Notes · StatisticsUK · A-Levels

Binomial distribution

This topic treats the binomial distribution B(n, p) as a model for the number of successes in a fixed number of independent trials. It covers the four modelling conditions, the calculation of exact and cumulative probabilities, the translation of worded phrases into inequalities, the mean np and variance np(1 - p), and the judgement of whether the model is appropriate.

5 sections~12 min reading time3 competenciesLevel Foundation 1 · Standard 3 · Advanced 1

T·0666 / 18

Exam profile

AO1 · Calculate binomial probabilities, mean and varianceAO2 · Interpret and justify the binomial modelling conditionsAO3 · Select the binomial where appropriate and solve problems in context

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basic level

The AS foundation expects the binomial model, its conditions, and probabilities via the formula and cumulative tables.

higher level

The full A-Level expects fluent cumulative work, the mean and variance, and a reasoned judgement of the model in context.

Depth

Text

Contents · 5 sections

Binomial distribution

The binomial model and its conditions#

FoundationPearson Edexcel 9ST0, Topic 5 (Paper 1)

The binomial distribution models the number of successes

X

in a fixed number

n

of trials. Four conditions must hold, and they are examined explicitly: there is a fixed number of trials

n

; each trial has only two outcomes, success or failure; the probability of success

p

is constant from trial to trial; and the trials are independent. When these hold we write

X \sim B(n, p)

, read as '

X

is binomially distributed with parameters

n

and

p

The two outcomes need only be a success/failure dichotomy imposed on the situation: a die roll becomes binomial if 'success' is defined as rolling a six, with

p = \tfrac{1}{6}

. The parameter

n

is the number of trials and

p

the probability of success on a single trial; identifying these correctly from the wording is the essential first step.

The independence and constant-probability conditions are the ones most often violated. Sampling without replacement from a small population changes

p

between draws and destroys independence, so the binomial is then only approximate; sampling with replacement, or from a very large population, preserves them. Being able to say why the conditions do or do not hold in a given context is an AO2/AO3 skill worth as much as the arithmetic.

Because each trial is a Bernoulli trial with the same

p

, the binomial is the distribution of the sum of

n

independent Bernoulli variables. This structural fact explains its mean and variance later and its symmetry when

p = 0.5

: the number of successes and the number of failures then have the same distribution.

X \sim B(n, p) \quad\text{with fixed } n,\; \text{constant } p,\; \text{independent two-outcome trials}

The binomial model

The four conditions that must be checked before the model is used.

Worked example

Is the binomial model appropriate?

Twenty cards are drawn one at a time without replacement from a standard pack, and $X$ is the number of hearts drawn. Explain whether $X$ is binomial.

Fixed trials and two outcomes
There are $n = 20$ draws, each a heart (success) or not (failure), so those two conditions hold.
Constant probability
Without replacement, $P(\text{heart})$ changes after each draw (it starts at $\tfrac{13}{52}$ and shifts), so $p$ is not constant.
Independence
The draws are dependent for the same reason.
Conclude
$X$ is NOT exactly binomial; a binomial would be appropriate only if the cards were replaced (or the pack were very large).

Result: The model fails the constant-probability and independence conditions, so $X$ is not binomial under sampling without replacement.

Typical mistakes

Applying the binomial when trials are not independent (sampling without replacement from a small population).
Confusing $n$ (number of trials) with the number of successes, or mislabelling $p$ .

Active revision

A biased coin has $P(\text{head}) = 0.3$ and is tossed 12 times. State the distribution of the number of heads and check the binomial conditions.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Calculating binomial probabilities#

StandardPearson Edexcel 9ST0, Topic 5 (Paper 1)

A symmetric binomial distribution

Fig. 1B(6, 0.5) is symmetric about its mean np = 3; the bars give P(X = r).

The probability of exactly

r

successes is

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

. The binomial coefficient

\binom{n}{r}

counts the number of ways of choosing which

r

of the

n

trials succeed;

p^{r}

is the probability that those succeed and

(1 - p)^{n - r}

that the remaining trials fail. Summed over all

r

from 0 to

n

, these probabilities total 1, since they are the terms of the binomial expansion of

(p + (1-p))^n

The shape of the distribution depends on

p

. When

p = 0.5

the distribution is symmetric about its mean, as the bars of

B(6, 0.5)

below show, peaking at

r = 3

. When

p \neq 0.5

the distribution is skewed — towards the right for small

p

and towards the left for large

p

— a point revisited when the mean is introduced.

For a single exact probability the formula is used directly. For example, with

X \sim B(20, 0.3)

P(X = 6) = \binom{20}{6}(0.3)^6(0.7)^{14} = 0.192

(3 s.f.), using

\binom{20}{6} = 38760

. Keeping full accuracy through the powers before rounding at the end avoids accumulated error.

In examinations most binomial probabilities are found with a calculator's binomial functions rather than by the formula, but the formula must be understood, because it explains where cumulative tables come from and is required whenever the calculator's built-in functions do not directly give the quantity asked for.

P(X = r) = \binom{n}{r} p^{r}(1 - p)^{n - r}, \quad r = 0, 1, \ldots, n

The binomial probability

Choose which $r$ of $n$ trials succeed, then multiply the success and failure probabilities.

Worked example

An exact binomial probability

A component has probability 0.3 of failing a stress test, independently. Twenty components are tested. Find the probability that exactly 6 fail.

State the distribution
Let $X$ be the number that fail, so $X \sim B(20, 0.3)$ .
Apply the formula
$P(X = 6) = \binom{20}{6}(0.3)^6(0.7)^{14}$ with $\binom{20}{6} = 38760$ .
Evaluate
$38760 \times (0.3)^6 \times (0.7)^{14} = 0.192$ (3 s.f.).
$P(X = 6) = 38760\,(0.3)^6(0.7)^{14} = 0.192$

Result: $P(X = 6) = 0.192$ (3 s.f.).

Typical mistakes

Swapping the exponents of $p$ and $(1 - p)$ .
Rounding the powers early, so the final probability is inaccurate.

Active revision

For $X \sim B(20, 0.3)$ , find $P(X = 6)$ using the binomial formula.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Cumulative probabilities and worded phrases#

StandardPearson Edexcel 9ST0, Topic 5 (Paper 1)

The cumulative function gives

P(X \leq r)

, and tables and calculators return it directly. Every other probability is obtained from it by subtraction:

P(X = r) = P(X \leq r) - P(X \leq r - 1)

P(X \geq r) = 1 - P(X \leq r - 1)

, and

P(a \leq X \leq b) = P(X \leq b) - P(X \leq a - 1)

. The whole difficulty lies in choosing the right cumulative value.

Translating the wording into the correct inequality is where marks are won and lost. 'At least

r

' means

X \geq r

; 'more than

r

' means

X \geq r + 1

; 'fewer than

r

' means

X \leq r - 1

; 'at most

r

' means

X \leq r

. A single shift of one in the index converts a wrong answer into a right one, so it is worth writing the inequality out explicitly before touching the calculator.

For

X \sim B(20, 0.3)

, 'at least 8 fail' means

P(X \geq 8) = 1 - P(X \leq 7)

. With

P(X \leq 7) = 0.772

from the cumulative distribution,

P(X \geq 8) = 1 - 0.772 = 0.228

(3 s.f.). Note how the complement turns an awkward upper-tail sum into a single table look-up.

The same technique answers questions about the smallest sample size or the critical number needed to achieve a stated probability, which are the seeds of hypothesis testing. There, the critical region is precisely the set of values of

X

whose cumulative probability falls below the significance level, so this cumulative fluency transfers directly.

P(X \geq r) = 1 - P(X \leq r - 1), \qquad P(a \leq X \leq b) = P(X \leq b) - P(X \leq a - 1)

Cumulative binomial identities

Every probability is built from $P(X \leq r)$ ; mind the shift of one at each endpoint.

Worked example

An upper-tail cumulative probability

For $X \sim B(20, 0.3)$ , find the probability that at least 8 components fail, given $P(X \leq 7) = 0.772$ .

Translate the phrase
'At least 8' means $X \geq 8$ .
Use the complement
$P(X \geq 8) = 1 - P(X \leq 7)$ .
Substitute
$1 - 0.772 = 0.228$ (3 s.f.).
$P(X \geq 8) = 1 - 0.772 = 0.228$
Interpret
There is about a 22.8% chance that 8 or more of the 20 components fail.

Result: $P(X \geq 8) = 0.228$ (3 s.f.).

Typical mistakes

Using $P(X \leq r)$ where $P(X \leq r - 1)$ is required for $P(X \geq r)$ (an off-by-one error).
Reading 'more than $r$ ' as $X \geq r$ instead of $X \geq r + 1$ .

Active revision

For $X \sim B(20, 0.3)$ with $P(X \leq 7) = 0.772$ , find $P(X \geq 8)$ and $P(X > 5)$ given $P(X \leq 5) = 0.416$ .

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Mean and variance of the binomial#

StandardPearson Edexcel 9ST0, Topic 5 (Paper 1)

A right-skewed binomial distribution

Fig. 2B(10, 0.2) has mean np = 2 and is right-skewed; probabilities for r = 0 to 8.

The binomial has mean

E(X) = np

and variance

\text{Var}(X) = np(1 - p)

. The mean follows from the structure of the distribution as a sum of

n

independent Bernoulli trials each with mean

p

; by the additivity of expectation the means add to

np

. Likewise the variance of each Bernoulli trial is

p(1 - p)

, and independence lets the variances add to

np(1 - p)

The variance is largest when

p = 0.5

, where

p(1 - p) = 0.25

is maximised, and shrinks as

p

moves towards 0 or 1. This matches intuition: a very reliable or very unreliable process is more predictable than a fifty-fifty one. The standard deviation is

\sqrt{np(1 - p)}

, in the units of a count of successes.

The distribution

B(10, 0.2)

below has mean

np = 2

and is right-skewed, with its peak at

r = 2

and a tail stretching to the right. Comparing it with the symmetric

B(6, 0.5)

makes concrete how

p

controls both the location (through

np

) and the asymmetry of the distribution.

These formulae are used both forwards, to summarise a known binomial, and backwards, to recover

n

p

from a stated mean and variance. Since

\frac{\text{Var}(X)}{E(X)} = 1 - p

, the mean and variance together determine

p

and then

n

, a manipulation that appears regularly in problem-solving questions.

E(X) = np, \qquad \text{Var}(X) = np(1 - p)

Mean and variance of B(n, p)

The variance is maximised at $p = 0.5$ ; the standard deviation is $\sqrt{np(1 - p)}$ .

Worked example

Recovering n and p

A binomial random variable has mean 6 and variance 4.2. Find $p$ and $n$ .

Use the ratio
$\frac{\text{Var}(X)}{E(X)} = \frac{np(1 - p)}{np} = 1 - p = \frac{4.2}{6} = 0.7$ .
Find p
$1 - p = 0.7$ , so $p = 0.3$ .
Find n
$np = 6$ gives $n = \frac{6}{0.3} = 20$ .
$n = \frac{E(X)}{p} = \frac{6}{0.3} = 20$
Check
$np(1 - p) = 20 \times 0.3 \times 0.7 = 4.2$ , as given.

Result: The distribution is $B(20, 0.3)$ , so $n = 20$ and $p = 0.3$ .

Typical mistakes

Using $np$ for the variance instead of $np(1 - p)$ .
Forgetting the square root when the standard deviation is required.

Active revision

A binomial variable has mean 6 and variance 4.2. Find $n$ and $p$ .

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Checking the model and problem solving#

AdvancedPearson Edexcel 9ST0, Topic 5 (Paper 1)

Deciding whether the binomial is appropriate is a modelling judgement that recurs throughout the paper. The checklist is the four conditions: a fixed number of trials, two outcomes, constant

p

and independence. Real contexts often satisfy them only approximately, and a good answer states which condition is in doubt and why, rather than asserting the model without comment.

A useful diagnostic is that a genuine binomial has variance

np(1 - p)

, which is always less than the mean

np

(since

1 - p < 1

). Count data whose variance markedly exceeds their mean (overdispersion) are a warning that the binomial, or later the Poisson, is a poor fit, perhaps because the trials are not independent or

p

varies between them.

Problem-solving questions typically combine the model with the earlier probability laws. A common type asks for the probability of an event over several independent groups, or conditions on a binomial outcome; here the binomial supplies one stage of a tree or a compound calculation. Setting the distribution up carefully, and defining the success clearly, prevents most errors.

Finally, conclusions must be returned to the context. A computed probability of 0.228 is only an answer once it is stated as 'there is about a 23% chance that at least 8 components fail', with any modelling caveats attached. The examination consistently rewards this honest, contextual interpretation over a bare numerical result.

\text{Var}(X) = np(1 - p) < np = E(X)

Binomial variance is below the mean

A count whose variance exceeds its mean is a sign the binomial model may be inappropriate.

Worked example

A binomial in context with a modelling comment

In a large population, 15% of people carry a certain gene. A random sample of 12 is taken and $X$ is the number of carriers. Find $P(X \geq 1)$ and comment on the model.

State the model
The population is large so sampling barely changes $p$ ; take $X \sim B(12, 0.15)$ .
Use the complement
$P(X \geq 1) = 1 - P(X = 0) = 1 - (0.85)^{12}$ .
Evaluate
$(0.85)^{12} = 0.142$ , so $P(X \geq 1) = 1 - 0.142 = 0.858$ (3 s.f.).
$P(X \geq 1) = 1 - (0.85)^{12} = 0.858$
Comment
The binomial is reasonable because the population is large, keeping $p$ approximately constant and the selections approximately independent.

Result: $P(X \geq 1) = 0.858$ (3 s.f.); the binomial is appropriate because the large population keeps $p$ essentially constant.

Typical mistakes

Asserting the binomial model without checking or naming the doubtful condition.
Leaving the answer as a number without a contextual interpretation.

Active revision

A quality inspector models the number of faulty items in a box of 24 as $B(24, 0.05)$ . Discuss one reason this model might fail and find the probability that a box contains no faulty items.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content (Statistics) (Department for Education / Ofqual)

References & sources

Sources

Pearson Edexcel

Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification

Department for Education / Ofqual

GCE AS and A level subject content (Statistics)

Binomial distribution

5 sections~12 min reading time3 competenciesLevel Foundation 1 · Standard 3 · Advanced 1

Binomial distribution

Key points

Is the binomial model appropriate?

Key points

An exact binomial probability

Key points

An upper-tail cumulative probability

Key points

Recovering n and p

Key points

A binomial in context with a modelling comment

Sources

Binomial distribution

Key points

Is the binomial model appropriate?

Key points

An exact binomial probability

Key points

An upper-tail cumulative probability

Key points

Recovering n and p

Key points

A binomial in context with a modelling comment

Sources