FRM Exam Part I · Random Variables

Common Probability Distributions for FRM Part I

Updated 11 October 2026 · Fact-checked

Common probability distributions are standard models for random outcomes. Bernoulli and binomial count successes in fixed trials, Poisson counts events in an interval, and uniform spreads probability evenly. To solve a question, identify the distribution from the wording, write its parameters, apply the mean, variance or probability formula, and compute.

Understand Common Probability Distributions

A probability distribution tells you every possible value of a random variable and how likely each one is. In risk work, a few standard shapes cover most situations, so you learn their formulas once and reuse them.

The Bernoulli distribution is one trial with two outcomes: success (probability p) or failure (probability 1 − p). A single loan either defaults or it does not. The binomial distribution adds up n independent Bernoulli trials with the same p. It answers: how many of 20 loans default?

The Poisson distribution counts events that happen at a constant average rate λ over a time period or area, such as the number of operational loss events in a month. It has no fixed upper limit. When n is large and p is small, the binomial is close to a Poisson with λ = np.

The uniform distribution gives equal likelihood across a range. The discrete version has equally likely integers (a fair die). The continuous version on [a, b] has constant density 1 ÷ (b − a). The normal distribution is covered on its own page.

Key link: for the Poisson, mean and variance are both λ. For the binomial, variance np(1 − p) is always less than the mean np. Use this to spot which model fits.

Key formulas to remember

Bernoulli mean and variance: E(X) = p; Var(X) = p(1 − p); X = 1 with probability p, X = 0 with probability 1 − p.
Binomial probability: P(X = k) = [n! ÷ (k!(n − k)!)] × p^k × (1 − p)^(n − k); k = 0, 1, ..., n. Trials independent, same p.
Binomial mean and variance: E(X) = np; Var(X) = np(1 − p); Standard deviation = √[np(1 − p)].
Poisson probability: P(X = k) = e^(−λ) × λ^k ÷ k!; k = 0, 1, 2, ... λ is the average number of events in the period.
Poisson mean and variance: E(X) = λ; Var(X) = λ; Scale λ to the period asked: a rate of 3 per month is 9 per quarter.
Continuous uniform on [a, b]: f(x) = 1 ÷ (b − a); E(X) = (a + b) ÷ 2; Var(X) = (b − a)² ÷ 12; P(c ≤ X ≤ d) = (d − c) ÷ (b − a) for a ≤ c ≤ d ≤ b.
Discrete uniform on integers 1 to n: E(X) = (n + 1) ÷ 2; Var(X) = (n² − 1) ÷ 12; Each value has probability 1 ÷ n.
Poisson approximation to binomial: λ = np; Reasonable when n is large and p is small.

How to solve Common Probability Distributions questions

Use this routine for any question on standard distributions.

1Read the wording. Fixed number of trials with success or failure points to binomial; counts over time or space with no upper limit points to Poisson; 'equally likely' or 'any value in a range' points to uniform.
2Write down the parameters: n and p, or λ, or a and b. Check the units and period match the question.
3Decide what is asked: an exact probability, a cumulative probability, a mean, a variance or a standard deviation.
4For a cumulative probability such as P(X ≤ k), add the individual terms. For 'at least one', use 1 − P(X = 0).
5Substitute into the formula and compute carefully. Use the e^x key for Poisson and the nCr key for binomial.
6Sense-check: the probability must lie between 0 and 1, and the mean should be near the centre of likely outcomes.
7State the answer in the form the options use (decimal or percentage).

Quickest way: Complement and parameter shortcut

When to use it: Use when the question asks for 'at least one' or for a mean or variance, or when you must choose between answer options quickly.

For 'at least one' events, compute 1 − P(X = 0). Binomial: 1 − (1 − p)^n. Poisson: 1 − e^(−λ).
For uniform probabilities, skip integration: probability = length of the interval ÷ total length.
For mean and variance, plug into np, np(1 − p), λ, (a + b) ÷ 2 or (b − a)² ÷ 12 directly.
Eliminate options by size: a binomial probability cannot exceed 1, and a uniform variance is never negative.
Remember the identity check: if a question gives a count with mean equal to variance, think Poisson.

Common mistakes in Common Probability Distributions

Using the wrong period for the Poisson rate
The question gives a monthly rate but asks about a year or a quarter.
Fix: Rescale λ first. A rate of 2 per month becomes λ = 6 for three months.
Forgetting the combination term in the binomial
Students compute p^k × (1 − p)^(n − k) and stop.
Fix: Always multiply by n! ÷ (k!(n − k)!). It counts the ways to place k successes.
Dividing by 12 incorrectly for uniform variance
Students use (b − a) ÷ 12 or forget to square the range.
Fix: Use (b − a)² ÷ 12. For [0, 10] that is 100 ÷ 12 = 8.33, not 0.83.
Treating P(X ≤ k) as P(X = k)
Skipping the wording 'at most' or 'no more than'.
Fix: Underline the inequality. Sum probabilities from 0 to k for 'at most k'.
Using the binomial when trials are not independent or p changes
The formula is memorised without its conditions.
Fix: Check for a fixed n, independent trials and constant p before using binomial.
Taking the variance of a binomial as np
Confusing it with the Poisson, where variance equals the mean.
Fix: Binomial variance is np(1 − p). Only the Poisson has variance equal to mean.

Worked examples

Example 1

A bank holds 10 independent loans, each with a 5% probability of default over a year. What is the probability that exactly 2 default? What is the expected number of defaults and the standard deviation?

Show the solution

Binomial with n = 10, p = 0.05, k = 2.
Combination: 10! ÷ (2! × 8!) = 45.
p^2 = 0.0025. (1 − p)^8 = 0.95^8 = 0.66342.
P(X = 2) = 45 × 0.0025 × 0.66342 = 0.07463.
Mean = np = 10 × 0.05 = 0.5.
Variance = np(1 − p) = 10 × 0.05 × 0.95 = 0.475. Standard deviation = √0.475 = 0.6892.

Answer: P(exactly 2 defaults) ≈ 7.46%; expected defaults = 0.5; standard deviation ≈ 0.689.

Example 2

A trading desk records an average of 3 operational errors per month, modelled as Poisson. What is the probability of at least one error in a given month? Separately, a loss severity is uniform between USD 2 million and USD 10 million. What are its mean and variance?

Show the solution

Poisson with λ = 3. P(X = 0) = e^(−3) = 0.049787.
P(at least one) = 1 − 0.049787 = 0.950213.
Uniform with a = 2, b = 10 (USD million).
Mean = (2 + 10) ÷ 2 = 6.
Variance = (10 − 2)² ÷ 12 = 64 ÷ 12 = 5.3333.

Answer: P(at least one error) ≈ 95.0%; severity mean = USD 6 million; variance ≈ 5.33 (USD million squared).

Exam tips

Identify the distribution from keywords before touching the calculator: 'fixed number of trials' is binomial, 'average rate per period' is Poisson, 'equally likely' is uniform.
Questions often test the mean and variance formulas rather than long probability sums. Know all of them cold.
Use the complement for 'at least one' questions. It saves several calculations.
Watch the units: rescale λ to the period asked and keep uniform variance in squared units.
Practise the nCr and e^x keys on your approved calculator so that binomial and Poisson answers take under a minute.

Practice questions from Random Variables

Common Probability Distributions in other exams

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

Common Probability Distributions: frequently asked questions

What is the difference between binomial and Poisson distributions?

Binomial counts successes in a fixed number of independent trials, so it has an upper limit n. Poisson counts events over a period at an average rate λ with no upper limit. Binomial variance is np(1 − p), while Poisson variance equals its mean λ.

How do I calculate a binomial probability?

Identify n, p and k. Compute the combination n! ÷ (k!(n − k)!), then multiply by p^k and (1 − p)^(n − k). For cumulative probabilities, add the terms for each value of k required.

What are the mean and variance of a uniform distribution?

For a continuous uniform on [a, b], the mean is (a + b) ÷ 2 and the variance is (b − a)² ÷ 12. The density is flat at 1 ÷ (b − a) across the range.

When can I approximate a binomial with a Poisson?

When n is large and p is small, the binomial is close to a Poisson with λ = np. This is common for rare events such as defaults in a large portfolio.