N1StatisticsCore

The Binomial Distribution

Q: Why does the variance involve p(1-p)?

Each trial contributes variance p(1-p), which is largest at p=0.5 where the outcome is least predictable, and independence lets the n trials’ variances add.

Q: When is the binomial coefficient not needed?

Only for r = 0 or r = n, where nr = 1 because there is exactly one arrangement. Every other case needs it.

Count the successes in a fixed number of independent trials, each with the same probability of success, and you have a $\textbf{binomial}$ random variable. It is the first named distribution in the course, and most of the marks come from knowing when it applies rather than from the arithmetic.

35 min Statistical Distributions

The big picture

The binomial matters because it is the bridge from probability to inference. Once you can compute the probability of an observed count under an assumed success rate, you can ask whether the observed count is surprising — and that question is exactly what a hypothesis test formalises in the next chapter. The four conditions are therefore not a checklist to recite: they are the assumptions the whole inference rests on, and when one fails the conclusion fails with it. Sampling without replacement is the case worth watching, because it breaks independence quietly.

What you'll be able to do

State the four conditions for a binomial model and test a scenario against them
Use the binomial probability formula for individual outcomes
Use cumulative binomial probabilities for "at most", "at least" and ranges
Calculate the mean and variance of a binomial distribution

The four conditions

A binomial model requires all four of the following. There must be a $\textbf{fixed number}$ of trials $n$ , decided in advance. Each trial has exactly $\textbf{two outcomes}$ , labelled success and failure. The trials are $\textbf{independent}$ . And the probability of success $p$ is $\textbf{constant}$ across trials.

Questions frequently describe a scenario and ask whether a binomial model is suitable. The answer is a check against these four, naming which one fails when it does.

The condition that most often fails is independence, and the standard cause is sampling $\textbf{without replacement}$ from a small population — each selection changes the composition for the next, so $p$ shifts as you go.

X \sim B(n, p)

Notation:

n

trials, success probability

p

Example 1A bag has 5 red and 5 blue counters. Ten are drawn without replacement and the number of reds is counted. Is a binomial model appropriate?

1Fixed number of trials: yes, 10.

2Two outcomes: yes, red or blue.

3Constant

p

and independence: no — after each draw the composition changes, so

p

is not constant.

AnswerNot binomial: drawing without replacement breaks independence and makes

p

vary between trials.

Tip — When a model is rejected, name the specific failing condition and why. "Not binomial because the trials are not independent — the counters are not replaced" is the full answer.

The probability formula

The probability of exactly $r$ successes has three factors. There are $\binom{n}{r}$ ways to choose which trials succeed; each of those arrangements has probability $p^{r}$ for the successes and $\left(1-p\right)^{n-r}$ for the failures.

The binomial coefficient is what accounts for order. Three successes in five trials can happen in $\binom{5}{3} = 10$ different arrangements, all equally likely, so a single arrangement’s probability is multiplied by 10.

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

Arrangements × successes × failures.

\binom{n}{r} = \dfrac{n!}{r!\left(n-r\right)!}

Number of arrangements.

Example 1A fair coin is tossed 6 times. Find the probability of exactly 4 heads.

n = 6

r = 4

p = 0.5

\binom{6}{4} = 15

P = 15 \times 0.5^{4} \times 0.5^{2} = 15 \times 0.5^{6} = \tfrac{15}{64}

Answer

\tfrac{15}{64} \approx 0.234

Example 2A component fails with probability

0.2

. In a batch of 8, find the probability exactly 2 fail.

\binom{8}{2} = 28

P = 28 \times 0.2^{2} \times 0.8^{6}

= 28 \times 0.04 \times 0.262144

Answer

0.294

(3 s.f.)

Why this works

The exponents must sum to $n$ : $r$ successes and $n-r$ failures account for every trial. If they do not add up, something has been miscounted.

Cumulative probabilities

Most questions ask about ranges rather than single values, and calculators provide the cumulative function $P(X \leq r)$ . Translating the wording into cumulative form is the skill being tested.

"At most $r$ " is $P(X \leq r)$ directly. "Fewer than $r$ " is $P(X \leq r-1)$ , since $X$ takes whole-number values. "At least $r$ " is $1 - P(X \leq r-1)$ , and "more than $r$ " is $1 - P(X \leq r)$ .

A range $a \leq X \leq b$ is $P(X \leq b) - P(X \leq a-1)$ — subtract everything strictly below the lower limit, which means using $a-1$ rather than $a$ .

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

At least: complement of at most

r-1

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

Ranges, with the

-1

that is easy to lose.

Example 1For

X \sim B(20, 0.3)

, express

P(X \geq 8)

and

P(5 \leq X \leq 9)

in cumulative form.

1"At least 8" excludes 0 to 7:

1 - P(X \leq 7)

2For the range, take everything up to 9 and remove everything up to 4.

Answer

P(X \geq 8) = 1 - P(X\leq7)

;

P(5\leq X\leq9) = P(X\leq9) - P(X\leq4)

Tip — Because the binomial is discrete, "at least 8" and "more than 8" differ. Write out which values are included before reaching for the calculator.

Mean and variance

The mean of a binomial is $np$ , which matches intuition: 20 trials at $p = 0.3$ should give about 6 successes.

The variance is $np\left(1-p\right)$ . It is largest when $p = 0.5$ and shrinks towards zero as $p$ approaches 0 or 1 — which makes sense, since an almost-certain or almost-impossible event produces very consistent counts.

The standard deviation is the square root, and is used to judge whether an observed count is unusual — the idea the hypothesis-testing chapter formalises.

E(X) = np

Mean.

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

Variance.

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

Standard deviation.

Example 1For

X \sim B(50, 0.4)

, find the mean and standard deviation.

1Mean:

50 \times 0.4 = 20

2Variance:

50 \times 0.4 \times 0.6 = 12

3Standard deviation:

\sqrt{12}

AnswerMean 20, standard deviation

3.46

(3 s.f.).

Why this works

Note the variance uses $p(1-p)$ , which peaks at $p = 0.5$ . Uncertainty about the outcome of each trial is greatest when success and failure are equally likely, and that feeds straight through to the spread of the total.

Think like an examiner

When asked whether a binomial model applies, check all four conditions and name any that fail.
Define the variable explicitly: "Let $X$ be the number of faulty components, $X \sim B(8, 0.2)$ ."
Translate the wording into cumulative form before calculating, minding the $-1$ for "at least" and ranges.
Give probabilities to 3 significant figures unless told otherwise.

Common misconceptions

Using

P(X \leq r)

for "at least

r

"At least

r

" is

1 - P(X \leq r-1)

Applying a binomial model to sampling without replacement.

That breaks independence and makes

p

vary. Say which condition fails.

Omitting the binomial coefficient.

\binom{n}{r}

counts the arrangements; without it you have only one ordering.

Writing the variance as

np

The mean is

np

; the variance is

np(1-p)

Binomial

X \sim B(n,p)

Notation.

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

Exact probability.

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

At least.

E(X) = np

Mean.

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

Variance.

Stretch yourself

A multiple-choice test has 15 questions, each with 4 options. A student guesses every answer. Find the probability they get at least 6 correct, and comment on whether guessing is a viable strategy for a pass mark of 6.

Hint — Identify $n$ and $p$ first, then convert "at least 6" into cumulative form.

Reveal solution

SolutionGuessing gives

p = \tfrac14 = 0.25

per question, and the questions are independent with a fixed

n = 15

, so

X \sim B(15, 0.25)

. "At least 6" is

1 - P(X \leq 5)

. From cumulative binomial tables or a calculator,

P(X \leq 5) \approx 0.8516

, so

P(X \geq 6) \approx 0.148

. The mean number correct is

np = 15 \times 0.25 = 3.75

, with standard deviation

\sqrt{15 \times 0.25 \times 0.75} = \sqrt{2.8125} \approx 1.68

. A score of 6 is therefore about

\tfrac{6 - 3.75}{1.68} \approx 1.34

standard deviations above the mean. Guessing succeeds roughly one time in seven — better than nothing, but it fails about 85% of the time, so it is not a viable strategy.

\blacksquare

Questions students ask

What if the population is large but sampling is without replacement?

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 a binomial model is an acceptable approximation. Say that you are approximating.

Does "success" have to be a good outcome?

No. Success simply labels the outcome you are counting. Counting faulty components makes "faulty" the success, however odd that sounds.

Why does the variance involve

p(1-p)

Each trial contributes variance

p(1-p)

, which is largest at

p=0.5

where the outcome is least predictable, and independence lets the

n

trials’ variances add.

When is the binomial coefficient not needed?

Only for

r = 0

r = n

, where

\binom{n}{r} = 1

because there is exactly one arrangement. Every other case needs it.

Key takeaways

A binomial needs fixed $n$ , two outcomes, independence, and constant $p$ .
Sampling without replacement breaks independence and is the usual reason a model fails.
$P(X=r) = \binom{n}{r}p^{r}(1-p)^{n-r}$ , where the coefficient counts arrangements.
Convert wording carefully: "at least $r$ " is $1 - P(X\leq r-1)$ .
Mean is $np$ and variance is $np(1-p)$ , greatest at $p = 0.5$ .