Year 12 Maths Extension 1 (2027) Statistical analysis

Binomial distributions

Q: What is a binomial random variable?

The number of successes in n independent trials with constant success probability p, written X is Bin(n, p).

Q: What is the binomial probability formula?

P of X equals r is n choose r times p to the r times one minus p to the n minus r.

Q: When is a situation binomial?

Fixed n, two outcomes, constant p, and independent trials.

Q: How do you find at least one?

Use the complement, one minus P of X equals zero.

Q: What is an expected frequency?

The number of experiments times the probability, N times P of X equals r.

20 practice questions 2 video lessons Theory + worked examples

Understand the binomial distribution for NSW Year 12 Mathematics Extension 1. It gives the probability of a set number of successes in repeated independent trials that each have the same chance of success — the model behind coin tosses, quality testing and sampling with replacement.

You will learn to recognise a binomial random variable, calculate the probability of exactly, at least or at most a given number of successes, and find expected frequencies — an essential statistical analysis skill for the HSC Extension 1 course.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

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Watch 2 video(s)

Binomial distribution introduction Watch
The binomial Probability Formula Watch

Theory

A binomial variable $X \sim Bin (n, p)$ counts successes in $n$ independent trials, with $P (X = r) = (\binom{n}{r}) p^{r} (1 - p)^{n - r}$ . This NSW Year 12 Mathematics Extension 1 topic is NESA outcome ME1-12-06.

A binomial random variable $X$ counts the number of successes in $n$ independent trials, each with the same success probability $p$ . We write $X \sim Bin (n, p)$ .

P (X = r) = (\binom{n}{r}) p^{r} (1 - p)^{n - r}, r = 0, 1, \dots, n .

Here $(\binom{n}{r}) =^{n} C_{r}$ counts the ways to place $r$ successes among $n$ trials. The four conditions: a fixed number of trials $n$ ; two outcomes per trial; a constant $p$ ; and independent trials.

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. Drawing cards with replacement is binomial; without replacement is not.

Bin (8, 0.5)

: symmetric about

r = 4

Bin (8, 0.25)

: skewed toward smaller

r

P (X = r) = (\binom{n}{r}) p^{r} (1 - p)^{n - r} .

P(X=r) = C(n,r) p^r (1-p)^(n-r)

Useful shortcuts: $P (X \geq 1) = 1 - P (X = 0)$ ; over $N$ experiments the expected frequency of $r$ successes is $N \cdot P (X = r)$ .

P(X >= 1) = 1 - P(X = 0); expected frequency = N times P(X = r)

Watch the exponent. The power on $(1 - p)$ is $n - r$ , not $n$ . For "at least one", use $1 - P (X = 0)$ instead of summing many terms.

How to use the binomial formula

Check the four conditions and identify $n$ and $p$ .
Apply $P (X = r) = (\binom{n}{r}) p^{r} (1 - p)^{n - r}$ .
For "at least one", use the complement $1 - P (X = 0)$ .
For an expected frequency, multiply the probability by the number of experiments.

Example 1 — Exactly two

A die is rolled

6

times. Find the probability of exactly two sixes.

Solution

$X \sim Bin (6, \frac{1}{6})$ , $r = 2$ .

P (X = 2)

=

(\binom{6}{2}) {(\frac{1}{6})}^{2} {(\frac{5}{6})}^{4} \approx 0.201

P(X=2) approx 0.201

$P (X = 2) \approx 0.201$ .

Example 2 — At least one

5 %

of items are defective. In a sample of

8

, find

P (at least one defective)

Solution

$X \sim Bin (8, 0.05)$ ; use the complement.

P (X \geq 1)

=

1 - (0.95)^{8} \approx 0.337

P(X >= 1) approx 0.337

$P (X \geq 1) \approx 0.337$ .

Example 3 — A given r

X \sim Bin (7, 0.4)

, find

P (X = 3)

Solution

P (X = 3)

=

(\binom{7}{3}) (0.4)^{3} (0.6)^{4} \approx 0.290

P(X=3) approx 0.290

$P (X = 3) \approx 0.290$ .

Example 4 — Expected frequency

500

families of

4

children, how many are expected to have exactly

3

boys? (Take

p = 0.5

Solution

$X \sim Bin (4, 0.5)$ , so $P (X = 3) = (\binom{4}{3}) (0.5)^{4} = 0.25$ .

500 \times 0.25

=

125

expected frequency = 125 families

About $125$ families.

Common pitfalls

Independence and constant $p$ . With replacement is binomial; without replacement is not.

Exponent. The power on

(1 - p)

n - r

, not

n

"At least one". Use

1 - P (X = 0)

rather than summing many terms.

Expected frequency. Multiply the probability by the number of experiments

N

Frequently asked questions

What is a binomial random variable?

The number of successes in $n$ independent trials with constant success probability $p$ , written $X \sim Bin (n, p)$ .

What is the binomial probability formula?

$P (X = r) = (\binom{n}{r}) p^{r} (1 - p)^{n - r}$ .

When is a situation binomial?

Fixed $n$ , two outcomes, constant $p$ , and independent trials.

How do you find 'at least one'?

Use the complement $1 - P (X = 0)$ .

What is an expected frequency?

The number of experiments times the probability, $N \cdot P (X = r)$ .

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Mean & variance of a binomial distribution

Binomial distributions

📖 Theory

How to use the binomial formula

Common pitfalls

Frequently asked questions

Theory