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WAMath MethodsSyllabus dot point

How do we model and analyse discrete random variables and the binomial distribution?

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

WACE Year 12 Mathematics Methods Unit 3 discrete random variables: probability distributions, expected value, variance and the binomial distribution with mean np and variance np(1-p), shown through worked SCSA-style examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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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 $np(1-p)$

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:

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

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.

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.

Exam-style practice questions

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

WACE 20218 marksCalculator-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 20235 marksCalculator-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 $\dfrac{3}{8}$ .

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 npnpnp and np(1−p)np(1-p)np(1−p)

Exam-style practice questions

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