The binomial distribution

The binomial distribution models the number of successes in a fixed number of repeated trials. It is appropriate only when each trial has two possible outcomes, trials are independent, and the probability of success stays constant.

If $X$ counts successes in $n$ trials and each trial has success probability $p$, write

$$\boxed{X\sim \operatorname{B}(n,p)}.$$

Then, for $x=0,1,2,\ldots,n$,

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

This lesson develops the model, the formula, cumulative probabilities, calculator methods, and the distribution’s mean and variance.

Prerequisites

You should be able to:

• use complements, mutually exclusive events and independence from probability;
• calculate combinations such as $\binom{n}{r}$;
• interpret a discrete random variable;
• distinguish a mathematical model from reality using probability modelling.

Recognising a binomial model

Check all four conditions before writing $X\sim\operatorname{B}(n,p)$.

Condition	What it means
Fixed number of trials	$n$ is decided in advance
Two outcomes per trial	Each trial is classified as success or failure
Independent trials	One result does not change the probabilities on another trial
Constant success probability	The same value of $p$ applies on every trial

The word success is only a label for the outcome being counted. A defective component, a missed penalty, or a patient experiencing a side effect may all be called a success in a binomial model.

A quick memory aid

Think fixed $n$, two outcomes, independent, constant $p$. Writing these conditions in context is stronger than merely naming them.

Example 1: deciding whether the model applies

For each situation, decide whether a binomial model is reasonable.

1. A fair coin is tossed $20$ times and $X$ counts heads.
2. Five cards are drawn from an ordinary pack without replacement and $Y$ counts aces.
3. A machine produces components independently, with probability $0.03$ that any component is defective. Of the next $50$ components, $D$ counts defects.
4. A die is rolled until the first six appears, and $T$ counts the number of rolls.

Situation 1: yes. There are $20$ fixed, independent trials, each with outcomes head or not head and constant probability $1/2$. Thus

$$X\sim\operatorname{B}\left(20,\frac12\right).$$

Situation 2: not exactly. Without replacement, the probability of an ace changes after each draw and the trials are dependent.

Situation 3: yes, under the stated independence and constant defect rate:

$$D\sim\operatorname{B}(50,0.03).$$

Situation 4: no. The number of trials is not fixed. The experiment stops at a random time.

Sampling without replacement

Drawing from a finite population without replacement is not exactly binomial because the probability changes. A binomial model can sometimes be a useful approximation when the sample is a small fraction of a large population, but that approximation must be justified by context.

Self-check 1

A bag contains $8$ red and $12$ blue counters. A counter is selected, its colour is recorded, and it is replaced. This is repeated $15$ times. Let $R$ count red selections. State a suitable distribution for $R$.

Answer

Replacement keeps the red probability constant and makes the trials independent. Since

$$p=\frac{8}{20}=0.4,$$

the model is

$$\boxed{R\sim\operatorname{B}(15,0.4)}.$$

Why the probability formula works

Suppose $X\sim\operatorname{B}(n,p)$ and let

$$q=1-p$$

be the failure probability.

One particular sequence containing $x$ successes and $n-x$ failures has probability

$$p^xq^{n-x},$$

because the trials are independent. There are

$$\binom{n}{x}$$

different positions in which the $x$ successes can occur. These sequences are mutually exclusive, so their probabilities add:

$$\operatorname{P}(X=x)=\binom{n}{x}p^xq^{n-x}.$$

The three factors have distinct jobs:

$$\underbrace{\binom{n}{x}}_{\text{arrangements}} \underbrace{p^x}_{\text{successes}} \underbrace{(1-p)^{n-x}}_{\text{failures}}.$$

Worked example 2: an exact probability

A basketball player scores a free throw with probability $0.8$. Assume successive attempts are independent. She takes $10$ free throws. Find the probability that she scores exactly $7$.

Let $X$ be the number scored. Then

$$X\sim\operatorname{B}(10,0.8).$$

Use $x=7$:

$$\begin{aligned} \operatorname{P}(X=7) &=\binom{10}{7}(0.8)^7(0.2)^3\\ &=120(0.8)^7(0.2)^3\\ &=0.201326592. \end{aligned}$$

Therefore

$$\boxed{\operatorname{P}(X=7)\approx0.201}.$$

The power of $0.2$ is $3$, not $7$, because exactly three of the ten attempts must be failures.

Worked example 3: success is defined by the question

A biased die has probability $0.3$ of showing a six. It is rolled $12$ times independently. Find the probability of exactly two results that are not sixes.

Define success as obtaining a result that is not a six. Its probability is

$$p=1-0.3=0.7.$$

If $Y$ counts results that are not sixes, then $Y\sim\operatorname{B}(12,0.7)$ and

$$\begin{aligned} \operatorname{P}(Y=2) &=\binom{12}{2}(0.7)^2(0.3)^{10}\\ &\approx\boxed{0.000190}. \end{aligned}$$

This very small answer is sensible: when a non-six has probability $0.7$, only two non-sixes in twelve rolls is unusual.

Self-check 2

For $X\sim\operatorname{B}(8,0.25)$, calculate $\operatorname{P}(X=3)$.

Answer

$$\begin{aligned} \operatorname{P}(X=3) &=\binom83(0.25)^3(0.75)^5\\ &=\frac{1701}{8192}\\ &\approx\boxed{0.2076}. \end{aligned}$$

Translating probability language

Because a binomial random variable is integer valued, boundary words must be translated carefully.

Words	Probability statement
exactly $r$	$\operatorname{P}(X=r)$
at most $r$	$\operatorname{P}(X\leq r)$
fewer than $r$	$\operatorname{P}(X<r)=\operatorname{P}(X\leq r-1)$
at least $r$	$\operatorname{P}(X\geq r)$
more than $r$	$\operatorname{P}(X>r)=\operatorname{P}(X\geq r+1)$
between $a$ and $b$ inclusive	$\operatorname{P}(a\leq X\leq b)$

Two facts are especially useful:

$$\operatorname{P}(X\geq r)=1-\operatorname{P}(X\leq r-1),$$

and

$$\operatorname{P}(a\leq X\leq b) =\operatorname{P}(X\leq b)-\operatorname{P}(X\leq a-1).$$

Notice the $r-1$ and $a-1$. These are common sources of lost marks.

Cumulative probabilities

For $X\sim\operatorname{B}(n,p)$,

$$\operatorname{P}(X\leq r) =\sum_{x=0}^{r}\binom nxp^x(1-p)^{n-x}.$$

For a short sum, calculate the terms directly. For a large sum, use the calculator’s binomial cumulative distribution function. Calculator menus differ, but they usually distinguish:

• binomial PDF or binomial probability, which returns $\operatorname{P}(X=x)$;
• binomial CDF or binomial cumulative probability, which returns $\operatorname{P}(X\leq x)$.

Always write the probability statement before using the calculator. It makes the required boundary explicit.

Worked example 4: at most

The probability that a seed germinates is $0.85$. Assume seeds germinate independently. Twelve seeds are planted. Find the probability that at most nine germinate.

Let $X$ count germinating seeds. Then

$$X\sim\operatorname{B}(12,0.85).$$

“At most nine” means $X\leq9$, so

$$\begin{aligned} \operatorname{P}(X\leq9) &=\sum_{x=0}^{9}\binom{12}{x}(0.85)^x(0.15)^{12-x}\\ &\approx\boxed{0.2642}. \end{aligned}$$

A calculator CDF call with $n=12$, $p=0.85$ and upper value $9$ gives the same result.

Worked example 5: at least

For the same seed model, find the probability that at least ten germinate.

“At least ten” means $X\geq10$. Its complement is $X\leq9$, not $X\leq10$. Hence

$$\begin{aligned} \operatorname{P}(X\geq10) &=1-\operatorname{P}(X\leq9)\\ &=1-0.264181\ldots\\ &\approx\boxed{0.7358}. \end{aligned}$$

The events $X\leq9$ and $X\geq10$ cover every possible value without overlap.

Worked example 6: an interval

Let $Y\sim\operatorname{B}(20,0.3)$. Find

$$\operatorname{P}(4<Y\leq8).$$

Since $Y$ is integer valued, the required values are $5,6,7,8$. Use two cumulative probabilities:

$$\begin{aligned} \operatorname{P}(4<Y\leq8) &=\operatorname{P}(Y\leq8)-\operatorname{P}(Y\leq4)\\ &\approx0.8867-0.2375\\ &\approx\boxed{0.6492}. \end{aligned}$$

Do not subtract $\operatorname{P}(Y\leq5)$, because that would remove the allowed value $Y=5$.

Self-check 3

Let $X\sim\operatorname{B}(15,0.4)$. Write each event in a form suitable for a calculator CDF.

1. $\operatorname{P}(X<6)$
2. $\operatorname{P}(X>6)$
3. $\operatorname{P}(3\leq X\leq7)$

Answer

$$\begin{aligned} \operatorname{P}(X<6)&=\operatorname{P}(X\leq5),\\ \operatorname{P}(X>6)&=1-\operatorname{P}(X\leq6),\\ \operatorname{P}(3\leq X\leq7)&=\operatorname{P}(X\leq7)-\operatorname{P}(X\leq2). \end{aligned}$$

Numerically, these are approximately $0.4032$, $0.3902$, and $0.7598$ respectively.

At least one success

Questions involving “at least one” are usually shortest by complement:

$$\boxed{\operatorname{P}(X\geq1)=1-\operatorname{P}(X=0)=1-(1-p)^n}.$$

Worked example 7: using a complement

A server request fails with probability $0.02$, independently of other requests. Find the probability that at least one of the next $40$ requests fails.

Let $X$ count failures, so $X\sim\operatorname{B}(40,0.02)$. Then

$$\begin{aligned} \operatorname{P}(X\geq1) &=1-\operatorname{P}(X=0)\\ &=1-\binom{40}{0}(0.02)^0(0.98)^{40}\\ &=1-(0.98)^{40}\\ &\approx\boxed{0.5543}. \end{aligned}$$

Although each request has only a $2\%$ failure probability, repeated exposure makes at least one failure more likely than not.

Worked example 8: finding the number of trials

A component has probability $0.1$ of failing during a test. Independent components are tested. Find the smallest number $n$ that must be tested for the probability of at least one failure to exceed $0.95$.

We need

$$1-(0.9)^n>0.95.$$

Rearrange:

$$(0.9)^n<0.05.$$

Taking logarithms gives

$$n\log(0.9)<\log(0.05).$$

Since $\log(0.9)<0$, dividing reverses the inequality:

$$n>\frac{\log(0.05)}{\log(0.9)}\approx28.43.$$

As $n$ is an integer, the smallest possible value is

$$\boxed{n=29}.$$

Check the boundary:

$$1-0.9^{28}\approx0.9477<0.95, \qquad 1-0.9^{29}\approx0.9529>0.95.$$

Mean, variance and standard deviation

If $X\sim\operatorname{B}(n,p)$, then

$$\boxed{\operatorname{E}(X)=np},$$ $$\boxed{\operatorname{Var}(X)=np(1-p)},$$

and therefore

$$\boxed{\operatorname{SD}(X)=\sqrt{np(1-p)}}.$$

The expectation $np$ has a direct interpretation. Across $n$ trials, each contributes an average of $p$ successes, so the expected total is $np$.

Expectation need not be a possible observed value. If $X\sim\operatorname{B}(5,0.3)$, then $\operatorname{E}(X)=1.5$, although $X$ can only be an integer. The mean describes the long run average over many repetitions of the whole five-trial experiment.

Why the formulas are natural

Let $I_i$ equal $1$ when trial $i$ succeeds and $0$ when it fails. Then

$$X=I_1+I_2+\cdots+I_n.$$

Each indicator has

$$\operatorname{E}(I_i)=p$$

and

$$\operatorname{Var}(I_i)=p(1-p).$$

Expectations add, so $\operatorname{E}(X)=np$. Because the trials are independent, their variances also add, giving $\operatorname{Var}(X)=np(1-p)$.

Worked example 9: interpreting the moments

A call centre resolves each enquiry at first contact with probability $0.72$. Assume independence. Let $X$ be the number resolved at first contact among $50$ enquiries.

Then

$$X\sim\operatorname{B}(50,0.72).$$

The mean is

$$\operatorname{E}(X)=50(0.72)=\boxed{36}.$$

The variance is

$$\operatorname{Var}(X)=50(0.72)(0.28)=\boxed{10.08},$$

so the standard deviation is

$$\operatorname{SD}(X)=\sqrt{10.08}\approx\boxed{3.17}.$$

Over many groups of $50$ enquiries, the average number resolved at first contact would approach $36$. This does not claim that every group contains exactly $36$.

Worked example 10: recovering the parameters

A binomial random variable $X$ has mean $12$ and variance $3$. Find $n$ and $p$.

Using the formulas,

$$np=12 \tag{1}$$

and

$$np(1-p)=3. \tag{2}$$

Divide (2) by (1):

$$1-p=\frac{3}{12}=\frac14.$$

Thus

$$p=\frac34.$$

Substitute into (1):

$$n\left(\frac34\right)=12,$$

so

$$\boxed{n=16,\qquad p=\frac34}.$$

The answer passes essential checks: $n$ is a positive integer and $0\leq p\leq1$.

Self-check 4

Let $Y\sim\operatorname{B}(80,0.15)$. Find its mean, variance and standard deviation.

Answer

$$\operatorname{E}(Y)=80(0.15)=\boxed{12},$$ $$\operatorname{Var}(Y)=80(0.15)(0.85)=\boxed{10.2},$$

and

$$\operatorname{SD}(Y)=\sqrt{10.2}\approx\boxed{3.19}.$$

Shape and parameter effects

The values of $n$ and $p$ determine the distribution’s location and spread.

• Increasing $n$ with $p$ fixed increases both the mean $np$ and variance $np(1-p)$.
• Replacing $p$ by $1-p$ reflects the distribution: if $X\sim\operatorname{B}(n,p)$, then $n-X\sim\operatorname{B}(n,1-p)$.
• When $p=0.5$, the distribution is symmetric about $n/2$.
• When $p<0.5$, probability tends to be concentrated near smaller values of $X$.
• When $p>0.5$, probability tends to be concentrated near larger values of $X$.

For fixed $n$, the variance $np(1-p)$ is largest when $p=0.5$. Outcomes are most variable when success and failure are equally likely. If $p$ is close to $0$ or $1$, results are more predictable.

Modelling assumptions in context

A calculation may be correct while its model is weak. Examine the assumptions behind $n$, independence and constant $p$.

Example 11: criticising a model

A footballer scores $75\%$ of penalties historically. A coach models the number scored from the next $10$ penalties as $X\sim\operatorname{B}(10,0.75)$.

Possible limitations include:

• the goalkeeper, venue and pressure may change the scoring probability;
• fatigue, confidence or learning may make successive penalties dependent;
• the historical rate may not represent the player’s current ability.

The model may still be useful, but its conclusions are conditional on treating the attempts as independent with constant probability $0.75$.

Population proportion and individual probability

A statement such as ”$4\%$ of items are defective” is often used as $p=0.04$ for a randomly selected item. This requires random sampling from a stable process. Clustering or changing production conditions can undermine independence or constant $p$.

Common misconceptions

Confusing $p$ with the more pleasant outcome

$p$ is the probability of whichever outcome $X$ counts. Define $X$ in words before substituting numbers.

Omitting the combination factor

$p^x(1-p)^{n-x}$ is the probability of one specific ordering. Multiply by $\binom nx$ to include every ordering with $x$ successes.

Using the wrong tail boundary

$$\operatorname{P}(X\geq r)=1-\operatorname{P}(X\leq r-1),$$

not $1-\operatorname{P}(X\leq r)$.

Calling dependent trials binomial

Two outcomes alone are not enough. Trials must also be independent, $p$ must remain constant, and $n$ must be fixed.

Confusing variance and standard deviation

The variance is $np(1-p)$. The standard deviation is its square root. Standard deviation has the same units as $X$.

Rounding too early

Keep full calculator values during intermediate steps. Round only the final answer, usually to the accuracy requested.

Mixed exam style example

A factory states that $6\%$ of its batteries fail an initial test. Assume battery results are independent. A random sample of $25$ batteries is tested, and $X$ is the number that fail.

1. State the distribution of $X$.
2. Find $\operatorname{P}(X=2)$.
3. Find $\operatorname{P}(X\geq3)$.
4. Find the expected number of failures and the standard deviation.
5. Give one reason why the model might not fit actual production.

1. Distribution

$$\boxed{X\sim\operatorname{B}(25,0.06)}.$$

2. Exactly two failures

$$\begin{aligned} \operatorname{P}(X=2) &=\binom{25}{2}(0.06)^2(0.94)^{23}\\ &\approx\boxed{0.2602}. \end{aligned}$$

3. At least three failures

Use the complement $X\leq2$:

$$\begin{aligned} \operatorname{P}(X\geq3) &=1-\operatorname{P}(X\leq2)\\ &\approx\boxed{0.1871}. \end{aligned}$$

4. Mean and standard deviation

$$\operatorname{E}(X)=25(0.06)=\boxed{1.5},$$ $$\operatorname{SD}(X)=\sqrt{25(0.06)(0.94)}\approx\boxed{1.19}.$$

5. Limitation

Batteries made in the same batch may share a fault, so their outcomes may not be independent. Alternatively, the failure probability may change when machinery or materials change.

Final self-check

1. State all conditions required for a binomial model.
2. If $X\sim\operatorname{B}(18,0.2)$, find $\operatorname{P}(X=4)$.
3. For the same distribution, express $\operatorname{P}(X\geq4)$ using one lower-tail cumulative probability.
4. If $Y\sim\operatorname{B}(n,0.4)$ and $\operatorname{E}(Y)=8$, find $n$ and $\operatorname{Var}(Y)$.
5. A trial succeeds with probability $0.35$. Find the probability of at least one success in six independent trials.

Answers

1. A fixed number of trials; two outcomes on each trial; independent trials; constant probability of success.

$$\operatorname{P}(X=4)=\binom{18}{4}(0.2)^4(0.8)^{14}\approx\boxed{0.2153}.$$

$$\boxed{\operatorname{P}(X\geq4)=1-\operatorname{P}(X\leq3)}.$$

4. From $0.4n=8$, $\boxed{n=20}$. Then

$$\operatorname{Var}(Y)=20(0.4)(0.6)=\boxed{4.8}.$$

$$\operatorname{P}(X\geq1)=1-(0.65)^6\approx\boxed{0.9246}.$$

Next steps

• Learn how evidence is judged using hypothesis testing language.
• Apply binomial tail probabilities in binomial hypothesis tests.
• Compare this model with alternatives in choosing a distribution.
• Study a continuous model in the normal distribution.