Actuarial Statistics · Basic univariate distributions and generating samples

Discrete Distributions: Binomial, Poisson and Geometric Explained

Updated 11 October 2026 · Fact-checked

These are discrete distributions for counting. Binomial counts successes in n fixed trials. Geometric and negative binomial count trials or failures until successes occur. Poisson counts events in a fixed interval. To solve a question, identify the setting, write the pmf, then use the standard mean and variance or sum probabilities.

Understand Discrete Distributions: Binomial, Poisson, Geometric

A discrete distribution gives the probability of each possible whole-number outcome. The pmf, written P(X = x), must be non-negative and sum to 1. In actuarial work these models count claims, deaths, defaults and lapses.

Start with the Bernoulli distribution. One trial, success with probability p, failure with probability q = 1 − p. X is 1 for success and 0 for failure. Everything else here is built from repeated Bernoulli trials or from a limit of them.

The binomial distribution adds up n independent Bernoulli trials with the same p. X is the number of successes, so X ranges from 0 to n. The geometric distribution asks a different question: how long until the first success? The negative binomial extends this to the k-th success. The key difference from the binomial is what is fixed. In the binomial, the number of trials is fixed and the successes are random. In the negative binomial, the number of successes is fixed and the number of trials is random.

The Poisson distribution counts events that occur at random, one at a time, at a constant average rate, in a fixed interval. It has one parameter, μ (the IAI notation is often μ; some texts use λ). Its mean and variance are both μ. If you have a binomial with large n and small p, it behaves like Poisson with μ = np. This is why Poisson is used for rare events such as claims on a large portfolio.

The IAI notation for geometric and negative binomial has two versions. One counts the number of failures before the first (or k-th) success, with X starting at 0. The other counts the total number of trials, with X starting at 1 (or k). Always check which version the question uses. The formulas for mean differ by exactly the shift between the two.

Key rules to remember

Bernoulli: P(X = 1) = p, P(X = 0) = q = 1 − p; E[X] = p; Var(X) = pq; Building block for the binomial and geometric.
Binomial pmf: P(X = x) = C(n, x) p^x q^(n−x), x = 0, 1, …, n; C(n, x) = n! ÷ (x!(n − x)!). Trials are independent with constant p.
Binomial mean and variance: E[X] = np; Var(X) = npq; Variance is less than the mean, since q < 1.
Poisson pmf: P(X = x) = e^(−μ) μ^x ÷ x!, x = 0, 1, 2, …; For a rate λ per unit time over t units, use μ = λt.
Poisson mean and variance: E[X] = μ; Var(X) = μ; Equal mean and variance is a signature of the Poisson.
Poisson recursion: P(X = x + 1) = P(X = x) × μ ÷ (x + 1); Fast way to build a table of probabilities.
Geometric (trials version): P(X = x) = q^(x−1) p, x = 1, 2, 3, …; E[X] = 1/p; Var(X) = q/p²; X is the trial number of the first success.
Geometric (failures version): P(X = x) = q^x p, x = 0, 1, 2, …; E[X] = q/p; Var(X) = q/p²; X is the number of failures before the first success. Variance is unchanged by the shift.
Geometric tail: Trials version: P(X > x) = q^x. Failures version: P(X ≥ x) = q^x; Handy for 'at least' and 'more than' questions.
Negative binomial (failures version): P(X = x) = C(x + k − 1, k − 1) p^k q^x, x = 0, 1, 2, …; E[X] = kq/p; Var(X) = kq/p²; X is the number of failures before the k-th success.
Negative binomial (trials version): P(N = n) = C(n − 1, k − 1) p^k q^(n−k), n = k, k+1, …; E[N] = k/p; Var(N) = kq/p²; N is the trial on which the k-th success occurs.
Poisson approximation to binomial: Bin(n, p) ≈ Poisson(np); Use when n is large and p is small. A common guide is n ≥ 50 and p ≤ 0.1 or so, but it is only a rule of thumb.
Sum of independent Poissons: If X ~ Poisson(μ1) and Y ~ Poisson(μ2) are independent, X + Y ~ Poisson(μ1 + μ2); Also true for binomials with the same p: Bin(n1, p) + Bin(n2, p) = Bin(n1 + n2, p).

How to solve Discrete Distributions: Binomial, Poisson, Geometric questions

Use this method for any question on these distributions.

1Read what is being counted. Decide whether it is successes in fixed trials, trials until a success, or events in an interval.
2Pick the distribution: binomial (fixed n), geometric or negative binomial (wait for success), Poisson (events at a constant rate).
3Write the parameters clearly, such as X ~ Bin(20, 0.1). For Poisson, adjust μ to match the interval length.
4For geometric and negative binomial, state which version you use: trials or failures. Write the pmf for that version.
5Convert the wording into a probability statement, such as P(X ≥ 3) or P(X ≤ 2). Use the complement when it is shorter.
6Calculate with the pmf, the recursion or the tail formula. Keep enough decimals, at least four.
7If asked for a mean or variance, use the standard result or derive it from the pmf if the question says 'show that'.
8State the answer in words, and mention any assumption such as independence or constant rate.

Quickest way: Complement and recursion shortcut

When to use it: Use when you need a cumulative probability such as 'at least one' or 'at most three' under time pressure.

For 'at least one', compute 1 − P(X = 0). For Poisson this is 1 − e^(−μ). For binomial it is 1 − q^n.
For a geometric tail, skip summing and use q^x directly.
For a Poisson or binomial cumulative probability, find P(X = 0) first, then build later terms by recursion. Poisson: multiply by μ/(x + 1). Binomial: multiply by (n − x)/(x + 1) × p/q.
If n is large and p small and an approximation is allowed, switch to Poisson(np) to cut the calculation.
Check the sum of your terms does not exceed 1.

Common mistakes in Discrete Distributions: Binomial, Poisson, Geometric

Mixing up the two versions of the geometric or negative binomial.
Textbooks and exam questions differ on whether X counts trials or failures.
Fix: Check the stated range of X. If it starts at 1, X counts trials and the mean is 1/p. If it starts at 0, X counts failures and the mean is q/p.
Using the binomial when trials are not fixed, or when p changes.
Students see 'successes' and reach for the binomial without checking the set-up.
Fix: Ask: is n fixed and are trials independent with the same p? If you stop at a success, it is geometric or negative binomial.
Forgetting to rescale the Poisson mean for a different time period.
The rate is given per year but the question asks about six months or a different portfolio size.
Fix: Set μ = rate × length of interval before writing the pmf. For a portfolio, multiply by the number of policies.
Wrong inequality when using 'at least' and 'more than'.
P(X ≥ 3) and P(X > 3) look similar, and the complement shifts by one term.
Fix: Write P(X ≥ 3) = 1 − P(X ≤ 2). Write the complement explicitly every time.
Applying the Poisson approximation when p is not small.
Students treat the approximation as always valid.
Fix: Check n is large and p is small. If p is moderate, use the exact binomial or a normal approximation.
Writing Var(X) = q/p for the geometric distribution.
Students confuse the variance with the mean of the failures version.
Fix: Remember Var(X) = q/p², the same for both versions. Only the mean changes.

Worked examples

Example 1

An insurer's portfolio gives claims that follow a Poisson distribution with mean 3 per month. (a) Find the probability of exactly 2 claims in a month. (b) Find the probability of at least 2 claims in a month. Give answers to 4 decimal places.

Show the solution

X ~ Poisson(3). Here e^(−3) = 0.049787.
(a) P(X = 2) = e^(−3) × 3² ÷ 2! = 0.049787 × 9 ÷ 2 = 0.224042.
(b) P(X ≥ 2) = 1 − P(X = 0) − P(X = 1).
P(X = 0) = 0.049787.
P(X = 1) = 0.049787 × 3 = 0.149361.
P(X ≥ 2) = 1 − 0.049787 − 0.149361 = 0.800852.

Answer: (a) 0.2240. (b) 0.8009.

Example 2

A fair-minded underwriter checks applications one by one. Each is independently approved with probability 0.2. Let X be the number of applications rejected before the first approval. (a) State the distribution and find E[X] and Var(X). (b) Find P(X ≥ 3). (c) Find the probability that the first approval occurs on the 4th application.

Show the solution

(a) Success is approval, p = 0.2, q = 0.8. X counts failures before the first success, so P(X = x) = 0.8^x × 0.2, x = 0, 1, 2, …
E[X] = q/p = 0.8 ÷ 0.2 = 4.
Var(X) = q/p² = 0.8 ÷ 0.04 = 20.
(b) P(X ≥ 3) = q³ = 0.8³ = 0.512.
(c) The first approval on the 4th application means 3 rejections first, so X = 3.
P(X = 3) = 0.8³ × 0.2 = 0.512 × 0.2 = 0.1024.

Answer: (a) Geometric (failures version) with E[X] = 4 and Var(X) = 20. (b) 0.512. (c) 0.1024.

Exam tips

Write down which version of the geometric or negative binomial you are using before any calculation. Examiners reward a clear statement of assumptions.
In MCQs, check whether the question gives a rate per year or per month before using the Poisson pmf. Many wrong options come from an unscaled mean.
For 'show that' questions on the mean or variance of the Poisson, be ready to sum the series using the expansion of e^μ. Know E[X(X − 1)] = μ² as the key step.
For Paper B, know the R functions dbinom, pbinom, dpois, ppois, dgeom and dnbinom. Note that R's geometric and negative binomial count failures, not trials.
Use the Poisson approximation only when the question allows it or when n is large and p small. State the parameter np clearly.

Practice questions from Basic univariate distributions and generating samples

Discrete Distributions: Binomial, Poisson, Geometric in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Discrete Distributions: Binomial, Poisson, Geometric: frequently asked questions

How do I derive the mean and variance of the Poisson distribution?

For the mean, E[X] = Σ x e^(−μ) μ^x ÷ x!. Cancel x with x!, take out μ, and the remaining sum is e^μ, so E[X] = μ. For the variance, find E[X(X − 1)] = μ² the same way, then Var(X) = E[X(X − 1)] + E[X] − (E[X])² = μ² + μ − μ² = μ.

What is the difference between binomial and negative binomial distributions?

The binomial fixes the number of trials and counts the successes. The negative binomial fixes the number of successes and counts the trials (or failures) needed. The geometric distribution is the negative binomial with k = 1.

When can I use the Poisson approximation to the binomial?

Use it when n is large and p is small, so np is moderate. Then Bin(n, p) is close to Poisson(np). It works because the binomial variance npq is close to np when q is near 1.

What is the mean and variance of the geometric distribution?

If X counts trials up to the first success, E[X] = 1/p and Var(X) = q/p². If X counts failures before the first success, E[X] = q/p and Var(X) = q/p². The variance is the same in both versions.