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How do we model a single trial with exactly two outcomes, and what are its mean and variance?

Use the Bernoulli distribution to model a single two-outcome trial and find its mean and variance

WACE Year 12 Mathematics Methods Unit 3 the Bernoulli distribution: a single success-or-failure trial, its probability function, mean p and variance p times one minus p, and its role as the building block of the binomial, with a worked example.

Generated by Claude Opus 4.87 min answerUpdated 2026-06-02

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What this dot point is asking
The Bernoulli model
Mean and variance
Where the variance is largest
Link to the binomial distribution

What this dot point is asking

SCSA Unit 3 introduces the Bernoulli distribution as the simplest discrete model: one trial, two outcomes. This dot point asks you to state its distribution and derive its mean and variance, which then generalise directly to the binomial distribution. It supports questions in both examination sections.

The Bernoulli model

A Bernoulli trial has exactly two outcomes, conventionally labelled success and failure. The random variable $X$ records $1$ for success and $0$ for failure. This $0$ -or- $1$ coding is what makes the algebra so clean: the mean of $X$ turns out to be exactly the success probability, and because squaring $0$ and $1$ leaves them unchanged, the variance simplifies neatly as well. The Bernoulli distribution is the simplest possible non-trivial probability distribution, which is why the whole of discrete probability in the course can be built up from it.

Examples include a single coin toss (success = heads), one inspected item (success = defective), or one penalty kick (success = goal). The choice of which outcome to call "success" is a labelling decision; swapping the labels simply replaces $p$ with $1-p$ . The variable is sometimes called an indicator variable, because it indicates whether the event of interest occurred, recording $1$ if it did and $0$ if it did not.

Mean and variance

Because $X$ takes only the values $0$ and $1$ , the mean and variance follow directly from the definitions of expected value and variance.

The mean is

E(X)=0\cdot(1-p)+1\cdot p = p.

For the variance, note that $X^{2}=X$ since $0^{2}=0$ and $1^{2}=1$ , so $E(X^{2})=E(X)=p$ . Then

\mathrm{Var}(X)=E(X^{2})-[E(X)]^{2}=p-p^{2}=p(1-p).

Bernoulli mean and variance

Step 1: State the mean using the Bernoulli formula

A component is defective with probability $p=0.05$ , and $X=1$ if defective. For any Bernoulli variable, the mean equals the success probability, so no calculation beyond substitution is needed:

E(X)=p=0.05.

Step 2: Compute the variance using $p(1-p)$

The Bernoulli variance formula $\mathrm{Var}(X)=p(1-p)$ follows from the shortcut $E(X^2) - [E(X)]^2 = p - p^2$ . Substituting $p=0.05$ :

\mathrm{Var}(X)=0.05\times 0.95 = 0.0475.

Step 3: Find the standard deviation

Taking the square root of the variance gives the standard deviation in the same units as $X$ :

\sigma=\sqrt{0.0475}\approx 0.218.

Final answer: $E(X)=0.05$ , $\mathrm{Var}(X)=0.0475$ , $\sigma\approx 0.218$

Where the variance is largest

The Bernoulli variance $p(1-p)$ is a downward parabola in $p$ , zero at $p=0$ and $p=1$ and largest at $p=0.5$ , where it equals $0.25$ . This matches intuition: a trial that is almost certain to succeed (or fail) carries little uncertainty, while a fair trial is the least predictable. The same shape carries through to the binomial, whose variance $np(1-p)$ is also maximised at $p=0.5$ for fixed $n$ .

Link to the binomial distribution

A binomial random variable counts the number of successes in $n$ independent Bernoulli trials, each with the same $p$ . Adding $n$ independent Bernoulli variables gives the binomial, which is why the binomial mean $np$ and variance $np(1-p)$ are simply $n$ times the Bernoulli values. Understanding the single-trial case makes the binomial formulae intuitive rather than memorised.

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 20214 marksCalculator-free. A single random voter supports a policy with probability

p=0.6

. Let

X=1

if the voter supports it and

X=0

otherwise. (a) State

E(X)

and

\mathrm{Var}(X)

. (b) Explain how this Bernoulli trial relates to the binomial distribution for a sample of

n

voters.

Show worked answer →

A definition-and-link question.

(a) $E(X)=p=0.6$ and $\mathrm{Var}(X)=p(1-p)=0.6\times 0.4=0.24$ .

(b) A sample of $n$ independent voters, each modelled by the same Bernoulli $(0.6)$ trial, gives a binomial variable counting total supporters. Summing $n$ independent Bernoulli variables produces $\mathrm{B}(n,0.6)$ with mean $np$ and variance $np(1-p)$ , that is $n$ times the single-trial values.

Markers reward stating $E(X)=p$ and $p(1-p)$ and explaining the binomial as a sum of $n$ Bernoulli trials.

WACE 20233 marksCalculator-free. For a Bernoulli random variable, show that the variance

p(1-p)

is greatest when

p=0.5

, and state the maximum value.

Show worked answer →

A short reasoning question about the variance.

The variance $V(p)=p(1-p)=p-p^{2}$ . Differentiating, $V'(p)=1-2p$ , which is zero at $p=0.5$ . Since $V''(p)=-2<0$ , this is a maximum.

The maximum value is $V(0.5)=0.5\times 0.5=0.25$ .

Markers reward differentiating $p-p^{2}$ , solving $V'(p)=0$ , confirming a maximum, and the value $0.25$ .