The Bernoulli distribution: WACE Year 12 Mathematics Methods Unit 3

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

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

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.