# The Binomial Distribution

> Count successes in a fixed number of yes-or-no trials. Bernoulli trials, the binomial formula, and the mean and variance np and np(1 - p).

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/binomial-distribution/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: Ten three-point shots
A basketball player makes 30% of her three-point shots. In tonight's game she will take 10. Her coach wonders: how likely is she to make exactly 3? At least 1? More than 5?

Each shot is the same kind of event repeated: it either goes in or it does not, and the chance stays the same each time. Whenever a situation looks like that, one formula answers every question about how many successes to expect.

**In short:** The binomial distribution counts successes in $n$ independent trials that each have the same probability $p$ of success. $P(X = r) = \binom{n}{r}p^r(1 - p)^{n - r}$, with mean $np$ and variance $np(1 - p)$.

## Starting point: what you already know

- **Probability basics:** $P(\text{not } A) = 1 - P(A)$, and for independent events $P(A \text{ and } B) = P(A) \times P(B)$.
- **Combinations** (Combinations chapter): $\dbinom{n}{r} = \dfrac{n!}{r!(n - r)!}$ counts the ways to choose $r$ things from $n$.
- **Discrete random variables** (Mathematics Advanced): a table of values and probabilities that add to 1, with mean $E(X) = \sum x\,P(X = x)$ and variance $\mathrm{Var}(X) = E(X^2) - \big[E(X)\big]^2$.
- **The binomial theorem:** $(q + p)^n = \sum \dbinom{n}{r}p^r q^{n - r}$.

## Bernoulli trials

A **Bernoulli trial** has exactly two outcomes, called success and failure. The probability of success is $p$, and of failure is $1 - p$, sometimes written $q$.

Let $X = 1$ for a success and $X = 0$ for a failure. Then:

- $E(X) = 0 \times (1 - p) + 1 \times p = p$
- $\mathrm{Var}(X) = E(X^2) - p^2 = p - p^2 = p(1 - p)$

## The binomial distribution

Now repeat a Bernoulli trial $n$ times and count the successes, $X$. This is a **binomial distribution**, written $X \sim \mathrm{Bin}(n, p)$, when:

- there are a **fixed number** $n$ of trials,
- each has only **two outcomes**,
- the trials are **independent**,
- $p$ is the **same** every time.

### Key idea: The binomial formula
$$P(X = r) = \binom{n}{r}p^r(1 - p)^{n - r} \qquad r = 0, 1, 2, \ldots, n$$

Why it works: one particular order of $r$ successes and $n - r$ failures has probability $p^r(1 - p)^{n - r}$. There are $\dbinom{n}{r}$ different orders.

### Worked example: The basketball player
$X \sim \mathrm{Bin}(10, 0.3)$.

**Exactly 3:** $P(X = 3) = \dbinom{10}{3}(0.3)^3(0.7)^7 = 120 \times 0.027 \times 0.0824 \approx 0.267$.

**At least 1:** use the complement. $P(X \ge 1) = 1 - P(X = 0) = 1 - 0.7^{10} \approx 0.972$.

*Diagram: Bar chart of the binomial distribution with n equals 10 and p equals 0.3. The tallest bar is at 3 successes with probability about 0.27, with bars falling away on both sides and almost nothing beyond 7.*
*Bin(10, 0.3). The most likely number of successes is 3, which is also the mean, 10 × 0.3.*

### Calculator tip: One line for each probability
$P(X = 7)$ for $n = 10$, $p = 0.3$ is `10 nCr 7 × 0.3 ^ 7 × 0.7 ^ 3` ≈ 0.0090. For "at least one", use the complement: `1 - 0.8 ^ 5`. Some calculators have a binomial menu; use it only to check.

## Mean and variance

### Key idea: For X ~ Bin(n, p)
$$E(X) = np \qquad \mathrm{Var}(X) = np(1 - p) \qquad \sigma = \sqrt{np(1 - p)}$$

This makes sense: $X$ is the sum of $n$ separate Bernoulli trials, each with mean $p$ and variance $p(1 - p)$.

For the basketball player: $E(X) = 3$ shots, $\mathrm{Var}(X) = 2.1$, $\sigma \approx 1.45$ shots.

### Watch out: Is it really binomial?
Drawing cards **without replacement** changes $p$ after each draw, so it is not binomial. Always check the four conditions before using the formula.

**Mind map**

```text
Binomial distribution
  Bernoulli trial
    Two outcomes
    Mean p
    Variance p(1 - p)
  Conditions
    Fixed n
    Two outcomes
    Independent
    Same p
  Formula
    nCr p^r (1 - p)^(n - r)
    nCr counts the orders
  Summary numbers
    Mean np
    Variance np(1 - p)
  Tricks
    At least one: 1 - P(none)
    Use complements
```

### Year 11: How many tries do you need?
A die is rolled $n$ times. How large must $n$ be for the chance of at least one six to be more than 0.9?

$1 - \left(\dfrac{5}{6}\right)^n > 0.9$, so $\left(\dfrac{5}{6}\right)^n < 0.1$.

Take logs: $n\ln\dfrac{5}{6} < \ln 0.1$. Dividing by $\ln\dfrac{5}{6}$, which is negative, flips the sign: $n > 12.6$. So at least 13 rolls.

### Year 12: Working backwards from the mean and variance
$X \sim \mathrm{Bin}(n, p)$ has mean 6 and variance 4.2. Find $n$ and $p$.

Divide: $\dfrac{np(1 - p)}{np} = \dfrac{4.2}{6}$, so $1 - p = 0.7$ and $p = 0.3$. Then $n = \dfrac{6}{0.3} = 20$.

**More facts from the mean and variance.** $E(X^2) = \mathrm{Var}(X) + [E(X)]^2$, so for $X \sim \mathrm{Bin}(10, 0.5)$, $E(X^2) = 2.5 + 25 = 27.5$. The variance of one Bernoulli trial, $p(1 - p)$, is largest at $p = 0.5$, where it is 0.25.

**The most likely value.** Compare neighbouring probabilities: $\dfrac{P(X = k + 1)}{P(X = k)} = \dfrac{n - k}{k + 1} \cdot \dfrac{p}{1 - p}$. The probabilities rise while this ratio is above 1. For $\mathrm{Bin}(10, 0.3)$ the most likely value is 3, close to the mean $np$.

**Two or more.** $P(X \geq 2) = 1 - P(X = 0) - P(X = 1)$.

**A normal approximation.** For large $n$, $\mathrm{Bin}(n, p)$ is close to a normal distribution with mean $np$ and standard deviation $\sqrt{np(1 - p)}$. For $\mathrm{Bin}(100, 0.5)$, the mean is 50 and the standard deviation 5, so getting 60 or more ($z = 2$) has a chance of about 2.5%.

### Year 12: Why the probabilities add to 1
By the binomial theorem, $\displaystyle\sum_{r=0}^{n}\binom{n}{r}p^r(1 - p)^{n - r} = \big(p + (1 - p)\big)^n = 1^n = 1$. This is where the distribution gets its name.

## Quick summary

- A Bernoulli trial has two outcomes, with mean $p$ and variance $p(1 - p)$.
- Binomial needs: fixed $n$, two outcomes, independent trials, same $p$.
- $P(X = r) = \dbinom{n}{r}p^r(1 - p)^{n - r}$.
- $E(X) = np$ and $\mathrm{Var}(X) = np(1 - p)$.
- "At least one" is $1 - P(X = 0)$.

## Quiz

Try these questions from the chapter quiz.

1. Which of these is a Bernoulli trial?
   - A. Rolling a die and recording the number
   - B. Tossing a coin and recording heads or not
   - C. Measuring a person's height
   - D. Counting cars in an hour
   - Answer: B. Tossing a coin and recording heads or not. It has exactly two outcomes.

2. A Bernoulli variable has $p = 0.4$. Its variance is:
   - A. 0.4
   - B. 0.24
   - C. 0.16
   - D. 0.6
   - Answer: B. 0.24. $p(1 - p) = 0.4 \times 0.6$.

3. Which situation is NOT binomial?
   - A. 8 coin tosses, count heads
   - B. 20 free throws with a steady 60% success rate, count goals
   - C. Draw 5 cards without replacement, count hearts
   - D. 10 seeds each with an 80% chance to sprout, count sprouts
   - Answer: C. Draw 5 cards without replacement, count hearts. Without replacement, $p$ changes after each draw.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/binomial-distribution/#quiz