NSWMaths Extension 1Syllabus dot point

What is the binomial distribution, and what are its mean and variance?

Define the binomial distribution $B(n, p)$ , state its probability mass function, and find its mean $n p$ and variance $n p (1 - p)$

A focused answer to the HSC Maths Extension 1 dot point on the binomial distribution. The pmf $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$ , mean $n p$ , variance $n p (1 - p)$ , and standard situations that fit the model.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to recognise scenarios that fit the binomial model (fixed number of independent Bernoulli trials with the same success probability), state the probability mass function, and compute the mean, variance and standard deviation.

The answer

Definition

Let $X$ be the number of successes in $n$ independent Bernoulli trials, each with success probability $p$ . Then $X$ has the binomial distribution $B(n, p)$ (or $\text{Bin}(n, p)$ ).

The parameters are:

IMATH_10 : the number of trials (a positive integer).
IMATH_11 : the success probability on each trial ( $0 \le p \le 1$ ).
IMATH_13 : the failure probability.

The probability mass function

For $k = 0, 1, 2, \dots, n$ ,

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

The binomial coefficient $\binom{n}{k}$ counts the number of ways to arrange $k$ successes among $n$ trials.

Mean and variance

By summing $n$ independent Bernoulli trials (each with mean $p$ and variance $p(1 - p)$ ):

E(X) = n p, \qquad \text{Var}(X) = n p (1 - p), \qquad \sigma = \sqrt{n p (1 - p)}.

Conditions for the binomial model

Four conditions must hold:

A fixed number of trials, $n$ .
Each trial has exactly two outcomes (success or failure).
The trials are independent.
The probability of success is the same for every trial.

If any of these fails, the distribution is not binomial.

Common scenarios

Number of heads in $n$ coin flips: $X \sim B(n, p)$ with $p$ the heads probability.
Number of items defective in a batch of $n$ (with replacement, or with a large enough population to treat as approximately independent).
Number of correct answers on a multiple-choice test if every question is guessed.
Number of successful free throws in a fixed number of attempts (independence is a simplifying assumption).

Symmetric and skewed cases

If $p = \tfrac{1}{2}$ , the distribution is symmetric around $\frac{n}{2}$ .
If $p < \tfrac{1}{2}$ , the distribution is right-skewed (long tail to the right).
If $p > \tfrac{1}{2}$ , it is left-skewed.

The skew decreases as $n$ grows; for large $n$ , the binomial approaches the normal (see the normal approximation dot point).

Worked example

Direct mean and variance

A coin lands heads with probability $0.6$ . It is flipped $20$ times. Find the mean and variance of the number of heads.

$X \sim B(20, 0.6)$ . $E(X) = 12$ . $\text{Var}(X) = 20 \cdot 0.6 \cdot 0.4 = 4.8$ .

pmf at a specific value

A multiple-choice quiz has $5$ questions, each with $4$ options. If you guess randomly, what is the probability of getting exactly $3$ correct?

$X \sim B(5, 0.25)$ . $P(X = 3) = \binom{5}{3} (0.25)^3 (0.75)^2 = 10 \cdot 0.015625 \cdot 0.5625 \approx 0.0879$ .

Conditions check

A urn has $5$ red and $5$ blue marbles. Draw $4$ without replacement. Is the number of red draws binomial?

No: without replacement, the probability of red changes each draw (decreases if you drew red, increases if you drew blue). This is the hypergeometric distribution, not binomial.

Verifying the model

In a factory, machines produce items independently. Each item has probability $0.02$ of being defective. In a sample of $100$ items, find the mean and standard deviation of the number defective.

Conditions: fixed $n = 100$ , two outcomes (defective or not), independent (given), constant $p = 0.02$ . Binomial.

$E(X) = 2$ . $\text{Var}(X) = 100 \cdot 0.02 \cdot 0.98 = 1.96$ . $\sigma = 1.4$ .

Two-trial product

Two independent flips of a coin with $p = 0.7$ . Find $P(X = 1)$ .

$X \sim B(2, 0.7)$ . $P(X = 1) = \binom{2}{1} (0.7)(0.3) = 2 \cdot 0.21 = 0.42$ .

Linear transformation

$Y = 3 X + 5$ where $X \sim B(10, 0.5)$ . Find $E(Y)$ and $\text{Var}(Y)$ .

$E(X) = 5$ , $\text{Var}(X) = 2.5$ .

$E(Y) = 3 \cdot 5 + 5 = 20$ .

$\text{Var}(Y) = 9 \cdot 2.5 = 22.5$ (variance scales by the square of the multiplier, constant shift has no effect).

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2023 HSC Q213 marksA biased coin lands heads with probability

0.4

. It is flipped

10

times. Let

X

be the number of heads. Find

E(X)

\text{Var}(X)

and the standard deviation.

Show worked answer →

$X \sim B(10, 0.4)$ .

$E(X) = n p = 10 \cdot 0.4 = 4$ .

$\text{Var}(X) = n p (1 - p) = 10 \cdot 0.4 \cdot 0.6 = 2.4$ .

Standard deviation: $\sigma = \sqrt{2.4} \approx 1.549$ .

Markers reward identifying the binomial model with $n = 10$ , $p = 0.4$ , applying the formulas, and the square root for $\sigma$ .