Discrete Distributions

August 21, 2025

An introduction to key discrete probability distributions: the uniform, Bernoulli, and binomial distributions, with examples, expectations, and variances.

ProbabilityBeginnerDiscrete Distribution

So far, we have seen that a probability distribution describes the relationship between the values a random variable can take and the probabilities assigned to them.
In this article, we focus on three fundamental discrete distributions:
the discrete uniform distribution, the Bernoulli distribution, and the binomial distribution.

For each, we will look at its definition, an example, expectation, and variance.

Discrete Uniform Distribution

Definition
If a random variable $X$ takes values in $\\{1,2,3,\\dots,n\\}$ with equal probability:

P(X=k) = \frac{1}{n}, \quad k = 1,2,\\dots,n

Example: A fair die
Rolling a six-sided die corresponds to a uniform distribution with $n=6$ :

P(X=k) = \frac{1}{6}, \quad k=1,2,3,4,5,6

Expectation and Variance

Expectation:

E[X] = \frac{1+2+\\dots+n}{n} = \frac{n+1}{2}

Variance:

V[X] = \frac{n^2-1}{12}

For a die ( $n=6$ ):

E[X] = 3.5, \quad V[X] = \frac{35}{12} \approx 2.92

Bernoulli Distribution

Definition
If a random variable $X$ represents a “success” (1) with probability $p$ and a “failure” (0) with probability $1-p$ :

P(X=1) = p, \quad P(X=0) = 1-p

Example: A coin toss
Consider tossing a coin once, with heads as “success.”
If $p=0.5$ :

P(X=1) = 0.5, \quad P(X=0) = 0.5

Expectation and Variance

Expectation:

E[X] = 0 \cdot (1-p) + 1 \cdot p = p

Variance:

V[X] = p(1-p)

For a fair coin ( $p=0.5$ ):

E[X] = 0.5, \quad V[X] = 0.25

Binomial Distribution

Definition
The binomial distribution describes the number of successes in $n$ independent Bernoulli trials, each with success probability $p$ .

P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}, \quad k=0,1,2,\\dots,n

Example: 10 coin tosses
If we toss a fair coin ( $p=0.5$ ) 10 times, the probability of getting $k$ heads is:

P(X=k) = \binom{10}{k} \left(\frac{1}{2}\right)^{10}

Expectation and Variance

Expectation:

E[X] = np

Variance:

V[X] = np(1-p)

For $n=10, p=0.5$ :

E[X] = 5, \quad V[X] = 2.5

Summary

Discrete Uniform Distribution: All outcomes equally likely (e.g., a die).
- $E[X]=\\tfrac{n+1}{2}, \quad V[X]=\\tfrac{n^2-1}{12}$
Bernoulli Distribution: Two outcomes, success (1) or failure (0) (e.g., a coin toss).
- $E[X]=p, \quad V[X]=p(1-p)$
Binomial Distribution: The number of successes in $n$ independent Bernoulli trials.
- $E[X]=np, \quad V[X]=np(1-p)$

These three are the basic building blocks of discrete probability distributions and form the foundation for learning more advanced distributions.

Interactive Discrete Distributions Demo

Number of outcomes (n):6

Uniform Distribution

Statistics

Expectation (Mean)

E[X] = 3.500

Variance

Var(X) = 2.917

Standard Deviation

σ = 1.708

Formulas:

E[X] = (n+1)/2

Var(X) = (n²-1)/12

About This Distribution

The discrete uniform distribution assigns equal probability to each of n possible outcomes. Like rolling a fair die, each outcome is equally likely. As you increase n, the variance increases but each individual probability decreases.