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FRM Exam Part I · Common Univariate Random Variables

Discrete Uniform and Bernoulli Distributions for FRM Part I

Updated 11 October 2026 · Fact-checked

A discrete uniform variable gives equal probability 1/n to each of n outcomes. A Bernoulli variable has two outcomes, 1 with probability p and 0 with probability 1 − p, so its mean is p and variance is p(1 − p). To solve questions, identify the outcomes, apply the PMF, then use the mean and variance formulas.

Understand Discrete Uniform and Bernoulli Distributions

A random variable assigns a number to each outcome of an uncertain event. A discrete random variable takes a countable set of values. Its probability mass function (PMF) gives the probability of each value. Probabilities must be non-negative and sum to 1.

The discrete uniform distribution is the simplest case. Every one of n possible values is equally likely, so each has probability 1/n. A fair six-sided die is the classic example. The values are 1 to 6 and each has probability 1/6. In FRM questions the values are usually consecutive integers from a to b, so n = b − a + 1.

The Bernoulli distribution models one trial with two outcomes: success (coded 1) or failure (coded 0). Success has probability p and failure has probability 1 − p. In risk work, think of a bond that defaults (1) or does not default (0) over a year, or a VaR exceedance (1) or no exceedance (0) on a day.

Because X is 0 or 1, X² = X. That is why E(X) = p and E(X²) = p, and the variance is p − p² = p(1 − p). The variance is largest at p = 0.5 and shrinks to zero as p approaches 0 or 1.

The Bernoulli is the building block of the binomial distribution. A binomial variable is the sum of n independent Bernoulli trials with the same p. So Bernoulli is a single trial, and binomial counts successes across many trials.

Key formulas to remember

Discrete uniform PMF
P(X = x) = 1 ÷ n, for each of n equally likely values
For consecutive integers a to b, n = b − a + 1.
Discrete uniform mean
E(X) = (a + b) ÷ 2
Valid for consecutive integers from a to b. It is the midpoint.
Discrete uniform variance
Var(X) = (n² − 1) ÷ 12
For consecutive integers, n = b − a + 1. Standard deviation is the square root.
Bernoulli PMF
P(X = 1) = p; P(X = 0) = 1 − p
Also written P(X = x) = p^x × (1 − p)^(1 − x) for x = 0 or 1.
Bernoulli mean
E(X) = p
The expected value equals the success probability.
Bernoulli variance
Var(X) = p(1 − p)
Maximum value is 0.25 at p = 0.5.
Binomial link
Y = X₁ + X₂ + … + Xₙ; E(Y) = np; Var(Y) = np(1 − p)
Holds for n independent Bernoulli trials with the same p.

How to solve Discrete Uniform and Bernoulli Distributions questions

Use this method for any question on these two distributions.

  1. 1Decide which distribution fits: many equally likely values means discrete uniform; one trial with two outcomes means Bernoulli.
  2. 2Write down the parameters: a and b (so n = b − a + 1) for uniform, or p for Bernoulli.
  3. 3Write the PMF and check that the probabilities sum to 1.
  4. 4For a single probability, read it directly: 1/n for uniform, p or 1 − p for Bernoulli.
  5. 5For a range of uniform values, count the favourable values and divide by n.
  6. 6Compute the mean and variance with the formulas. Take the square root only if asked for standard deviation.
  7. 7If the question has several independent trials, switch to the binomial: mean np, variance np(1 − p).
  8. 8Check the answer: probabilities between 0 and 1, variance not negative, Bernoulli variance at most 0.25.

Quickest way: Shortcut formulas and sanity checks

When to use it: Use when you have little time and the question asks for a mean, variance or simple probability.

  1. For consecutive integers a to b, the mean is the midpoint (a + b) ÷ 2. No summing needed.
  2. Variance is (n² − 1) ÷ 12 with n the count of values. For a die, (36 − 1) ÷ 12 = 35/12.
  3. For Bernoulli, mean is p. Variance is p × (1 − p). Do this mentally.
  4. Count favourable values for a uniform probability: favourable ÷ n.
  5. Eliminate options with a Bernoulli variance above 0.25 or a probability above 1.

Common mistakes in Discrete Uniform and Bernoulli Distributions

  • Using (b − a)² ÷ 12 or n²÷12 for discrete uniform variance

    It is mixed up with the continuous uniform variance (b − a)² ÷ 12.

    Fix: For the discrete case use (n² − 1) ÷ 12, where n = b − a + 1.

  • Taking n = b − a instead of b − a + 1

    Students forget that both end points count.

    Fix: Values 0 to 9 give n = 10, not 9. Always count the values.

  • Writing Bernoulli variance as p² or p(1 + p)

    The formula is memorised loosely.

    Fix: Remember it as p × q, with q = 1 − p. Derive it: E(X²) − E(X)² = p − p².

  • Calling a multi-trial count a Bernoulli variable

    The two-outcome nature of each trial hides the fact that the sum is binomial.

    Fix: One trial is Bernoulli. The number of successes in n trials is binomial.

  • Treating the Bernoulli mean as a possible outcome

    Students assume the mean must be a value the variable can take.

    Fix: The mean p is a probability-weighted average. It lies between 0 and 1 and the variable only takes 0 or 1.

Worked examples

Example 1

A random variable X is equally likely to take any integer value from 1 to 10. Find the mean, the variance, and P(X ≥ 8).

Show the solution
  1. Here a = 1, b = 10, so n = 10 − 1 + 1 = 10.
  2. Mean = (1 + 10) ÷ 2 = 5.5.
  3. Variance = (n² − 1) ÷ 12 = (100 − 1) ÷ 12 = 99 ÷ 12 = 8.25.
  4. Favourable values for X ≥ 8 are 8, 9, 10, so 3 values.
  5. P(X ≥ 8) = 3 ÷ 10 = 0.30.

Answer: Mean 5.5, variance 8.25, P(X ≥ 8) = 0.30.

Example 2

A bond has a 4% probability of default over one year. Let X = 1 if it defaults and 0 otherwise. Find E(X), Var(X) and the standard deviation. Then find the mean and variance of the number of defaults in a portfolio of 50 such independent bonds.

Show the solution
  1. X is Bernoulli with p = 0.04.
  2. E(X) = p = 0.04.
  3. Var(X) = 0.04 × 0.96 = 0.0384.
  4. Standard deviation = √0.0384 ≈ 0.1960.
  5. For 50 independent bonds the count is binomial with n = 50, p = 0.04.
  6. Mean = 50 × 0.04 = 2.
  7. Variance = 50 × 0.0384 = 1.92.

Answer: E(X) = 0.04, Var(X) = 0.0384, standard deviation ≈ 0.196. For 50 bonds, mean 2 defaults and variance 1.92.

Exam tips

  • Expect direct formula questions: give the mean or variance from stated parameters. Memorise the formulas cold.
  • Watch the wording on the range of a uniform variable. Check if it is 0 to n or 1 to n, since that changes n and the mean.
  • Bernoulli questions often appear inside default or VaR exceedance contexts. Spot the 0/1 indicator and use p.
  • If options include a Bernoulli variance above 0.25, drop them at once.
  • Use the calculator only for square roots. The rest is quick mental arithmetic.

Practice questions from Common Univariate Random Variables

Discrete Uniform and Bernoulli Distributions in other exams

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

Discrete Uniform and Bernoulli Distributions: frequently asked questions

What is the mean and variance of a discrete uniform distribution?

For consecutive integers from a to b, the mean is (a + b) ÷ 2. The variance is (n² − 1) ÷ 12, where n = b − a + 1 is the number of values.

What is the Bernoulli distribution formula?

A Bernoulli variable equals 1 with probability p and 0 with probability 1 − p. Its mean is p and its variance is p(1 − p). For example, with p = 0.3 the variance is 0.3 × 0.7 = 0.21.

What is the difference between Bernoulli and binomial distribution?

A Bernoulli variable describes one trial with two outcomes. A binomial variable counts the successes in n independent trials with the same p. The binomial is the sum of n Bernoulli variables.

When is Bernoulli variance the largest?

It is largest at p = 0.5, where it equals 0.25. As p moves toward 0 or 1, outcomes become more certain and variance falls toward zero.