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The binomial distribution

What it models

The binomial distribution counts the number of successes in a fixed number of trials.
Write it as:

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

where $n$ is the number of trials and $p$ is the probability of success on each trial.
"Success" simply means the outcome being counted — it need not be a good thing. Counting faulty items makes "faulty" the success.

The four conditions

These must all hold, and questions very often turn on one of them failing. Learn them as a checklist:

Condition	What it means	Fails when
Fixed $n$	The number of trials is decided in advance	You keep going until something happens
Two outcomes	Each trial is a success or a failure	There are three or more categories
Constant $p$	The probability is the same on every trial	Conditions change; sampling without replacement from a small population
Independence	Trials do not affect each other	A shared cause links the trials

Stating which condition fails, and why, is worth more than any calculation. This is the most reliable place to earn Merit and Excellence marks in this topic.

The formula and the calculator

$P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}$

$\binom{n}{x}$ counts the number of ways to get $x$ successes among $n$ trials.
In practice you use your calculator's binomial pdf for $P(X = x)$ and binomial cdf for $P(X \leqslant x)$ . The specification expects a graphing calculator.

Wanted	Calculator
$P(X = 4)$	binompdf $(n, p, 4)$
$P(X \leqslant 4)$	binomcdf $(n, p, 4)$
$P(X < 4)$	binomcdf $(n, p, 3)$
$P(X \geqslant 4)$	$1 -$ binomcdf $(n, p, 3)$
$P(X > 4)$	$1 -$ binomcdf $(n, p, 4)$
$P(2 \leqslant X \leqslant 5)$	binomcdf $(n,p,5) -$ binomcdf $(n,p,1)$

Mean and standard deviation

$\mu = np \qquad \sigma = \sqrt{np(1-p)}$

Both are given in the Formulae and Tables Booklet.
Use $\mu = np$ as a sanity check on every binomial answer. If $n = 40$ and $p = 0.1$ , the mean is 4 — so a calculated $P(X = 12)$ ought to be small. An answer that contradicts this signals a data-entry error.

Shape

$p < 0.5$ → right-skewed. $p = 0.5$ → symmetric. $p > 0.5$ → left-skewed.
The larger $n$ becomes, the more the shape approaches a normal curve — which is why a normal model can approximate a binomial with large $n$ .

Sampling without replacement

Strictly, drawing without replacement makes $p$ change between trials, so the binomial does not apply.
In practice, if the sample is a small fraction of the population (a common rule of thumb is under 10%), the change in $p$ is negligible and the binomial is an acceptable approximation.
Say this explicitly when it applies: "Sampling 20 from 5 000 changes $p$ negligibly, so a binomial model is reasonable."

Tips and Tricks

Define the variable and state the distribution first. "Let $X$ = number of defective components in the sample; $X \sim \text{Bin}(25, 0.08)$ ." One line, and it makes everything after it checkable.

Convert the inequality before touching the calculator. Write $P(X \geqslant 3) = 1 - P(X \leqslant 2)$ on the page. Nearly every lost binomial mark is an off-by-one on the cdf.

Check against $\mu = np$ . It takes two seconds and catches transposed parameters.

When asked whether a binomial model is appropriate, go through all four conditions and say something about each. Naming the one that fails, with the reason from the context, is what the question is testing — a bare "yes it is binomial" earns almost nothing.

Interpret in context with the number of trials named: "the probability that more than 3 of the 25 components are defective is 0.089."

Worked ExampleBinomial calculations and checking the conditions

A vineyard's bottling line has a 7% rate of under-filled bottles. A quality inspector takes a random sample of 30 bottles from a day's production of about 8 000.

Let $X$ be the number of under-filled bottles in the sample.

(a) State the distribution of $X$ and justify that a binomial model is appropriate. (b) Find the probability that exactly 2 bottles are under-filled. (c) Find the probability that at least 4 bottles are under-filled. (d) Find the mean and standard deviation of $X$ , and comment on whether finding 6 under-filled bottles would be surprising. (e) The inspector instead decides to keep testing bottles until she finds 3 under-filled ones, and record how many she tested. Explain why a binomial model does not apply.

(a) The distribution and its conditions

$X \sim \text{Bin}(n = 30,\ p = 0.07)$

Check all four conditions:

Fixed $n$ ✓ — the inspector decided in advance to test 30 bottles.
Two outcomes ✓ — each bottle is either under-filled or not.
Constant $p$ ✓ (approximately) — the sample of 30 is drawn without replacement, but from about 8 000 bottles, so the sample is under 0.4% of production and the probability changes negligibly from bottle to bottle.
Independence ✓ (assumed) — whether one bottle is under-filled does not affect another, provided the filling machine's behaviour is stable across the day.

(b) Exactly 2 under-filled

$P(X = 2) = \binom{30}{2}(0.07)^2(0.93)^{28} = \text{binompdf}(30, 0.07, 2)$

$= 0.26374...$

(c) At least 4 under-filled

Step 1 — Convert the inequality.

$P(X \geqslant 4) = 1 - P(X \leqslant 3)$

Step 2 — Compute.

$P(X \leqslant 3) = \text{binomcdf}(30, 0.07, 3) = 0.89305...$

$P(X \geqslant 4) = 1 - 0.89305... = 0.10694...$

Note the off-by-one: using binomcdf at 4 would have given $P(X \geqslant 5)$ and an answer of 0.037 — a very different figure.

(d) Mean, standard deviation, and judging 6

Step 1 — Mean.

$\mu = np = 30 \times 0.07 = 2.1$

Step 2 — Standard deviation.

$\sigma = \sqrt{np(1-p)} = \sqrt{30 \times 0.07 \times 0.93} = \sqrt{1.953} = 1.3975...$

Step 3 — Judge 6 under-filled bottles.

$\frac{6 - 2.1}{1.3975} = 2.79 \text{ standard deviations above the mean}$

$P(X \geqslant 6) = 1 - \text{binomcdf}(30, 0.07, 5) = 0.01185...$

Step 4 — Interpret for the vineyard.

(e) Why the redesigned sampling is not binomial

Step 1 — Test the conditions.

Step 2 — Identify what is now the random variable.

Step 3 — Conclude.