Derivation · Statistics

The binomial and Poisson distributions, derived

Q: What is the formula for the binomial distribution?

If X counts the successes in n independent trials, each with chance p of success, then P(X = k) = C(n, k) p^k (1 − p)^(n − k) for k = 0, 1, …, n. The mean is np and the variance is np(1 − p).

Q: Why is there a binomial coefficient in the formula?

Every particular sequence with k successes and n − k failures has the same chance, p^k q^(n − k), because independent probabilities multiply. The coefficient C(n, k) counts how many different sequences there are, that is, how many ways to choose which k of the n trials are the successes.

Q: How is the Poisson distribution derived from the binomial?

Cut the interval into n tiny pieces, each with chance p = λ/n of an event, so the count is binomial. Let n grow with λ fixed. The binomial probability tends to e^(−λ) λ^k / k!, because the product of falling factors over n^k tends to 1 and (1 − λ/n)^n tends to e^(−λ).

Q: Why are the mean and variance of a Poisson distribution equal?

The mean works out to λ by summing k times the probability. The variance is E[X(X − 1)] + E[X] − (E[X])² = λ² + λ − λ² = λ. It also follows from the limit, since the binomial variance np(1 − p) tends to λ as p becomes small.

Q: When can I approximate a binomial by a Poisson?

When n is large and p is small, with np moderate, say under about 10. Then the Poisson with λ = np has nearly the same probabilities. A normal approximation is used instead when np and n(1 − p) are both large.

Q: What is the largest the variance of a binomial can be?

The variance np(1 − p) is largest when p = 1/2, where it equals n/4. It is never larger than n/4 for any p.

The binomial counts successes in a fixed number of tries. The Poisson counts events that arrive at random with no fixed number of tries. They look unrelated and are the same thing seen at two scales. This page derives the binomial by counting, gets its mean and variance without a single long sum, and then lets the number of tries grow until the Poisson appears. 14 past-paper questions at the end.

A question to hold on to

If X is binomial with parameters n and p, E(X) = 2 and Var(X) = 4/3, what are n and p: n = 2 and p = 1/6; n = 6 and p = 1/3; n = 2 and p = 2/3; or n = 6 and p = 2/3?

JNU 2019Solved at the end ↓

SymbolsReading the maths on this pageΣ, x̄, E[ ] and friends: how to say them and what they tell you to do.

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The idea

Two stories about counting

Story one. You make n independent tries. Each one succeeds with the same chance p. How many succeed? That count is binomial. Ten coin tosses, twenty customers who each buy with probability 0.25, thirty parts off a line that each fail with probability 0.02.

Story two. Events arrive at random over a stretch of time or space, at an average rate λ, and there is no fixed number of tries. How many arrive? That count is Poisson. Calls to a helpline in a minute, misprints on a page, accidents on a stretch of road in a week.

The link is that story two is story one with the tries made so small and so many that you cannot count them. That is how the Poisson is derived below.

Pause and predict

A binomial count has 12 tries with a success chance of 0.5. Its mean is 6. Is its variance more than, equal to, or less than 6?

Derivation

The binomial, derived

Derive it

Where the binomial formula comes from

We want to show

The number of successes in n independent trials, each with chance p, has

P (X = k) = (k n) p^{k} (1 - p)^{n - k}

, with mean np and variance np(1 − p).

Start with one trial.
A Bernoulli trial has two outcomes, success with chance p and failure with chance q = 1 − p. Count success as 1 and failure as 0, and call the count I. Then E[I] = 0 × q + 1 × p = p, and since I² = I its average is p too, so Var(I) = p − p² = pq.
Chain n independent trials and write down one sequence.
Independence means probabilities multiply. The sequence S S F S F S (n = 6, with 4 successes and 2 failures) has chance p · p · q · p · q · p = p⁴q². The order does not change the product: any sequence with k successes has chance p^kq^n−k.
Count the sequences with exactly k successes.
To build one, choose which k of the n positions hold the successes. The number of ways to choose k places out of n is
$(k n) = \frac{n !}{k ! ( n - k )!}$
Try the lab. List all 2ⁿ sequences, mark those with k successes, and watch that the count is the binomial coefficient and that every marked sequence has the same chance.
Add them up.
The sequences are different outcomes that cannot happen together, so their chances add. There are C(n, k) of them, each with chance p^kq^n−k:
$P (X = k) = (k n) p^{k} q^{n - k}$
Check with n = 4, p = 0.3, k = 2: 6 × 0.09 × 0.49 = 0.2646. The whole distribution adds to 1, because (p + q)ⁿ = 1ⁿ = 1 and the binomial theorem says that (p + q)ⁿ is exactly the sum of these terms.
Say it in words
“The chance of k successes is the number of ways to choose where they fall, times p to the k for the successes, times q to the n minus k for the failures.”
Get the mean and variance without summing anything.
Write X as a sum of n indicators, X = I₁ + I₂ + … + I_n, one per trial. The mean of a sum is the sum of the means, so E[X] = n · p. The trials are independent, so the covariances between them are zero and the variances add: Var(X) = n · pq.
$E [X] = n p, Var (X) = n p (1 - p)$
For n = 10 and p = 0.3 the mean is 3 and the variance 2.1. The variance is always smaller than the mean, since 1 − p < 1, and it is largest at p = 1/2, where it is n/4 (UOH 2017). A past paper that gives mean 2 and variance 4/3 is telling you q = (4/3) ÷ 2 = 2/3, so p = 1/3 and n = 6.

Step 1 of 5

The binomial distribution

P (X = k) = (k n) p^{k} q^{n - k}, E [X] = n p, Var (X) = n pq

n independent trials, each with chance p. Here q = 1 − p.

Derivation

The Poisson, derived

Derive it

The Poisson distribution as the limit of the binomial

We want to show

For events arriving at an average rate λ over an interval,

P (X = k) = \frac{e ^{- λ} λ ^{k}}{k !}

, and both the mean and the variance equal λ.

Cut the interval into tiny pieces.
Say a call centre gets 2 calls a minute on average, and you want the chance of exactly 3 in a minute. Split the minute into n equal pieces so short that a piece holds at most one call. Each piece then has a call with chance p = λ ÷ n, since n pieces must add up to the average λ. The count is a binomial with n trials and p = λ ÷ n.
Write the binomial and rearrange it.
$P (X = k) = \frac{n ( n - 1 ) \dots ( n - k + 1 )}{k !} (\frac{λ}{n})^{k} (1 - \frac{λ}{n})^{n - k}$
Pull the pieces apart. The k factors of n on top pair with the k factors of n in (λ ÷ n)^k:
$P (X = k) = \frac{λ ^{k}}{k !} \cdot A \frac{n ( n - 1 ) \dots ( n - k + 1 )}{n ^{k}} \cdot B (1 - \frac{λ}{n})^{n} \cdot C (1 - \frac{λ}{n})^{- k}$
Let the pieces get smaller and smaller.
Keep k and λ fixed and let n grow. Piece A is a product of k fractions (n ÷ n)((n − 1) ÷ n) …, each going to 1, so A → 1. Piece C is (1 − λ ÷ n) raised to a fixed power, so C → 1. Piece B is the famous limit
$(1 - \frac{λ}{n})^{n} ⟶ e^{- λ}$
That leaves P(X = k) → e^−λλ^k ÷ k!. The numbers agree already at modest n. For λ = 2 and k = 3, the binomial gives 0.2013 at n = 10, 0.1823 at n = 100, 0.1806 at n = 1000, heading for the Poisson value 0.1804.
Check the probabilities add to 1.
The sum of λ^k ÷ k! over k = 0, 1, 2, … is the series for e^λ. So the total is e^−λ × e^λ = 1, as it must be.
Find the mean and variance.
The k = 0 term of the mean contributes nothing, and k ÷ k! = 1 ÷ (k − 1)!, so
$E [X] = k \geq 1 \sum k \frac{e ^{- λ} λ ^{k}}{k !} = λ k \geq 1 \sum \frac{e ^{- λ} λ ^{k - 1}}{( k - 1 )!} = λ \cdot 1 = λ$
In the same way E[X(X − 1)] = λ², so E[X²] = λ² + λ (JNU 2020) and Var(X) = λ² + λ − λ² = λ. This also follows from the limit: the binomial mean np stays at λ, and the variance npq = λ(1 − λ ÷ n) tends to λ because q → 1. Equal mean and variance is the Poisson signature (JNU 2017), and its coefficient of variation is √λ ÷ λ = 1 ÷ √λ, which is 0.2 for λ = 25 (GATE 2026).
Say it in words
“Slice the interval into n pieces, each with chance lambda over n. The binomial chance of k events tends to e to the minus lambda, times lambda to the k, over k factorial.”

Step 1 of 5

The Poisson distribution

P (X = k) = \frac{e ^{- λ} λ ^{k}}{k !}, E [X] = Var (X) = λ

λ is the average number of events in the interval. The mean and the variance are both λ.

Choosing

Which one do I use?

Three questions settle it.

Is there a fixed number of tries, each a success or a failure? Then binomial. If you can say “out of n”, it is binomial.
Are you counting events over time or space, with no natural “out of”? Then Poisson, and the information given is an average rate.
Is n large and p small? Then either works, and the Poisson with λ = np is the easier sum. CUET 2024 shows the pattern: a large-n binomial is approximated by a Poisson.

A quick test in the data: compare the mean and variance. Equal points to Poisson. A variance clearly below the mean points to binomial. For n = 50 and p = 0.04, the exact binomial chance of 2 successes is 0.2762and the Poisson with λ = 2 gives 0.2707, close but not equal. The discrete distributions lesson applies them to the full range of paper questions.

The one-line difference

Binomial: a fixed number of tries, count the successes, variance below the mean. Poisson: no fixed number of tries, count the events, variance equal to the mean.

Exam focus

The mistakes examiners are fishing for

Watch out for

The mistakes examiners are fishing for

Writing the variance as np. That is the mean. The binomial variance is npq, and it equals np only as p shrinks toward 0.
Forgetting the binomial coefficient. p^kq^n−k is the chance of one sequence. You need to multiply by how many sequences there are.
Using a binomial with a probability outside its range. P(X = 11) is 0 when n = 10 (DSE 2017). Find n from the mean and variance first.
Calling it Poisson because the numbers are small. A fixed number of tries with a success chance is binomial, even if small.
Mixing the Poisson with the exponential. The Poisson counts events in an interval. The exponential is the waiting time between them. For parameter 1 they share a mean (JNU 2017).
Subtracting from the wrong total. “At least one” is 1 − P(0), not P(1) plus a guess.

Back to the puzzle

The opening question, solved

JNU 2019 · easy

X is binomial, E(X) = 2 and Var(X) = 4/3. Find n and p.

Divide variance by mean
npq ÷ np = q = (4/3) ÷ 2 = 2/3, so p = 1/3.
Then n
n = E(X) ÷ p = 2 ÷ (1/3) = 6.

Answer

n = 6 and p = 1/3.

DSE 2017 · moderate

X is binomial with mean 7 and variance 2.1. What is P(X = 11)?

The same division
q = 2.1 ÷ 7 = 0.3, so p = 0.7 and n = 7 ÷ 0.7 = 10.
Read the range
X takes the values 0 to 10 only.

Answer

0. The long expressions in the options are distractors.

JNU 2012 · easy

Printing mistakes per page are Poisson with mean 3, and e⁻³ = 0.049787. What is the chance of some mistake?

Complement
1 − P(0) = 1 − e⁻³ × 3⁰ ÷ 0! = 1 − 0.049787.

Answer

0.950212.

Revision

The page on one screen

The lesson on one screen

Binomial formula: C(n, k) p^kq^n−k: the number of sequences times the chance of each.
Why a coefficient: Independent trials multiply, so every sequence with k successes is equally likely, and C(n, k) counts them.
Binomial moments: Mean np, variance npq, from adding n independent indicators. Variance at most n/4.
Poisson as a limit: Cut the interval into n pieces with p = λ/n and let n grow: the binomial becomes e^−λλ^k/k!.
Poisson moments: Mean and variance both λ. E[X²] = λ² + λ.
Neighbour rule: P(x + 1) = λ ÷ (x + 1) × P(x).
Choosing: Fixed tries: binomial. Rate over an interval: Poisson. Large n, small p: either, and the Poisson is easier.

Practice

Check yourself

0 of 11 correct so far

Numbers

Give answers to four decimal places where they are not whole numbers.

Counting sequences

In 6 independent trials, how many different sequences contain exactly 2 successes?

Sequences

A binomial probability

X is binomial with n = 5 and p = 0.4. Find P(X = 2).

P(X = 2)

Mean and variance

X is binomial with n = 20 and p = 0.25. What is its variance?

Variance

Finding the parameters

A binomial has mean 6 and variance 1.5. What is n?

A Poisson probability

Mistakes on a page are Poisson with mean 3, and e⁻³ = 0.049787. What is the chance of at least one mistake?

P(X ≥ 1)

A ratio of neighbours

X is Poisson with λ = 2. What is P(X = 4) ÷ P(X = 3)?

Ratio

Concepts

1The variance of a binomial with parameters n and p is

2For a Poisson distribution

3For a binomial distribution, the variance can never exceed

4A binomial has n = 1000 and p = 0.002. The best simple approximation is

5If X is binomial with mean 7 and variance 2.1, P(X = 11) is

Past papers

Past papers, by level

Foundation

Definitions, directions and one-step sums. Most CUET, UGC NET and Hyderabad questions sit here.

8 questionsPractise →

Standard

Classification traps, multi-step numericals and statement questions, up to the hardest JNU, JAM and GATE items.

6 questionsPractise →

Delhi School level

The Delhi School entrance asks this topic harder than any other exam. These questions are kept apart from the ordinary sets.

2 questionsTry them →

All 14 past-paper questions on the binomial and Poisson distributions, easiest first: practise the full set →

These are questions from the TestStop bank. A free account answers 10 a day, each with its worked solution; All-Access opens every set with no daily limit.

Quick answers

What is the formula for the binomial distribution?

If X counts the successes in n independent trials, each with chance p of success, then P(X = k) = C(n, k) p^k (1 − p)^(n − k) for k = 0, 1, …, n. The mean is np and the variance is np(1 − p).

Why is there a binomial coefficient in the formula?

Every particular sequence with k successes and n − k failures has the same chance, p^k q^(n − k), because independent probabilities multiply. The coefficient C(n, k) counts how many different sequences there are, that is, how many ways to choose which k of the n trials are the successes.

How is the Poisson distribution derived from the binomial?

Cut the interval into n tiny pieces, each with chance p = λ/n of an event, so the count is binomial. Let n grow with λ fixed. The binomial probability tends to e^(−λ) λ^k / k!, because the product of falling factors over n^k tends to 1 and (1 − λ/n)^n tends to e^(−λ).

Why are the mean and variance of a Poisson distribution equal?

The mean works out to λ by summing k times the probability. The variance is E[X(X − 1)] + E[X] − (E[X])² = λ² + λ − λ² = λ. It also follows from the limit, since the binomial variance np(1 − p) tends to λ as p becomes small.

When can I approximate a binomial by a Poisson?

When n is large and p is small, with np moderate, say under about 10. Then the Poisson with λ = np has nearly the same probabilities. A normal approximation is used instead when np and n(1 − p) are both large.

What is the largest the variance of a binomial can be?

The variance np(1 − p) is largest when p = 1/2, where it equals n/4. It is never larger than n/4 for any p.

The binomial and Poisson distributions, derived

Two stories about counting

The binomial, derived

The Poisson, derived

Which one do I use?

The mistakes examiners are fishing for

The mistakes examiners are fishing for

The opening question, solved

The page on one screen

Check yourself

Numbers

Concepts

Past papers, by level

Foundation

Standard

Delhi School level

Quick answers

What is the formula for the binomial distribution?

Why is there a binomial coefficient in the formula?

How is the Poisson distribution derived from the binomial?

Why are the mean and variance of a Poisson distribution equal?

When can I approximate a binomial by a Poisson?

What is the largest the variance of a binomial can be?

Read next

The binomial and Poisson distributions, derived

Two stories about counting

The binomial, derived

The Poisson, derived

Which one do I use?

The mistakes examiners are fishing for

The mistakes examiners are fishing for

The opening question, solved

The page on one screen

Check yourself

Numbers

Concepts

Past papers, by level

Foundation

Standard

Delhi School level

Quick answers

What is the formula for the binomial distribution?

Why is there a binomial coefficient in the formula?

How is the Poisson distribution derived from the binomial?

Why are the mean and variance of a Poisson distribution equal?

When can I approximate a binomial by a Poisson?

What is the largest the variance of a binomial can be?

Read next