What is the binomial distribution?

It is the distribution of the random variable counting successes in a Bernoulli scheme. It is written B(n, p), where n is the number of trials and p the probability of success in one of them. The formula for P(X = k) is exactly the Bernoulli formula of the previous lesson — what is new is that we look at the whole family of these probabilities at once, as the distribution of a single variable.

Where does the formula E(X) = np come from?

From splitting the variable into a sum. Each single trial is a variable taking 1 with probability p and 0 with probability 1 − p, and its expected value is simply p. The number of successes is the sum of n such variables, and the expected value of a sum is the sum of the expected values — hence np, with no binomial coefficients involved.

How does changing p affect the shape?

It shifts the distribution and changes its symmetry. The peak sits near np, so it moves with p; at p = 1/2 the distribution is symmetric, for p below one half it has a longer right tail, and for p above one half a longer left tail. The number of trials n governs the width instead: more trials make the distribution wider in absolute terms but narrower relative to its own mean.

What is the Poisson distribution?

The distribution of a count of rare events occurring independently in a fixed window of time, length or area. It is described by a single number λ, the average number of events in that window. Failures per month, calls per hour, typos per page and raisins per cake are the standard examples.

How does the Poisson distribution follow from the binomial one?

Through a limit in which the number of trials n grows to infinity, the probability p falls to zero, and their product np stays fixed and equal to λ. The binomial coefficient and the powers then converge to λ to the k times e to the minus λ, divided by k factorial. In practice the approximation is already very good for n in the tens and p in the hundredths.

Why does the Poisson mean equal its variance?

Because both come out of the same limit from the binomial formulas. The expected value np is λ by definition, and the variance np(1 − p) tends to np as p falls to zero, since the factor 1 − p converges to one. Hence a practical hint: if the spread in your data is markedly larger than the mean, a Poisson model does not fit it.

Does a Poisson count have an upper bound?

No. A binomial distribution stops at n because that is how many trials there are, but a Poisson distribution assigns positive probability to every whole number. Those probabilities fall very fast — for λ = 2 the value at k = 10 is already under 4 in 100,000 — but they never become zero.

Advanced level

The binomial and Poisson distributions

Q: When is the variance of a binomial distribution largest?

At p = 1/2. The product p(1 − p) then reaches its maximum of 1/4, so the variance is n/4. The meaning is plain: an outcome is hardest to predict when success and failure are equally likely, while for p near 0 or 1 the result is nearly certain and the spread shrinks to zero.

The number of successes in a Bernoulli scheme is a random variable, and its distribution is called binomial. It has two parameters and two short formulas for its mean and its spread. When the trials are very many and the success very rare, that distribution turns into the Poisson one — the model of breakdowns, calls and typos.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

Where this is used

Real situations where you count exactly the way this lesson teaches:

Overbooking a flight
An airline sells 210 tickets for 200 seats because a passenger shows up with probability 0.95. The number who show up is binomial with E(X) = 210 · 0.95 = 199.5 and variance 210 · 0.95 · 0.05 = 9.975, that is a standard deviation of 3.16 passengers. On average half a seat stays free, but a spread of three passengers means overbooking happens on a sizeable share of flights — and that is the number the airline is knowingly buying.
Server failures per month
If a server fails on average 3 times a month, the count of failures is Poisson with λ = 3. A month with no failure at all occurs with probability e to the minus three, that is 0.0498 — under 5 per cent of months. At most two failures has probability 0.0498 · (1 + 3 + 4.5) = 0.423, so more than half of all months bring at least three.
Typos in a book
A proofreader estimates that 0.5 typos per page survive their work. The count on a random page is Poisson with λ = 0.5, so a clean page occurs with probability e to the minus one half, that is 0.6065. Across a 300-page book that gives about 300 · 0.3935 = 118 pages carrying at least one typo — a realistic forecast rather than pessimism.
Prize caps on bottles
A bottle cap hides a prize with probability 1/50. Buying 20 bottles wins nothing with probability 0.98 to the twentieth power, that is 0.668, so the chance of at least one prize is 33 per cent. To push past one half you have to buy 35 — because 0.98 to the thirty-fifth power is already 0.493.

All formulas

Binomial distribution
$X \sim B(n, p), \qquad P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
the formula derived in the Bernoulli lesson
Expected value
$E(X) = np$
trials times the chance of success
Variance
$\sigma^2(X) = np(1-p)$
largest at p = 1/2
At least one success
$P(X \geq 1) = 1 - (1-p)^n$
through the complement
Poisson distribution
$X \sim \text{Poiss}(\lambda), \qquad P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$
rare events in a fixed window
Poisson parameters
$E(X) = \sigma^2(X) = \lambda$
one number describes the whole distribution
The limiting passage
$\lambda = np$
n large, p small, the product fixed

In the Bernoulli scheme we computed the probability of exactly $k$ successes in $n$ trials. The previous lesson gave us the notion of a random variable. Put the two together: the number of successes is a random variable, and the set of its probabilities is its distribution.

The binomial distribution

The variable $X$ counting successes in a Bernoulli scheme is said to have a binomial distribution, written

X \sim B(n, p), \qquad P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

The formula is not derived a second time — it is exactly the one from before, binomial coefficient counting paths and all. What is new is the viewpoint: instead of asking about one $k$ , we look at the whole table at once.

Ten trials, two different success probabilities. The wide bars are p = 0.5 — symmetric about 5. The narrow ones are p = 0.2, leaning left with a peak at 2. Both have exactly eleven bars, from k = 0 to k = 10.

The drawing shows two things at once. The peak sits near $np$ : at $5$ for $p = 0.5$ , at $2$ for $p = 0.2$ . And symmetry is the exception rather than the rule — it belongs to $p = \tfrac{1}{2}$ alone; for smaller $p$ the distribution has a longer right tail.

The two parameters

Rather than compute an expected value from a table of eleven terms, we have short formulas:

E(X) = np, \qquad \sigma^2(X) = np(1-p)

The first has a one-line justification with no binomial coefficient in it. A single trial is a variable taking $1$ with probability $p$ and $0$ with probability $q$ , so its expected value is

1 \cdot p + 0 \cdot q = p

The number of successes is the sum of $n$ such variables, and the expected value of a sum is the sum of the expected values — hence $np$ .

The variance carries an observation of its own: the factor $p(1-p)$ is largest at $p = \tfrac{1}{2}$ , where it equals $\tfrac{1}{4}$ . An outcome is hardest to predict when success and failure are equally likely; for $p$ near zero or one almost everything is settled and the spread disappears.

We inspect 200 units from a production run in which 3% are faulty. How many faults do we expect to find, and with what spread?

At least once

The question asked most often about a binomial distribution is not “how many exactly” but “will it happen at all”. The answer goes through the complement and fits on one line:

P(X \geq 1) = 1 - P(X = 0) = 1 - (1-p)^n

A bottle cap hides a prize with probability 1/50. How many bottles must you buy for the chance of any prize to exceed 50%?

The Poisson distribution

Now take a situation where the trials are very many and the success very rare: millions of ad impressions and a few dozen purchases; 2.6 million seconds in a month and three failures. There $n$ and $p$ separately stop being known or meaningful — what stays meaningful is their product, the average number of events.

A variable with a Poisson distribution has one parameter $\lambda$ and the distribution

P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}, \qquad k = 0, 1, 2, \dots

The Poisson distribution with λ = 2. The two tallest bars, at k = 1 and k = 2, are exactly equal — a property of this distribution at whole λ. To the right the tail never ends, it only shrinks fast.

The parameters here are remarkably short:

E(X) = \sigma^2(X) = \lambda

The equality of mean and variance is a practical test of the model: if the spread in the data is clearly larger than the mean, the events are not independent and a Poisson distribution does not fit them.

The limiting passage

Where does the formula with $e$ come from? From the binomial distribution, in the limit $n \to \infty$ , $p \to 0$ , with the product held fixed:

\lambda = np

You need not take that on faith — the numbers are enough.

A thousand tickets are sold and each wins with probability 0.002. Compare the exact probability of three wins with the Poisson approximation.

In practice the approximation is used for $n \geq 50$ and $np \leq 10$ . This is exactly why the Poisson distribution is called the distribution of rare events: not because the events are few, but because each single opportunity for one is negligible while the opportunities are very many.

Exercises

The set asks three things and does not repeat the question about $P(X = k)$ for the binomial — that is the same formula you practised in the Bernoulli lesson.

The $P(X \geq 1)$ prompt gives $n$ and $p$ and asks about at least one success; the answer is a fraction in lowest terms. The $E(X)$ or $\sigma^2(X)$ prompt gives the same two parameters and asks for the expected value or the variance — the answer is a plain number, because that is a quantity on the variable’s own scale rather than a probability. The $P(X = k)$ prompt with a parameter $\lambda$ is about the Poisson distribution: there $e^{-\lambda}$ is irrational, so the answer is given rounded to four decimal places.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

P(X ≥ 1): n = 3, p = 1/2

The generator does not ask you to judge whether a Poisson approximation is allowed here — that is a verdict, not a number. The criterion is stated above: many trials, small $p$ , a fixed product.

Common mistakes

Reading $np$ as the outcome rather than the average — $E(X) = 6$ does not mean there will be six faults; it means six on average, with a deviation of $2.41$ .
Applying the binomial distribution to dependent trials — drawing without replacement from a small set is not a Bernoulli scheme, so none of these formulas holds.
Dropping the factor $e^{-\lambda}$ — without it the probabilities do not add up to one and can exceed one as early as $k = 2$ .
Substituting $n$ and $p$ into the Poisson formula instead of $\lambda$ — the parameter is their product $np$ , and it goes into both the power and the exponent.
Approximating with Poisson at large $p$ — the limit needs small $p$ ; at $p = 0.3$ the approximation already parts company in the second decimal.
Expecting the Poisson variance to be smaller than the mean — both equal $\lambda$ , and a spread larger than the mean in real data is a sign that the model is wrong.

Formula card

Topic: Binomial and Poisson

Binomial distribution
$X \sim B(n, p), \qquad P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$
the formula derived in the Bernoulli lesson
Expected value
$E(X) = np$
trials times the chance of success
Variance
$\sigma^2(X) = np(1-p)$
largest at p = 1/2
At least one success
$P(X \geq 1) = 1 - (1-p)^n$
through the complement
Poisson distribution
$X \sim \text{Poiss}(\lambda), \qquad P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$
rare events in a fixed window
Poisson parameters
$E(X) = \sigma^2(X) = \lambda$
one number describes the whole distribution
The limiting passage
$\lambda = np$
n large, p small, the product fixed

Two binomial distributions with the same number of trials and different p. At p = 0.5 the distribution is symmetric about 5; at p = 0.2 it leans left with its peak at 2 — because E(X) = np moves with p.

The Poisson distribution with λ = 2. The peak sits at 1 and 2, and the tail runs right without end — a count of rare events has no upper bound, only ever smaller probabilities.

Frequently asked questions

Luminous histogram bars settling into the soft bell of a distribution, one outlying point brighter than the rest, against a cosmic nebula in violet and magenta

Statistics and probability

The average person does not exist

All of descriptive statistics rests on a single trade: you give up information about a crowd and receive one number in return. The trade is often worth making, but it is never free — and it pays to know exactly what gets lost in it. From the mortality tables of 1662, through Quetelet's average man, to the aircraft nobody ever saw.

16 min