Recognise the binomial distribution $X \sim \mathrm{Bin}(n, p)$ as the count of successes in $n$ independent Bernoulli trials, apply the binomial probability formula and use CAS, and use the formulas $E(X) = np$ and $\mathrm{Var}(X) = np(1 - p)$

What this dot point is asking

QCAA wants you to recognise when a count is binomially distributed, calculate binomial probabilities (by hand for small $n$, by CAS for larger $n$), and apply the mean and variance formulas. The binomial distribution is the most heavily examined probability model in QCE Mathematical Methods Unit 3, and it appears in IA1 PSMTs, IA2 short and extended response, and most EA Paper 2 probability questions.

The answer

When is $X$ binomial: the BINS conditions

A random variable $X$ has a binomial distribution if all four conditions hold.

• B (Binary): Each trial has two outcomes labelled success or failure.
• I (Independent): Trials are independent of one another.
• N (Number fixed): The number of trials $n$ is fixed in advance.
• S (Same probability): The probability of success $p$ is the same on every trial.

If all four hold, write $X \sim \mathrm{Bin}(n, p)$, where $X$ is the number of successes in the $n$ trials.

The binomial probability formula

For $X \sim \mathrm{Bin}(n, p)$ and $k = 0, 1, 2, \ldots, n$,

The binomial coefficient $\binom{n}{k} = \dfrac{n!}{k! \, (n - k)!}$ counts the number of ways to arrange $k$ successes among $n$ trials.

Paper 1 expects this formula by hand for small $n$ (typically $n \leq 6$) with values that simplify to clean fractions.

Paper 2 expects you to use CAS for any $n$ above about 6, calling functions named $\mathrm{binomPdf}(n, p, k)$ for $P(X = k)$ and $\mathrm{binomCdf}(n, p, a, b)$ for $P(a \leq X \leq b)$. Check the exact syntax for your approved CAS model.

Mean and variance

For $X \sim \mathrm{Bin}(n, p)$,

Standard deviation: $\sigma = \sqrt{n p (1 - p)}$.

These come from the fact that a binomial is the sum of $n$ independent Bernoulli trials, each with mean $p$ and variance $p (1 - p)$, and the rules $E(X + Y) = E(X) + E(Y)$ and $\mathrm{Var}(X + Y) = \mathrm{Var}(X) + \mathrm{Var}(Y)$ for independent variables.

Cumulative probabilities

QCAA commonly asks for $P(X \leq k)$, $P(X \geq k)$ or $P(a \leq X \leq b)$ rather than a single $P(X = k)$.

• IMATH_34 .
• IMATH_35 (use the complement to save work).
• IMATH_36 .

For $n$ above about 6, do all four with CAS. For Paper 1 with small $n$, set up the sums explicitly and evaluate by hand.

Worked examples

Paper 1: small $n$ by hand

A student answers 4 multiple-choice questions at random, each with 4 options. Let $X$ be the number correct. Find $P(X = 3)$.

$X \sim \mathrm{Bin}(4, 0.25)$.

$P(X = 3) = \binom{4}{3} (0.25)^3 (0.75)^1 = 4 \cdot \dfrac{1}{64} \cdot \dfrac{3}{4} = \dfrac{12}{256} = \dfrac{3}{64} \approx 0.0469.$

Paper 2: CAS-supported cumulative

A coin biased so that $P(\text{heads}) = 0.6$ is tossed 50 times. Find the probability of getting between 28 and 35 heads inclusive.

$X \sim \mathrm{Bin}(50, 0.6)$.

$P(28 \leq X \leq 35) = \mathrm{binomCdf}(50, 0.6, 28, 35) \approx 0.7411.$

By hand this would require 8 terms of the formula; CAS is required for IA2 and Paper 2 efficiency.

Mean and variance

A daily quality test on 100 items has $p = 0.02$ probability of any item being faulty. Find the expected number of faulty items and the standard deviation.

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

You would expect 2 faulty items per day, with results typically within $\pm 1.4$ of the mean.

Using the complement

A test has 20 multiple-choice questions, each with 5 options. A student answers at random. What is the probability of getting at least one correct?

$X \sim \mathrm{Bin}(20, 0.2)$.

$P(X \geq 1) = 1 - P(X = 0) = 1 - (0.8)^{20} \approx 1 - 0.0115 = 0.9885.$

The complement saves you from summing 20 binomial terms.

Recognition (not binomial)

A bag contains 5 red and 5 blue marbles. Three marbles are drawn without replacement. Let $X$ be the number of red marbles drawn. Is $X$ binomial?

No. The trials are not independent (drawing without replacement changes $p$ between draws). This is a hypergeometric situation, outside Methods. The BINS conditions fail at I (independence) and S (same $p$).

Common traps

Misidentifying $n$ and $p$. The number of trials $n$ is the count of opportunities for success, not the number of items in any other sense. The probability $p$ is the chance of success on one trial.

Forgetting the binomial coefficient. $P(X = k) = p^k (1 - p)^{n - k}$ alone is the probability of one specific sequence with $k$ successes. The coefficient $\binom{n}{k}$ counts how many sequences have that many successes.

Wrong direction for $P(X \geq k)$. It is $1 - P(X \leq k - 1)$, not $1 - P(X \leq k)$. The boundary value $k$ belongs in the "at least" event.

Using BINS for a without-replacement scenario. Without replacement violates independence. Methods only handles binomial probability for with-replacement (or effectively independent) trials.

Computing $E(X)$ as $p$ instead of $np$. Forgetting the $n$ factor is the most common Paper 1 mistake on the mean formula.

Wrong CAS syntax. Different calculators use slightly different function names (binomPdf, binomialPdf, etc.) and argument orders. Practise the exact syntax on your model before the IA2.

In one sentence

The binomial distribution $X \sim \mathrm{Bin}(n, p)$ models the number of successes in $n$ independent Bernoulli trials each with success probability $p$, with $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$, mean $E(X) = n p$ and variance $\mathrm{Var}(X) = n p (1 - p)$, and CAS support is the standard tool for cumulative probabilities on Paper 2.