Define a Bernoulli random variable, compute its mean and variance, and recognise scenarios that fit the model

What this dot point is asking

NESA wants you to recognise a Bernoulli trial (a single experiment with two outcomes), state its probability mass function, compute its mean and variance, and see it as the unit of a sequence used to build the binomial distribution.

The answer

Definition

A Bernoulli trial is a random experiment with exactly two possible outcomes, conventionally labelled "success" (value $1$) and "failure" (value $0$), where success occurs with probability $p$ and failure with probability $q = 1 - p$.

If $X$ is a Bernoulli random variable with parameter $p$ (written $X \sim B(p)$ or $\text{Bern}(p)$):

The whole distribution is just two bars: a bar of height $q = 1 - p$ at $x = 0$ and a bar of height $p$ at $x = 1$. Because something must happen, the two heights add to $1$, which is the only consistency check a Bernoulli variable needs.

Mean (expected value)

The expected value of a single Bernoulli trial is its success probability.

Variance

Equivalently, $\sigma = \sqrt{p (1 - p)}$.

Standard examples

• Flipping a fair coin with $X = 1$ if heads: $p = \tfrac{1}{2}$.
• Rolling a die and checking for a $6$: $p = \tfrac{1}{6}$.
• Asking a random voter if they support a policy with $p$ unknown.
• A medical test giving a positive result on someone with the condition (sensitivity).

Why this matters

A Bernoulli trial is the simplest non-trivial random variable. The binomial distribution counts the number of successes in a fixed number of independent Bernoulli trials.

A sequence of $n$ independent identically distributed Bernoulli trials, where $X_i = 1$ if the $i$-th trial is a success and $0$ otherwise, has total $S = X_1 + X_2 + \dots + X_n \sim B(n, p)$.

By linearity of expectation, $E(S) = n p$. By independence and the variance addition rule, $\text{Var}(S) = n p (1 - p)$.

Recognising a Bernoulli model

The examinable judgement is whether a described situation fits the Bernoulli model at all. Three conditions must hold: there is a single trial (or you are looking at one trial in a sequence), there are exactly two outcomes, and a fixed probability $p$ is attached to success. If the experiment has three or more outcomes, or a continuous range of outcomes, it is not Bernoulli. Where an outcome with many categories is reframed as "the event of interest happened, or it did not", a Bernoulli model is recovered. This reframing (success versus everything-else) is how almost every binomial question is built.

From one trial to many

The reason this small model matters is that it is the atom from which the binomial distribution is built. Each of $n$ independent, identically distributed Bernoulli trials contributes its mean $p$ and variance $p(1 - p)$, and because expectation is linear and variance adds for independent variables, the totals are $np$ and $np(1 - p)$. Understanding that the binomial mean and variance are just $n$ copies of the Bernoulli mean and variance removes any need to memorise them as separate facts.

Where the variance is largest, stage by stage

A favourite short proof question asks you to show that $\text{Var}(X) = p(1 - p)$ is greatest at $p = \tfrac{1}{2}$. The variance is a function of $p$, and seeing its shape makes the result obvious before you differentiate anything. The three panels build that picture up.

Stage 1, pin down the certain cases. At $p = 0$ failure is guaranteed and at $p = 1$ success is guaranteed; either way the outcome is certain, so there is no spread and $\text{Var}(X) = 0(1 - 0) = 0$ and $1(1 - 1) = 0$. The variance curve must therefore touch zero at both ends of the interval $[0, 1]$.

Stage 2, fill in the shape between the ends. Because $\text{Var}(X) = p - p^2$ is a quadratic with a negative leading coefficient, it is a downward parabola. Pinned to $0$ at each end and opening downward, it must bulge up in the middle: uncertainty is small near the certain ends and grows as the two outcomes become more evenly matched.

Stage 3, locate the peak. By symmetry of a parabola the maximum sits exactly halfway between the roots, at $p = \tfrac{1}{2}$. Differentiating confirms it: $\frac{d}{dp}(p - p^2) = 1 - 2p = 0$ gives $p = \tfrac{1}{2}$, and the second derivative $-2 < 0$ makes it a maximum. The peak value is $\tfrac{1}{2}\cdot\tfrac{1}{2} = \tfrac{1}{4}$, so a fair trial ($p = \tfrac{1}{2}$) is the most uncertain Bernoulli trial of all.

How exam questions ask about Bernoulli trials

Bernoulli rarely gets a question to itself; it is usually the setup line of a binomial question, or a short part that tests whether you know the parameters and formulas. The wordings to recognise:

• "A Bernoulli random variable has $P(X = 1) = \ldots$" or "a single trial succeeds with probability $p$": state $E(X) = p$ and $\text{Var}(X) = p(1 - p)$ and substitute. These are usually 1 to 2 mark recall parts.
• "Show that the variance is greatest / maximised when $p = \tfrac{1}{2}$": differentiate $V(p) = p - p^2$, set $V'(p) = 0$, and confirm a maximum with $V''(p) = -2 < 0$ before stating $V(\tfrac{1}{2}) = \tfrac{1}{4}$.
• "Does this situation fit a Bernoulli model?" or "explain why X is (not) a Bernoulli variable": check the three conditions out loud, that there is one trial, exactly two outcomes, and a fixed $p$. Continuous quantities (a time, a height) and three-or-more-outcome experiments are not Bernoulli unless reframed as "event happened or it did not".
• "$X$ counts successes in $n$ trials": this is the cue that you have left a single Bernoulli trial and entered the binomial; the mean becomes $np$ and the variance $np(1 - p)$.