Discrete random variables and the binomial distribution: WACE Year 12 Mathematics Methods Unit 3

Syllabus dot point

“Construct discrete probability distributions, calculate expected value and variance, and apply the binomial distribution to repeated independent trials”

9 min readUpdated 2 Jun 2026

Quick answer

Build a discrete probability distribution so probabilities sum to one, compute $E(X)=\sum x\,P(X=x)$ and variance, and use the binomial distribution $P(X=k)=\binom{n}{k}p^{k}(1-p)^{n-k}$ with mean $np$ and variance $np(1-p)$ .

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What this dot point is asking
Discrete probability distributions
Expected value and variance
The binomial distribution
Cumulative probabilities
Recognising when the binomial applies
Why mean and variance are $np$ and $np(1-p)$
Exam-style questions

What this dot point is asking

SCSA Unit 3 introduces discrete random variables: variables that take countable, separate values each with an attached probability. You must construct distributions, compute summary measures, and recognise and apply the binomial distribution for a fixed number of independent success-or-failure trials.

Discrete probability distributions

A discrete random variable $X$ has a probability distribution listing each value and its probability. Two conditions must hold:

Distribution conditions

0 \le P(X = x) \le 1 \quad \text{for every } x, \qquad \sum_x P(X = x) = 1.

The second condition is the most common way SCSA asks you to find an unknown probability: set the sum equal to one and solve.

Expected value and variance

Mean and variance of a discrete variable

E(X) = \mu = \sum_x x\,P(X = x)

\mathrm{Var}(X) = \sigma^2 = \sum_x (x - \mu)^2 P(X = x) = E(X^2) - [E(X)]^2

\mathrm{SD}(X) = \sigma = \sqrt{\mathrm{Var}(X)}

The form $\mathrm{Var}(X) = E(X^2) - [E(X)]^2$ is usually faster by hand because it avoids squaring deviations.

Mean and variance from a table

A spinner produces $X$ with distribution $P(X=1)=0.2$ , $P(X=2)=0.5$ , $P(X=3)=0.3$ . We find $E(X)$ and $\mathrm{Var}(X)$ .

Step 1: Check the distribution is valid

Before calculating summary measures, verify that the probabilities sum to one.

0.2 + 0.5 + 0.3 = 1. \quad\checkmark

This confirms the distribution is valid and that the expected value will be computed from a complete set of probabilities.

Step 2: Calculate the expected value

Multiply each value by its probability and sum the products.

E(X) = 1(0.2) + 2(0.5) + 3(0.3) = 0.2 + 1.0 + 0.9 = 2.1.

The expected value represents the long-run average outcome across many spins of the spinner.

Step 3: Calculate $E(X^2)$

Square each value, weight by its probability, and sum. This is needed for the shortcut variance formula.

E(X^2) = 1^2(0.2) + 2^2(0.5) + 3^2(0.3) = 0.2 + 2.0 + 2.7 = 4.9.

Step 4: Calculate the variance

Apply the shortcut $\mathrm{Var}(X)=E(X^2)-[E(X)]^2$ , which avoids squaring deviations from the mean.

\mathrm{Var}(X) = 4.9 - 2.1^2 = 4.9 - 4.41 = 0.49.

Final answer: $E(X)=2.1$ , $\mathrm{Var}(X)=0.49$ , $\sigma=\sqrt{0.49}=0.7$ . Markers reward the validity check, correct weighted sums, and use of the $E(X^2)-\mu^2$ shortcut.

The binomial distribution

A binomial random variable counts the number of successes in $n$ independent trials, each with the same probability $p$ of success. The requirements are: a fixed number of trials, two outcomes per trial, constant $p$ , and independence.

Binomial probability, mean and variance

For $X \sim \mathrm{B}(n, p)$ ,

P(X = k) = \binom{n}{k} p^{k} (1-p)^{n-k}, \quad k = 0,1,\dots,n,

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

Here $\binom{n}{k} = \dfrac{n!}{k!\,(n-k)!}$ counts the number of ways to arrange $k$ successes among $n$ trials.

A binomial calculation

A multiple-choice test has $10$ questions and each answer is guessed independently with $p=0.25$ . We find the probability of exactly $3$ correct and the mean number correct.

Step 1: Identify the binomial parameters

The number of correct answers $X$ counts successes in $n=10$ independent trials, each with constant success probability $p=0.25$ , so $X\sim\mathrm{B}(10,0.25)$ . Stating the distribution first makes explicit which formula and which parameters apply.

Step 2: Apply the binomial formula for $P(X=3)$

Substitute $n=10$ , $k=3$ , $p=0.25$ and $1-p=0.75$ into the binomial formula.

P(X = 3) = \binom{10}{3}(0.25)^{3}(0.75)^{7}.

Compute each factor separately to keep the arithmetic manageable.

\binom{10}{3} = 120, \qquad (0.25)^3 = 0.015625, \qquad (0.75)^7 \approx 0.133484.

P(X = 3) = 120 \times 0.015625 \times 0.133484 \approx 0.2503.

Step 3: State the mean

Use the binomial mean formula, which follows directly from the parameters.

E(X) = np = 10 \times 0.25 = 2.5, \qquad \mathrm{Var}(X)=np(1-p)=10\times 0.25\times 0.75=1.875.

Final answer: $P(X=3)\approx 0.2503$ , so there is about a $25\%$ chance of exactly $3$ correct. The mean is $E(X)=2.5$ . Markers reward the correct binomial coefficient, the matching powers of $p$ and $1-p$ , and a sensible interpretation.

Cumulative probabilities

"At least" and "at most" questions require summing several terms, or using the complement. For example $P(X \ge 1) = 1 - P(X = 0)$ is far faster than adding ten terms. In the calculator-assumed section, binomial cumulative functions handle these directly.

Recognising when the binomial applies

Before any calculation, confirm the four binomial conditions: a fixed number of trials $n$ , exactly two outcomes per trial, a constant success probability $p$ , and independent trials. The independence condition is the one most often broken. Sampling without replacement from a small population changes $p$ from trial to trial, so it is not binomial; sampling from a very large population (or with replacement) keeps $p$ effectively constant and the binomial model is valid. SCSA questions frequently test this judgement by describing a scenario and asking whether the binomial is appropriate, so state the conditions explicitly when justifying your model.

Why mean and variance are $np$ and $np(1-p)$

A binomial variable is the sum of $n$ independent Bernoulli trials, each with mean $p$ and variance $p(1-p)$ . Because expectation is additive, the means add to $np$ ; because the trials are independent, the variances also add, giving $np(1-p)$ . This is why understanding the single-trial Bernoulli case makes the binomial summary measures intuitive rather than something to memorise.

Common errors

Probabilities not summing to one: Always check $\sum P(X=x)=1$ before computing a mean; an unknown is often found from this condition.
Swapping $p$ and $1-p$ powers: In $\binom{n}{k}p^{k}(1-p)^{n-k}$ the exponent on $p$ is the number of successes $k$ , and the exponent on $1-p$ is the number of failures $n-k$ .
Using $np(1-p)$ as the standard deviation: That expression is the variance; the standard deviation is its square root, $\sqrt{np(1-p)}$ .
Misreading "at least" and "at most": $P(X \ge 2)$ includes $X = 2$ ; use the complement $1 - P(X \le 1)$ to avoid long sums.

Exam-style questions

Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

WACE 2021-style8 marks

Calculator-assumed. A factory's items are independently defective with probability

p=0.04

. A quality inspector tests

25

items. Let

X

be the number defective. (a) State the distribution of

X

. (b) Find

E(X)

and

\mathrm{Var}(X)

. (c) Find

P(X=0)

. (d) Find

P(X\ge 2)

Show worked answer →

A full binomial question with summary measures and cumulative probabilities.

(a) $X\sim\mathrm{B}(25,0.04)$ : $25$ independent trials, two outcomes, constant $p=0.04$ .

(b) $E(X)=np=25\times 0.04=1$ ; $\mathrm{Var}(X)=np(1-p)=25\times 0.04\times 0.96=0.96$ .

(d) Use the complement: $P(X\ge 2)=1-P(X\le 1)=1-\bigl[(0.96)^{25}+25(0.04)(0.96)^{24}\bigr]\approx 1-(0.360+0.375)=0.265$ .

Markers reward the stated distribution, $np$ and $np(1-p)$ , the exact $P(X=0)$ , and the complement in (d).

WACE 2023-style5 marks

Calculator-free. A fair coin is tossed

4

times. Let

X

be the number of heads. (a) Find

E(X)

and

\mathrm{Var}(X)

. (b) Find

P(X=2)

Show worked answer →

A calculator-free binomial with $p=0.5$ .

(a) $X\sim\mathrm{B}(4,0.5)$ , so $E(X)=np=4(0.5)=2$ and $\mathrm{Var}(X)=np(1-p)=4(0.5)(0.5)=1$ .

(b) $P(X=2)=\binom{4}{2}(0.5)^{2}(0.5)^{2}=6\times\dfrac{1}{16}=\dfrac{6}{16}=\dfrac{3}{8}=0.375$ .

Markers reward $np$ and $np(1-p)$ and the exact probability $\dfrac{3}{8}$ .

Discrete random variables and the binomial distribution: WACE Year 12 Mathematics Methods Unit 3

What this dot point is asking

Discrete probability distributions

Expected value and variance

Step 1: Check the distribution is valid

Step 2: Calculate the expected value

Step 3: Calculate $E(X^2)$

Step 4: Calculate the variance

The binomial distribution

Step 1: Identify the binomial parameters

Step 2: Apply the binomial formula for $P(X=3)$

Step 3: State the mean

Cumulative probabilities

Recognising when the binomial applies

Why mean and variance are $np$ and $np(1-p)$

Exam-style questions

Practise this

Sources & how we know this

What this dot point is asking

Discrete probability distributions

Expected value and variance

Step 1: Check the distribution is valid

Step 2: Calculate the expected value

Step 3: Calculate E(X2)E(X^2)E(X2)

Step 4: Calculate the variance

The binomial distribution

Step 1: Identify the binomial parameters

Step 2: Apply the binomial formula for P(X=3)P(X=3)P(X=3)

Step 3: State the mean

Cumulative probabilities

Recognising when the binomial applies

Why mean and variance are npnpnp and np(1−p)np(1-p)np(1−p)

Exam-style questions

Practise this

Sources & how we know this

Step 3: Calculate $E(X^2)$

Step 2: Apply the binomial formula for $P(X=3)$

Why mean and variance are $np$ and $np(1-p)$