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The binomial distribution (QCE Mathematical Methods Unit 3)

Syllabus dot point

“Recognise the binomial distribution X∼Bin(n,p)X \sim \mathrm{Bin}(n, p) as the count of successes in nn independent Bernoulli trials, apply the binomial probability formula and use a graphics calculator, and use the formulas E(X)=npE(X) = np and Var(X)=np(1−p)\mathrm{Var}(X) = np(1 - p)”

9 min read

Quick answer

The binomial distribution X∼Bin(n,p)X \sim \mathrm{Bin}(n, p) models the number of successes in nn independent Bernoulli trials each with success probability pp, with P(X=k)=(nk)pk(1−p)n−kP(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}, mean E(X)=npE(X) = n p and variance Var(X)=np(1−p)\mathrm{Var}(X) = n p (1 - p), and a graphics calculator is the standard tool for cumulative probabilities on Paper 2.

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  1. What this dot point is asking
  2. The answer
  3. Exam-style questions

What this dot point is asking

QCAA wants you to recognise when a count is binomially distributed, calculate binomial probabilities (by hand for small nn, by graphics calculator for larger nn), 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 XX binomial: the BINS conditions

A random variable XX 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 nn is fixed in advance.
  • S (Same probability): The probability of success pp is the same on every trial.

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

The binomial probability formula

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

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

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

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

Paper 2 expects you to use your graphics calculator for any nn above about 6, calling functions usually named binompdf\mathrm{binompdf} for P(X=k)P(X = k) and binomcdf\mathrm{binomcdf} for P(X≤k)P(X \leq k) (some models take a lower and upper bound instead). Check the exact syntax for your calculator. QCAA allows a graphics calculator with no CAS functionality (and/or a scientific calculator) in Paper 2, so confirm your model against QCAA's graphics calculator list; Paper 1 allows no calculator at all.

Mean and variance

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

E(X)=np,Var(X)=np(1−p).E(X) = n p, \qquad \mathrm{Var}(X) = n p (1 - p).

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

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

Cumulative probabilities

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

  • P(X≤k)=∑i=0kP(X=i)P(X \leq k) = \sum_{i = 0}^{k} P(X = i).
  • P(X≥k)=1−P(X≤k−1)P(X \geq k) = 1 - P(X \leq k - 1) (use the complement to save work).
  • P(a≤X≤b)=P(X≤b)−P(X≤a−1)P(a \leq X \leq b) = P(X \leq b) - P(X \leq a - 1).

For nn above about 6, do all four with your graphics calculator. For Paper 1 with small nn, set up the sums explicitly and evaluate by hand.

Worked example

Paper 1: small nn by hand

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

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

P(X=3)=(43)(0.25)3(0.75)1=4⋅164⋅34=12256=364≈0.0469.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: calculator-supported cumulative

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

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

P(28≤X≤35)=P(X≤35)−P(X≤27)=binomcdf(50,0.6,35)−binomcdf(50,0.6,27)≈0.7121.P(28 \leq X \leq 35) = P(X \leq 35) - P(X \leq 27) = \mathrm{binomcdf}(50, 0.6, 35) - \mathrm{binomcdf}(50, 0.6, 27) \approx 0.7121.

By hand this would require 8 terms of the formula; the calculator saves time in Paper 2.

Mean and variance

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

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

You would expect 2 faulty items per day, with results typically within ±1.4\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∼Bin(20,0.2)X \sim \mathrm{Bin}(20, 0.2).

P(X≥1)=1−P(X=0)=1−(0.8)20≈1−0.0115=0.9885.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 XX be the number of red marbles drawn. Is XX binomial?

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

Common traps
Misidentifying nn and pp
The number of trials nn is the count of opportunities for success, not the number of items in any other sense. The probability pp is the chance of success on one trial.
Forgetting the binomial coefficient
P(X=k)=pk(1−p)n−kP(X = k) = p^k (1 - p)^{n - k} alone is the probability of one specific sequence with kk successes. The coefficient (nk)\binom{n}{k} counts how many sequences have that many successes.
Wrong direction for P(X≥k)P(X \geq k)
It is 1−P(X≤k−1)1 - P(X \leq k - 1), not 1−P(X≤k)1 - P(X \leq k). The boundary value kk 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)E(X) as pp instead of npnp. Forgetting the nn factor is the most common Paper 1 mistake on the mean formula.

Wrong calculator syntax. Different calculators use slightly different function names (binompdf, binomialpdf, etc.) and argument orders. Practise the exact syntax on your model, and check it is an approved non-CAS model, before the exam.

Exam-style questions

Questions in the style of QCAA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice3 marks
A fair six-sided die is rolled 5 times. Let XX be the number of sixes. Find P(X=2)P(X = 2) exactly.
Show worked answer →

X∼Bin(5,16)X \sim \mathrm{Bin}(5, \frac{1}{6}) because rolling a six is a single success in each independent trial.

P(X=2)=(52)(16)2(56)3=10⋅136⋅125216=12507776=6253888.P(X = 2) = \binom{5}{2} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^3 = 10 \cdot \frac{1}{36} \cdot \frac{125}{216} = \frac{1250}{7776} = \frac{625}{3888}.

As a decimal, ≈0.1608\approx 0.1608.

Markers reward the correct identification of n=5n = 5 and p=16p = \frac{1}{6}, the binomial coefficient (52)=10\binom{5}{2} = 10, the powers of pp and 1−p1 - p matched to successes and failures, and the simplified exact fraction. Paper 1 wants exact form.

Practice5 marks
A quality control process tests batches of 20 components from a production line. The probability that any one component is faulty is 0.050.05, independently of the others. Let XX be the number of faulty components in a batch. (a) Find P(X=0)P(X = 0). (b) Find P(X≥2)P(X \geq 2). (c) Find the expected number of faulty components per batch and the standard deviation.
Show worked answer →

A 5-mark answer needs the distribution statement, both probabilities (with the complement trick for part (b)), and the descriptive statistics.

X∼Bin(20,0.05)X \sim \mathrm{Bin}(20, 0.05).

(a) P(X=0)=(0.95)20≈0.3585P(X = 0) = (0.95)^{20} \approx 0.3585.

(b) Use the complement: P(X≥2)=1−P(X=0)−P(X=1)P(X \geq 2) = 1 - P(X = 0) - P(X = 1).

P(X=1)=(201)(0.05)(0.95)19=20⋅0.05⋅(0.95)19≈0.3774P(X = 1) = \binom{20}{1} (0.05)(0.95)^{19} = 20 \cdot 0.05 \cdot (0.95)^{19} \approx 0.3774.

P(X≥2)=1−0.358486−0.377354≈0.2642P(X \geq 2) = 1 - 0.358486 - 0.377354 \approx 0.2642.

(Graphics calculator check: 1−binomcdf(20,0.05,1)≈0.26421 - \mathrm{binomcdf}(20, 0.05, 1) \approx 0.2642.)

(c) E(X)=np=20⋅0.05=1E(X) = n p = 20 \cdot 0.05 = 1. Var(X)=np(1−p)=20⋅0.05⋅0.95=0.95\mathrm{Var}(X) = n p (1 - p) = 20 \cdot 0.05 \cdot 0.95 = 0.95. Standard deviation ≈0.9747\approx 0.9747.

Markers reward stating the distribution explicitly, the complement-trick setup for P(X≥2)P(X \geq 2) (rather than summing 19 terms), the calculator-friendly numerical answers to 4 decimal places, and the mean-variance formulas applied cleanly.

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