QLDMath MethodsSyllabus dot point

Topic 3: Discrete random variables

Define a discrete random variable and its probability distribution, calculate the expected value $E(X)$ and the variance $\mathrm{Var}(X)$ and standard deviation, and recognise the Bernoulli distribution as the single-trial case

A focused answer to the QCE Mathematical Methods Unit 3 dot point on discrete random variables. Covers the probability distribution and its conditions ( $p_i \geq 0$ and $\sum p_i = 1$ ), the calculation of $E(X)$ and $\mathrm{Var}(X)$ from a distribution table, and the Bernoulli distribution as the single-trial case, with QCAA IA2-style worked examples.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-13

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

QCAA wants you to define a discrete random variable, identify and work with its probability distribution, compute the expected value and variance, and recognise the Bernoulli distribution as the simplest case (a single yes or no trial). This dot point underpins all of Topic 3 and feeds directly into the binomial distribution.

The answer

Random variables

A random variable $X$ assigns a numerical value to each outcome of a probability experiment. A random variable is discrete if its possible values form a countable set (typically a list of integers).

The probability distribution of $X$ is the list of possible values together with their probabilities, often shown as a table.

$x$	$x_1$	$x_2$	$\ldots$	$x_n$
$P(X = x)$	$p_1$	$p_2$	$\ldots$	$p_n$

For this to be a valid distribution, the probabilities must satisfy two conditions.

Non-negativity. $p_i \geq 0$ for every $i$ .
Normalisation. $\sum_i p_i = 1$ .

QCAA Paper 1 questions frequently give a partial distribution with an unknown $k$ and ask you to use the normalisation condition to solve for $k$ .

Expected value

The expected value (or mean) of $X$ is the probability-weighted average of its values:

E(X) = \mu = \sum_i x_i \, P(X = x_i).

Interpretation: the long-run average of $X$ over many independent repetitions.

For any constants $a$ and $b$ :

E(a X + b) = a E(X) + b.

Variance and standard deviation

The variance measures spread around the mean.

\mathrm{Var}(X) = \sigma^2 = E\bigl( (X - \mu)^2 \bigr) = \sum_i (x_i - \mu)^2 \, P(X = x_i).

The computational shortcut (the version you should use on Paper 1 and Paper 2) is

\mathrm{Var}(X) = E(X^2) - [E(X)]^2,

where $E(X^2) = \sum_i x_i^2 \, P(X = x_i)$ .

The standard deviation is $\sigma = \sqrt{\mathrm{Var}(X)}$ .

For any constants $a$ and $b$ :

\mathrm{Var}(a X + b) = a^2 \, \mathrm{Var}(X).

The $b$ disappears (a shift does not change spread) and the $a$ squares (a scale multiplies variance by $a^2$ ).

The Bernoulli distribution

A Bernoulli trial has exactly two outcomes, labelled success ( $X = 1$ ) and failure ( $X = 0$ ), with probability $p$ of success.

P(X = 1) = p, \qquad P(X = 0) = 1 - p

Write $X \sim \mathrm{Bern}(p)$ . Then

E(X) = p, \qquad \mathrm{Var}(X) = p (1 - p).

The Bernoulli distribution is the building block of the binomial: a binomial $\mathrm{Bin}(n, p)$ is the sum of $n$ independent Bernoulli trials with the same $p$ .

Worked example

Four tasks building from a validity check through to Bernoulli, using the distribution $P(X = 1) = 0.4$ , $P(X = 2) = 0.4$ , $P(X = 5) = 0.2$ .

Step 1: Check whether a claimed distribution is valid

A student proposes $P(X = 1) = 0.2$ , $P(X = 2) = 0.5$ , $P(X = 3) = 0.4$ . The normalisation condition requires all probabilities to sum to exactly $1$ .

0.2 + 0.5 + 0.4 = 1.1 \neq 1

This is not a valid distribution. The probabilities exceed $1$ in total, violating the normalisation condition.

Step 2: Compute the expected value and variance from a table

For the distribution $P(X = 1) = 0.4$ , $P(X = 2) = 0.4$ , $P(X = 5) = 0.2$ , the expected value is the probability-weighted sum of values.

E(X) = 1 \times 0.4 + 2 \times 0.4 + 5 \times 0.2 = 0.4 + 0.8 + 1.0 = 2.2

To find the variance, we first compute $E(X^2)$ by weighting each squared value.

E(X^2) = 1^2 \times 0.4 + 2^2 \times 0.4 + 5^2 \times 0.2 = 0.4 + 1.6 + 5.0 = 7.0

The variance shortcut avoids computing $(x_i - \mu)^2$ for each term.

\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = 7.0 - (2.2)^2 = 7.0 - 4.84 = 2.16

Final answer: $E(X) = 2.2$ , $\mathrm{Var}(X) = 2.16$ , and $\sigma = \sqrt{2.16} \approx 1.47$ .

Step 3: Apply a linear transformation

Using the results from Step 2, find $E(3X + 1)$ and $\mathrm{Var}(3X + 1)$ . The linearity rule $E(aX + b) = aE(X) + b$ applies, and for variance, the additive constant $b$ disappears while the scale factor is squared.

E(3X + 1) = 3 \times 2.2 + 1 = 7.6

\mathrm{Var}(3X + 1) = 3^2 \times 2.16 = 9 \times 2.16 = 19.44

Step 4: Identify and use the Bernoulli distribution

A Bernoulli trial has a single success (with probability $p$ ) or failure. For $X \sim \mathrm{Bern}(0.4)$ , the standard formulas $E(X) = p$ and $\mathrm{Var}(X) = p(1 - p)$ apply directly.

E(X) = 0.4, \qquad \mathrm{Var}(X) = 0.4 \times 0.6 = 0.24

Common traps

Forgetting the normalisation check: A discrete distribution must have probabilities that sum to exactly 1. Forgetting to set this up costs the first mark when QCAA gives an unknown $k$ .
Confusing $E(X^2)$ and $[E(X)]^2$: The variance shortcut is $E(X^2) - [E(X)]^2$ , where the first term squares each $x$ before weighting and the second squares after weighting. They are not equal.
Dropping the square on the scale factor: $\mathrm{Var}(a X + b) = a^2 \, \mathrm{Var}(X)$ , not $a \, \mathrm{Var}(X)$ .
Computing standard deviation but stopping at variance: If QCAA asks for standard deviation, take the square root.
Naming Bernoulli when you mean binomial: A Bernoulli is a single trial. A binomial is $n$ trials. Mixing the two is a frequent IA2 short response error.
Using negative probabilities: A probability cannot be negative. If your algebra produces a negative $p$ , you have set up the problem wrongly.

Exam-style practice questions

Practice questions written in the style of QCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 QCAA-style P15 marksA discrete random variable

X

has the probability distribution shown below, where

k

is a constant. (a) Find the value of

k

. (b) Find

E(X)

. (c) Find

\mathrm{Var}(X)

. |

x

| 0 | 1 | 2 | 3 | |-----|---|---|---|---| |

P(X = x)

0.1

0.3

k

0.2

Show worked answer →

A 5-mark answer needs the normalisation, the expected value, and the variance.

(a) Probabilities sum to 1: $0.1 + 0.3 + k + 0.2 = 1 \implies k = 0.4$ .

(b) $E(X) = \sum x \, P(X = x) = 0 \cdot 0.1 + 1 \cdot 0.3 + 2 \cdot 0.4 + 3 \cdot 0.2 = 0 + 0.3 + 0.8 + 0.6 = 1.7$ .

$\mathrm{Var}(X) = E(X^2) - [E(X)]^2 = 3.7 - (1.7)^2 = 3.7 - 2.89 = 0.81$ .

Markers reward the normalisation step, the correct $E(X)$ calculation, use of the shortcut $\mathrm{Var}(X) = E(X^2) - [E(X)]^2$ (rather than the longer first-principles formula), and a numerical answer to a reasonable number of decimal places.

2022 QCAA-style P23 marksA biased coin lands heads with probability

0.7

. Let

X = 1

if the coin lands heads,

X = 0

if it lands tails. State the distribution of

X

and find

E(X)

and

\mathrm{Var}(X)

Show worked answer →

$X$ is a Bernoulli random variable with parameter $p = 0.7$ , written $X \sim \mathrm{Bern}(0.7)$ .

$E(X) = p = 0.7$ and $\mathrm{Var}(X) = p (1 - p) = 0.7 \cdot 0.3 = 0.21$ .

To verify: $E(X) = 0 \cdot 0.3 + 1 \cdot 0.7 = 0.7$ and $E(X^2) = 0 \cdot 0.3 + 1 \cdot 0.7 = 0.7$ , so $\mathrm{Var}(X) = 0.7 - 0.49 = 0.21$ .

Markers reward naming the distribution explicitly (Bernoulli with $p = 0.7$ ), the standard formulas $E = p$ and $\mathrm{Var} = p (1 - p)$ , and a check by direct calculation. Stating $X \sim \mathrm{Bern}(p)$ in shorthand notation also earns recognition.