that the sum of all probabilities equals 1:
2
$P(X=1) + P(X=2) + P(X=3) = 1 \\ k(1) + k(2) + k(3) = 1$
3
Simplify and solve for $k$ :
4
$6k = 1 \implies k = \frac{1}{6}$

Exam tip:

Always check that your probabilities sum to 1 after finding an unknown constant, this catches simple arithmetic errors.

2. Expectation and Variance★★★☆☆⏱ 6 min

📘 Definition

Expected Value & Variance

$E [X]$ , $V a r (X)$

Expected value is the long-run average of the distribution. Variance is a measure of spread around the expected value.

Example:

For discrete distributions: $E [X] = \sum x P (X = x)$ , $V a r (X) = E [X^{2}] - (E [X])^{2}$

📐 Worked Example

For the distribution $P (X = x) = \frac{x}{6}$ for $x = 1, 2, 3$ , find $E [X]$ and $V a r (X)$ .

1
Calculate $E [X]$ by summing $x \cdot P (X = x)$ :
2
$E[X] = \left(1 \cdot \frac{1}{6}\right) + \left(2 \cdot \frac{2}{6}\right) + \left(3 \cdot \frac{3}{6}\right) = \frac{1 + 4 + 9}{6} = \frac{7}{3}$
3
Calculate $E [X^{2}]$ for the variance shortcut formula:
4
$E[X^2] = \left(1^2 \cdot \frac{1}{6}\right) + \left(2^2 \cdot \frac{2}{6}\right) + \left(3^2 \cdot \frac{3}{6}\right) = \frac{36}{6} = 6$
5
Substitute into the variance formula:
6
$Var(X) = E[X^2] - (E[X])^2 = 6 - \left(\frac{7}{3}\right)^2 = \frac{5}{9}$

Exam tip:

Always use the shortcut formula $E [X^{2}] - (E [X])^{2}$ for variance, it is much faster than expanding $E [(X - E [X])^{2}]$ in exams.

3. Probability Generating Functions (PGFs)★★★★☆⏱ 7 min

📘 Definition

Probability Generating Function

$G_{X} (t) = E [t^{X}] = \sum_{x = 0}^{\infty} P (X = x) t^{x}$

A function that encodes the full probability distribution of a non-negative integer-valued discrete random variable.

$G_{X} (1) = 1$
$E [X] = G_{X}^{'} (1)$ (first derivative at $t = 1$ )
$V a r (X) = G_{X}^{''} (1) + G_{X}^{'} (1) - (G_{X}^{'} (1))^{2}$
For independent $X, Y$ : $G_{X + Y} (t) = G_{X} (t) G_{Y} (t)$

📐 Worked Example

Find the PGF of the distribution $X$ from the previous example, and use it to confirm $E [X]$ .

1
Write the PGF using the definition:
2
$G_X(t) = P(X=1)t^1 + P(X=2)t^2 + P(X=3)t^3 = \frac{t + 2t^2 + 3t^3}{6}$
3
Differentiate $G_{X} (t)$ with respect to $t$ :
4
$G_X'(t) = \frac{1 + 4t + 9t^2}{6}$
5
Evaluate at $t = 1$ to get $E [X]$ :
6
$E[X] = G_X'(1) = \frac{1 + 4 + 9}{6} = \frac{14}{6} = \frac{7}{3}$
7
This matches our earlier calculation, so the result is confirmed.

Exam tip:

PGFs only work for non-negative integer-valued discrete random variables. Do not use them for variables that can take negative values.

4. Common Discrete Distributions★★★☆☆⏱ 4 min

Distribution	PMF	$E [X]$	$V a r (X)$	PGF
Discrete Uniform $(1.. n)$	$P (X = k) = \frac{1}{n}$	$\frac{n + 1}{2}$	$\frac{n ^{2} - 1}{12}$	$\frac{t ( 1 - t ^{n} )}{n ( 1 - t )}$
Bernoulli $(p)$	$P (1) = p, P (0) = 1 - p$	$p$	$p (1 - p)$	$1 - p + pt$
Binomial $(n, p)$	$(k n) p^{k} (1 - p)^{n - k}$	$n p$	$n p (1 - p)$	$(1 - p + pt)^{n}$
Poisson $(λ)$	$\frac{e ^{- λ} λ ^{k}}{k !}$	$λ$	$λ$	$e^{λ (t - 1)}$

5. Common Pitfalls

Wrong move:

Forgetting to check that the sum of probabilities equals 1 after finding an unknown constant

Why:

Arithmetic errors are common when solving for constants, and skipping the check leads to unnecessary lost marks

Correct move:

Always add up your final probabilities to confirm they sum to 1, this takes 10 seconds and catches simple mistakes

Wrong move:

Writing the variance formula as $V a r (X) = E [X^{2}] - E [X]$ instead of $E [X^{2}] - (E [X])^{2}$

Why:

This is a common algebraic slip when writing formulas quickly under exam pressure

Correct move:

Memorize the formula as 'expectation of X squared minus (expectation of X) all squared' and double-check the exponent

Wrong move:

Using a probability generating function for a discrete random variable that takes negative values

Why:

PGFs are only defined for non-negative integer-valued random variables, so results will be incorrect

Correct move:

Use direct calculation or moment generating functions for discrete variables that can take negative values

Wrong move:

Multiplying PGFs for dependent random variables to get the PGF of their sum

Why:

The product rule for PGFs of sums only holds for independent random variables

Correct move:

Only use $G_{X + Y} (t) = G_{X} (t) G_{Y} (t)$ if the question confirms X and Y are independent, otherwise calculate directly

Wrong move:

Confusing the probability mass function (PMF) with the cumulative distribution function (CDF)

Why:

Students often mix these up when a question asks for one specifically

Correct move:

Remember PMF = $P (X = x)$ , CDF = $P (X \leq x)$ , label your answer clearly to avoid confusion

6. Quick Reference Cheatsheet

Concept	Key Formula
Core Distribution Property	$\sum P (X = x_{i}) = 1$
Expectation	$E [X] = \sum x P (X = x)$
Variance	$V a r (X) = E [X^{2}] - (E [X])^{2}$
PGF Definition	$G_{X} (t) = \sum P (X = x) t^{x}$
PGF Expectation	$E [X] = G_{X}^{'} (1)$
PGF Variance	$V a r (X) = G_{X}^{''} (1) + G_{X}^{'} (1) - (G_{X}^{'} (1))^{2}$
Sum of Independent Variables	$G_{X + Y} (t) = G_{X} (t) G_{Y} (t)$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 1
Find unknown constant and expectation
2023 · 2
PGF for sum of independent variables
2024 · 1
Calculate variance of discrete distribution

Going deeper

SyllabusCIE 9231 Further Mathematics SyllabusOfficial syllabus reference

What's Next

Discrete probability distributions are the foundation for all further topics in CIE 9231 Further Probability & Statistics. The core concepts of expectation, variance and generating functions you learned here are tested in nearly every exam paper, and are required for all subsequent topics in statistics. PGF properties for sums of independent variables are especially common in combined exam questions that connect multiple concepts. Mastering the core rules here will make learning named discrete distributions, continuous distributions and statistical inference much more straightforward, and help you secure easy marks on foundational exam questions.