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 4.79 min answerUpdated 2026-05-29

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What this dot point is asking
Discrete probability distributions
Expected value and variance
The binomial distribution
Cumulative probabilities

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

Find $E(X)$ and $\mathrm{Var}(X)$

A spinner gives $X$ with distribution $P(X=1)=0.2$ , $P(X=2)=0.5$ , $P(X=3)=0.3$ .

Check the total. $0.2 + 0.5 + 0.3 = 1$ . Valid.

Expected value.

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

Compute $E(X^2)$ .

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

Variance.

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

So $\sigma = \sqrt{0.49} = 0.7$ . 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 test has $10$ questions, each guessed with $p = 0.25$ . Find $P(X = 3)$ and the mean.

This is binomial with $n = 10$ , $p = 0.25$ .

Probability of exactly $3$ correct.

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

Compute each part. $\binom{10}{3} = 120$ , $(0.25)^3 = 0.015625$ , $(0.75)^7 \approx 0.133484$ .

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

So about a $25\%$ chance of exactly $3$ correct. The mean number correct is

E(X) = np = 10 \times 0.25 = 2.5,

with variance $np(1-p) = 10 \times 0.25 \times 0.75 = 1.875$ . 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.