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Year 12 Maths Extension 1 (2027) Statistical analysis

Bernoulli distributions

20 practice questions 2 video lessons Theory + worked examples

Meet the Bernoulli distribution in NSW Year 12 Mathematics Extension 1. It models a single trial with just two outcomes — success or failure — and is the building block of the binomial distribution.

You will learn to define a Bernoulli random variable, assign the probabilities of success and failure, and use its mean and variance — the starting point for the binomial and sampling work that follows in the Extension 1 course.

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Practice questions

Every question with a fully worked solution.

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  • Introduction to the Bernoulli Distribution Watch
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Theory

A Bernoulli trial has two outcomes: success (X=1) with probability p, failure (X=0) with probability q=1p. Its mean is p and variance p(1p). This NSW Year 12 Mathematics Extension 1 topic is NESA outcome ME1-12-06.

A Bernoulli trial is a single experiment with exactly two outcomes — success (X=1) or failure (X=0). The random variable X that records the result has a Bernoulli distribution.

For success probability p (and q=1p): P(X=1)=p, P(X=0)=q. Its mean and variance are E(X)=μ=p and Var(X)=p(1p)=pq.

Over n repeated independent trials, the expected number of successes is np.

NESA link. Part of the Year 12 The binomial distribution and sampling distribution of the mean focus area, outcome ME1-12-06 ("solves problems involving binomial distributions, sampling distribution of the mean and the central limit theorem") with MAO-WM-01.

The Bernoulli distributionA two-bar probability distribution: failure at x equals zero with probability q, and success at x equals one with probability p.xy01qpP(X=x)x
Two outcomes: P(X=0)=q, P(X=1)=p.
Variance of a Bernoulli trialThe variance p times one minus p is a downward parabola in p, greatest at p equals one half where it equals 0.25.xy0.51max 0.25Var = p(1−p)p
Var(X)=p(1p) is greatest at p=0.5.
P(X=1)=p,P(X=0)=q=1p.
P(X=1) = p, P(X=0) = q = 1 - p
E(X)=μ=p,Var(X)=p(1p)=pq.
E(X) = p, Var(X) = p(1-p) = pq

Single trial. A Bernoulli trial is one trial; several such trials form a binomial experiment. The variance p(1p) is largest at p=0.5.

How to model a Bernoulli trial

  1. Identify the two outcomes (success and failure).
  2. Read off p, the probability of success; then q=1p.
  3. Mean and variance: E(X)=p, Var(X)=pq.
  4. Over n trials, expected successes =np (if the trials are independent with the same p).
Example 1 — Rolling a six
A die is rolled and success is a six. Write the distribution and find E(X) and Var(X).
Solution

p=16, q=56.

E(X)=p=16
Var(X)=1656=536
E(X) = 1/6, Var(X) = 5/36

E(X)=16, Var(X)=536.

Example 2 — Given p
A Bernoulli trial has p=0.2. Find the mean and variance.
Solution
E(X)=0.2
Var(X)=0.2×0.8=0.16
mean 0.2, variance 0.16

Mean 0.2, variance 0.16.

Example 3 — Expected successes
A spinner lands on red with probability 0.25. If it is spun 60 times, how many reds are expected?
Solution
np=60×0.25=15
expected reds = 15

About 15 reds are expected.

Example 4 — Independence check
Two balls are drawn without replacement from a bag of 3 red and 2 blue. Is each draw a Bernoulli trial with the same p?
Solution

A single draw (red or not) is a Bernoulli trial. But without replacement the probability of red changes after the first draw, so the two draws are not identical, independent trials.

no - draws without replacement are not identically distributed

No — the draws are not identically distributed.

Common pitfalls

One trial only. A Bernoulli trial is a single trial; several form a binomial experiment.
Independence. Repeated trials must be independent with the same p — drawing without replacement breaks this.
Variance form. Var(X)=pq, largest at p=0.5.
Mean is just p. For a single Bernoulli trial E(X)=p, not np.

Frequently asked questions

What is a Bernoulli trial?

A single experiment with two outcomes, success (X=1) with probability p and failure (X=0) with probability q=1p.

What are the mean and variance?

E(X)=p and Var(X)=p(1p)=pq.

When is the variance largest?

At p=0.5, where p(1p)=0.25.

How many successes are expected over n trials?

np, if the trials are independent with the same p.

Why does 'without replacement' matter?

It changes p between draws, so the trials are no longer identically distributed.