FRM Exam Part I · Common Univariate Random Variables

Binomial and Poisson Distributions for FRM Part I

Updated 11 October 2026 · Fact-checked

The binomial distribution gives the probability of k successes in n independent trials with the same success probability p. The Poisson distribution gives the probability of k events in a fixed interval when events occur at an average rate λ. To solve, identify the setup, pick the formula, and plug in the numbers.

Understand Binomial and Poisson Distributions

Both distributions count things. They are discrete random variables that take whole-number values 0, 1, 2 and so on. In risk work, the thing you count is often a default, a loss event or a breach of a limit.

The binomial distribution fits a fixed number of trials, n. Each trial has two outcomes, success or failure, with the same probability p. Trials must be independent. Example: 20 loans, each with a 3% chance of default over one year. The number of defaults is binomial with n = 20 and p = 0.03. Each trial is a Bernoulli trial, and a binomial variable is the sum of n independent Bernoulli variables.

The Poisson distribution fits events that arrive at random over time or space, with no fixed upper limit on the count. One parameter, λ, is the average number of events in the interval. Example: a bank averages 2 operational loss events a month. The count in a month is Poisson with λ = 2. In a Poisson distribution, the mean and the variance are both equal to λ.

The two are linked. When n is large and p is small, a binomial with mean np is well approximated by a Poisson with λ = np. This is the Poisson approximation to the binomial. It is handy for large loan portfolios with low default rates.

Key difference: binomial has a fixed n and a ceiling on the count (it cannot exceed n). Poisson has no ceiling. Binomial variance np(1 − p) is less than its mean np. Poisson variance equals its mean.

Key formulas to remember

Binomial probability: P(X = k) = C(n, k) × p^k × (1 − p)^(n − k); C(n, k) = n! ÷ [k! × (n − k)!]. k runs from 0 to n. Trials independent with constant p.
Binomial mean: E(X) = np; Expected number of successes.
Binomial variance: Var(X) = np(1 − p); Standard deviation is √[np(1 − p)].
Poisson probability: P(X = k) = e^(−λ) × λ^k ÷ k!; k = 0, 1, 2, ... λ is the average count in the interval.
Poisson mean and variance: E(X) = λ and Var(X) = λ; Both equal λ. If the interval scales, λ scales in proportion.
Poisson approximation to binomial: λ = np; Works well when n is large and p is small. It is an approximation, not exact.
At least one event: P(X ≥ 1) = 1 − P(X = 0); Binomial: 1 − (1 − p)^n. Poisson: 1 − e^(−λ).

How to solve Binomial and Poisson Distributions questions

Use this routine for any binomial or Poisson question.

1Decide the model. Fixed n, two outcomes, constant p, independent trials means binomial. Random arrivals at an average rate with no fixed maximum means Poisson.
2Write down the parameters: n and p for binomial, or λ for Poisson. Adjust λ if the time interval differs from the one given.
3Translate the question into probabilities: exactly k, at most k, at least k, or between.
4For 'at least' questions, use the complement, for example P(X ≥ 1) = 1 − P(X = 0).
5For 'at most k', add P(0) through P(k). Compute each term with the formula.
6Compute the term values carefully, keeping enough decimals, then add or subtract.
7If asked for mean or variance, use np and np(1 − p), or λ.
8Check the answer is between 0 and 1 and makes sense against the mean.

Quickest way: Complement and recursion shortcut

When to use it: Use it when the question asks for 'at least one' or when you need several Poisson terms.

For 'at least one', compute only P(0) and subtract from 1.
For Poisson, find P(0) = e^(−λ). Then P(1) = P(0) × λ, P(2) = P(1) × λ ÷ 2, P(3) = P(2) × λ ÷ 3.
For binomial, P(0) = (1 − p)^n. Then P(k + 1) = P(k) × [(n − k) ÷ (k + 1)] × [p ÷ (1 − p)].
Use the calculator's e^x and y^x keys. Many approved calculators also have combination (nCr) keys.
If n is large and p is small, switch to Poisson with λ = np to save effort, but only if the options are far enough apart.

Common mistakes in Binomial and Poisson Distributions

Using binomial when trials are not independent or p varies
Students apply the formula to any count of defaults without checking assumptions.
Fix: State the assumptions: fixed n, constant p, independence. If they fail, the binomial does not apply.
Forgetting the combination term C(n, k)
Students compute p^k × (1 − p)^(n − k) only, which is the probability of one specific sequence.
Fix: Always multiply by C(n, k), the number of ways to choose which k trials succeed.
Not rescaling λ to the interval asked
The rate is given per month but the question asks about a year or a quarter.
Fix: Multiply λ by the length of the new interval. 2 events a month becomes λ = 6 for three months.
Computing 'at least' by adding many terms
Students sum P(1), P(2), and so on, which is slow and error-prone.
Fix: Use 1 − P(X = 0), or 1 − P(X ≤ k − 1) for 'at least k'.
Confusing the binomial variance with np
Students recall that Poisson variance equals the mean and carry it over.
Fix: Binomial variance is np(1 − p), which is smaller than np. Only the Poisson has variance equal to mean.
Using the Poisson approximation when p is not small
Students see a large n and approximate without checking p.
Fix: Use it only when n is large and p is small. Otherwise compute the binomial directly.

Worked examples

Example 1

A bank holds 10 independent corporate loans, each with a one-year default probability of 5%. What is the probability that exactly 2 loans default, and what is the probability that at least one defaults?

Show the solution

Binomial with n = 10 and p = 0.05.
P(X = 2) = C(10, 2) × 0.05^2 × 0.95^8.
C(10, 2) = 10! ÷ (2! × 8!) = 45.
0.05^2 = 0.0025. 0.95^8 = 0.66342.
P(X = 2) = 45 × 0.0025 × 0.66342 = 0.07463.
P(X = 0) = 0.95^10 = 0.59874.
P(X ≥ 1) = 1 − 0.59874 = 0.40126.

Answer: P(exactly 2 defaults) ≈ 7.46%. P(at least one default) ≈ 40.13%.

Example 2

A bank records an average of 3 operational loss events per month, modelled as Poisson. What is the probability of exactly 2 events in a month, and of at most 1 event in a month?

Show the solution

Poisson with λ = 3.
e^(−3) = 0.049787.
P(X = 0) = 0.049787.
P(X = 1) = 0.049787 × 3 = 0.149361.
P(X = 2) = 0.049787 × 3^2 ÷ 2 = 0.049787 × 4.5 = 0.224042.
P(X ≤ 1) = 0.049787 + 0.149361 = 0.199148.

Answer: P(exactly 2 events) ≈ 22.40%. P(at most 1 event) ≈ 19.91%.

Exam tips

Read the setup first. A fixed number of obligors points to binomial. 'Average number per period' points to Poisson.
Questions on 'at least one' almost always reward the complement. Do not add terms.
Remember that Poisson mean equals variance. Exam options often test this directly.
Check that λ matches the time period in the question before you compute.
Practise computing e^(−λ) and powers quickly on your approved calculator so you save time on the 100-question paper.

Practice questions from Common Univariate Random Variables

Binomial and Poisson Distributions in other exams

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

Binomial and Poisson Distributions: frequently asked questions

What is the difference between binomial and Poisson distribution?

The binomial counts successes in a fixed number n of independent trials with constant probability p, so the count cannot exceed n. The Poisson counts events in an interval at average rate λ, with no upper limit. Binomial variance is np(1 − p), while Poisson variance equals its mean λ.

When can I use the Poisson approximation to the binomial?

Use it when n is large and p is small, so np is moderate. Set λ = np. It is an approximation, so exact binomial values will differ slightly.

How do I calculate binomial probability for loan defaults?

Set n as the number of loans and p as the default probability. Assume defaults are independent. Then use P(X = k) = C(n, k) × p^k × (1 − p)^(n − k). For at least one default, use 1 − (1 − p)^n.

Where does Poisson appear in finance?

It models counts of random events such as operational loss events, defaults in a large portfolio, or jumps in a price process. The parameter λ is the expected number of events in the period.