Define and apply the binomial distribution to model the number of successes in $n$ independent Bernoulli trials, including computing probabilities, expected value $np$ and variance $np(1-p)$

What this dot point is asking

VCAA wants you to recognise binomial situations, write the probability mass function, and use the standard formulas for the expected value and variance.

Bernoulli trial

A single experiment with two possible outcomes (success or failure) and probability $p$ of success. Examples: coin toss, single quality-control inspection, single penalty kick.

Binomial distribution

If $n$ independent Bernoulli trials are run with the same success probability $p$, then $X$ = number of successes is binomially distributed: $X \sim \text{Bin}(n, p)$.

Probability of exactly $k$ successes:

where $\binom{n}{k} = ^nC_k = \dfrac{n!}{k!(n - k)!}$.

Conditions

To apply the binomial:

• IMATH_11 is fixed in advance.
• Each trial has two outcomes (success / failure).
• Probability of success is the same on every trial.
• Trials are independent.

Expected value and variance

DMATH_1
DMATH_2

The mean $np$ is intuitive: in $10$ tosses of a fair coin, expect $5$ heads. The variance peaks at $p = 0.5$ (most uncertain outcome).

Cumulative probabilities

$P(X \le k) = \sum_{j=0}^k \binom{n}{j} p^j (1 - p)^{n - j}$.

CAS calculators give cumulative probabilities directly via BinomialCDF(n, p, k).

Worked example

A multiple-choice quiz has $20$ questions, each with $4$ options, and a student guesses randomly. $X$ = number correct.

$X \sim \text{Bin}(20, 0.25)$.

$E[X] = 20 \cdot 0.25 = 5$.

$\text{Var}(X) = 20 \cdot 0.25 \cdot 0.75 = 3.75$.

$P(X = 10) = \binom{20}{10} (0.25)^{10} (0.75)^{10} \approx 0.00992$.

About $1$% chance of getting half by random guessing.

Common traps

Misidentifying $p$. $p$ is the probability of "success" as defined. If the question asks for the probability of $k$ failures, redefine $p$ as the failure probability or use $\binom{n}{n-k} p^{n-k}(1-p)^k$.

Treating dependent trials as binomial. Drawing without replacement breaks independence. (For "without replacement" use the hypergeometric distribution, beyond Unit 2 scope.)

Using $E[X] = p$ instead of $np$. Multiply by $n$.

Computing $\binom{n}{k}$ incorrectly. $\binom{n}{k} = \binom{n}{n-k}$. CAS calculators give exact values.

In one sentence

The binomial distribution $X \sim \text{Bin}(n, p)$ models the number of successes in $n$ independent Bernoulli trials with success probability $p$, with $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$, expected value $E[X] = np$ and variance $\text{Var}(X) = np(1 - p)$.