Discrete Probability Distributions

Learning Objectives

  • Understand the concepts of probability mass functions (PMF) and cumulative distribution functions (CDF).
  • Calculate the expected value (mean) and variance of a discrete random variable.
  • Apply the Binomial and Poisson distributions to model relevant civil engineering scenarios.
  • Apply the Geometric and Negative Binomial distributions for scenarios involving trials to achieve successes.
  • Use the Hypergeometric distribution to model probabilities when sampling without replacement from finite batches.

Discrete probability distributions are foundational for modeling scenarios where outcomes are countable, such as the number of structural defects, rainfall events, or equipment failures. This lesson covers key distributions including Binomial, Poisson, Geometric, Negative Binomial, and Hypergeometric, and how to apply them to solve practical engineering problems.

Introduction to Random Variables

When analyzing engineering data, we often deal with variables whose outcomes are determined by chance. A random variable is a numerical description of the outcome of an experiment. A discrete random variable can take on a countable number of distinct values (e.g., the number of potholes on a 10km stretch of road, the number of defective bricks in a pallet, or the number of extreme weather events in a year). Understanding these variables allows engineers to quantify risk and predict the frequency of specific events.

Probability Mass Functions and Mathematical Expectation

Probability Functions and Expectation

The foundational math behind discrete random variables.

Probability Mass Function (PMF), f(x)f(x) or P(X=x)P(X=x)

A function that assigns a probability to each possible value of a discrete random variable.

PMF Conditions

  • f(x)≥0f(x) \ge 0 for all xx.
  • ∑xf(x)=1\sum_{x} f(x) = 1.

Cumulative Distribution Function (CDF), F(x)F(x)

The probability that the random variable XX will take a value less than or equal to xx.

Cumulative Distribution Function

Formula for calculating the CDF of a discrete random variable.

F(x)=P(X≤x)=∑t≤xf(t)F(x) = P(X \le x) = \sum_{t \le x} f(t)

Variables

SymbolDescriptionUnit
F(x)F(x)Cumulative Distribution Function-
XXDiscrete random variable-
xxSpecific value of the random variable-
f(t)f(t)Probability Mass Function evaluated at t-

Mathematical Expectation

Mathematical Expectation and Variance

The expected value represents the theoretical mean of the random variable.

Expected Value (Mean), μ\mu or E[X]E[X]

The long-run average value of the random variable over infinitely many trials. It is the center of the probability distribution.

Expected Value (Mean)

Formula for calculating the expected value.

μ=E[X]=∑xx⋅f(x)\mu = E[X] = \sum_{x} x \cdot f(x)

Variables

SymbolDescriptionUnit
μ\muExpected value or mean-
E[X]E[X]Expected value of random variable X-
xxSpecific value of the random variable-
f(x)f(x)Probability Mass Function evaluated at x-

Variance, σ2\sigma^2 or V(X)V(X)

A measure of the dispersion or spread of the probability distribution around the mean.

Variance

Formula for calculating the variance.

σ2=E[(X−μ)2]=∑x(x−μ)2⋅f(x)\sigma^2 = E[(X - \mu)^2] = \sum_{x} (x - \mu)^2 \cdot f(x)

Variables

SymbolDescriptionUnit
σ2\sigma^2Variance-
EEExpected value operator-
XXDiscrete random variable-
μ\muExpected value or mean-
xxSpecific value of the random variable-
f(x)f(x)Probability Mass Function evaluated at x-

Computational Formula for Variance

An alternative, easier formula for calculating variance.

σ2=E[X2]−(E[X])2\sigma^2 = E[X^2] - (E[X])^2

Variables

SymbolDescriptionUnit
σ2\sigma^2Variance-
E[X2]E[X^2]Expected value of the square of X-
E[X]E[X]Expected value of X-

Common Discrete Distributions in Engineering

Common Discrete Distributions

Specific models used to describe common engineering scenarios.

The Binomial Distribution

Introduction to the Binomial Distribution

The Binomial distribution models the number of successes in a fixed number of independent Bernoulli trials with the same success probability. In civil engineering, an example is the number of independently selected concrete cylinders that exceed a specified strength threshold in a fixed sample. Repeated days affected by weather should not be treated as independent trials unless that assumption is justified.

Binomial Distribution

A distribution that models the probability of exactly xx successes in nn independent trials.

Binomial Distribution Conditions

  • There are a fixed number of trials (nn).
  • Each trial has only two possible outcomes (Success or Failure).
  • The probability of success (pp) remains constant for each trial.
  • The trials are mutually independent.

Binomial Distribution Formula

Formula to calculate the probability of exactly x successes in n trials.

P(X=x)=(nx)px(1−p)n−xfor x=0,1,…,nP(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \quad \text{for } x = 0, 1, \dots, n

Variables

SymbolDescriptionUnit
P(X=x)P(X = x)Probability of exactly x successes-
nnTotal number of trials-
xxNumber of successes-
ppProbability of success on a single trial-
(1−p)(1-p)Probability of failure on a single trial-

Binomial Mean and Variance

Formulas for the mean and variance of a Binomial distribution.

μ=npandσ2=np(1−p)\mu = np \quad \text{and} \quad \sigma^2 = np(1-p)

Variables

SymbolDescriptionUnit
μ\muExpected value or mean-
σ2\sigma^2Variance-
nnTotal number of trials-
ppProbability of success on a single trial-

The Poisson Distribution

Introduction to the Poisson Distribution

The Poisson distribution models a count over a defined time, length, area, volume, or other exposure when the count process is adequately described by a Poisson model. For a homogeneous Poisson process, the event rate is stable over the stated exposure and counts in disjoint intervals are independent. Rarity alone does not establish that a Poisson model is appropriate.

Poisson Distribution

Models the nonnegative integer count XX in a specified exposure when its distribution is Poisson with mean count λ\lambda. Under a homogeneous Poisson process with rate rr per unit exposure EE, λ=rE\lambda=rE. This is a count model with stated assumptions, not a rule that every rare event follows a Poisson distribution. Separately, a Poisson distribution may approximate a Binomial count when the fixed-trial model and approximation accuracy support it.

Poisson Distribution Formula

Formula to calculate the probability of exactly x events occurring in a given interval.

P(X=x)=λxe−λx!for x=0,1,2,…P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!} \quad \text{for } x = 0, 1, 2, \dots

Variables

SymbolDescriptionUnit
P(X=x)P(X = x)Probability of exactly x events-
λ\lambdaExpected event count over the specified exposure-
eeEuler's number (approximately 2.71828)-
xxNumber of events-
Poisson Counts Scale with Exposure

Analytical schematic of a homogeneous Poisson count model: for a stable rate r per unit exposure, the expected count over exposure E is λ=rE. Event marks are illustrative locations, not measured data, and an individual count is random.

A stable event rate r over exposure E and twice that exposure, with expected counts λ=rE and 2λ=r(2E).

Poisson Mean and Variance

Formulas for the mean and variance of a Poisson distribution.

μ=λandσ2=λ\mu = \lambda \quad \text{and} \quad \sigma^2 = \lambda

Variables

SymbolDescriptionUnit
μ\muExpected value or mean-
σ2\sigma^2Variance-
λ\lambdaAverage rate of occurrence in the interval-

Unique Property of Poisson Distribution

A unique property of the Poisson distribution is that its mean equals its variance.

The Geometric and Negative Binomial Distributions

Geometric and Negative Binomial Models

The Geometric and Negative Binomial distributions model the number of trials needed to achieve a specific number of successes. For instance, testing identical soil samples until the first one fails a compaction test (Geometric), or testing until exactly three samples fail (Negative Binomial).

Geometric Distribution

Models the number of independent trials XX needed to get the first success. (e.g., How many times must we test a newly designed joint until we observe the first failure, assuming a constant failure probability pp?)

Geometric Distribution Formula

Formula for calculating the probability of getting the first success on the x-th trial.

P(X=x)=(1−p)x−1pfor x=1,2,3,…P(X = x) = (1-p)^{x-1}p \quad \text{for } x = 1, 2, 3, \dots

Variables

SymbolDescriptionUnit
P(X=x)P(X = x)Probability of first success on the x-th trial-
ppProbability of success on a single trial-
xxTrial number on which the first success occurs-

Geometric Mean and Variance

Formulas for calculating the mean and variance of a geometric distribution.

μ=1pandσ2=1−pp2\mu = \frac{1}{p} \quad \text{and} \quad \sigma^2 = \frac{1-p}{p^2}

Variables

SymbolDescriptionUnit
μ\muExpected value or mean-
σ2\sigma^2Variance-
ppProbability of success on a single trial-

Negative Binomial Distribution

A generalization of the geometric distribution. It models the number of independent trials XX needed to get exactly rr successes.

Negative Binomial Distribution Formula

Formula for calculating the probability of getting the r-th success on the x-th trial.

P(X=x)=(x−1r−1)pr(1−p)x−rfor x=r,r+1,…P(X = x) = \binom{x-1}{r-1} p^r (1-p)^{x-r} \quad \text{for } x = r, r+1, \dots

Variables

SymbolDescriptionUnit
P(X=x)P(X = x)Probability of r-th success on the x-th trial-
xxTrial number on which the r-th success occurs-
rrTotal number of successes desired-
ppProbability of success on a single trial-

Negative Binomial Mean and Variance

Formulas for calculating the mean and variance of a negative binomial distribution.

μ=rpandσ2=r(1−p)p2\mu = \frac{r}{p} \quad \text{and} \quad \sigma^2 = \frac{r(1-p)}{p^2}

Variables

SymbolDescriptionUnit
μ\muExpected value or mean-
σ2\sigma^2Variance-
rrTotal number of successes desired-
ppProbability of success on a single trial-

The Hypergeometric Distribution

Sampling Without Replacement

The Hypergeometric Distribution models sampling without replacement from a finite population. This is critical in quality control where inspecting an item destroys it or removes it from the batch, changing the probability for subsequent selections.

Hypergeometric Distribution

Unlike the Binomial distribution where pp is constant (sampling with replacement), the Hypergeometric distribution is used when sampling without replacement from a finite population of size NN, containing exactly KK successes. (e.g., Selecting 5 concrete cylinders from a batch of 50, where 3 are known to be defective).

Hypergeometric Distribution Formula

Formula for calculating hypergeometric probabilities.

P(X=x)=(Kx)(N−Kn−x)(Nn)P(X = x) = \frac{\binom{K}{x} \binom{N-K}{n-x}}{\binom{N}{n}}

Variables

SymbolDescriptionUnit
P(X=x)P(X = x)Probability of exactly x successes in the sample-
NNTotal population size-
KKTotal number of successes in the population-
nnSample size-
xxNumber of successes in the sample-

Hypergeometric Mean and Variance

Formulas for the mean and variance of a hypergeometric distribution.

μ=n(KN)andσ2=n(KN)(N−KN)(N−nN−1)\mu = n \left(\frac{K}{N}\right) \quad \text{and} \quad \sigma^2 = n \left(\frac{K}{N}\right) \left(\frac{N-K}{N}\right) \left(\frac{N-n}{N-1}\right)

Variables

SymbolDescriptionUnit
μ\muExpected value or mean-
σ2\sigma^2Variance-
NNTotal population size-
KKTotal number of successes in the population-
nnSample size-
Replacement Changes the Next-Draw Probability

Illustrative finite batch with 3 marked items among 8. With replacement, the next-draw probability remains 3/8; without replacement it becomes 2/7 after a marked item or 3/7 after an unmarked item. These values explain why finite sampling without replacement follows a Hypergeometric model rather than independent Binomial trials.

In an illustrative lot with three defective items among eight, the next-draw defect probability changes after a draw without replacement but stays 3/8 with replacement.

Interactive Simulation

Interact with the simulation below to visualize various discrete probability distributions.

Engineering Data Analysis

Discrete Probability Distributions Explorer

Loading chart...
P(X=x)=(10x)(0.50)x(0.50)10−xP(X=x) = \binom{10}{x} (0.50)^x (0.50)^{10-x}
10
0.50

Theoretical Properties

Mean (μ\mu):5.00
Variance (σ2\sigma^2):2.50
Std Dev (σ\sigma):1.58

Interactive Simulation

Compare the geometric and hypergeometric distributions under different parameters to see how sampling without replacement alters success probabilities.

Engineering Data Analysis • Topic 5

Discrete Probability Distributions Sandbox

Success Prob (pp)0.30
Loading chart...
Mean (Expected Value)3.333
Variance7.778
Select a Discrete Count Model

Match the random variable and sampling mechanism to a discrete probability model before calculating probabilities.

Use Hypergeometric for a finite population sampled without replacement, with N, K, and n identified. Use Binomial for a fixed number of independent Bernoulli trials with common p. A Poisson approximation to a Binomial count is a separate option only when its accuracy is adequate for the intended use; no universal rarity cutoff establishes that. A Poisson count over exposure E with mean λ = rE requires a suitable count process; a homogeneous Poisson process has a stable rate r and independent increments. Geometric and Negative Binomial models count independent constant-p trials to the first or r-th success. If the data-generation process is unclear or these conditions are unsupported, hold and establish an appropriate model.

Select a Discrete Count ModelMatch the random variable and sampling mechanism to a discrete probability model before calculating probabilities.. Use Hypergeometric for a finite population sampled without replacement, with N, K, and n identified. Use Binomial for a fixed number of independent Bernoulli trials with common p. A Poisson approximation to a Binomial count is a separate option only when its accuracy is adequate for the intended use; no universal rarity cutoff establishes that. A Poisson count over exposure E with mean λ = rE requires a suitable count process; a homogeneous Poisson process has a stable rate r and independent increments. Geometric and Negative Binomial models count independent constant-p trials to the first or r-th success. If the data-generation process is unclear or these conditions are unsupported, hold and establish an appropriate model.. Define the count, sampling plan, and exposure → Finite population sampled without replacement?; Finite population sampled without replacement? — Yes → Hypergeometric count; identify N, K, n; Finite population sampled without replacement? — No → Fixed n independent Bernoulli trials with common p?; Fixed n independent Bernoulli trials with common p? — Yes → Binomial count; Poisson approximation only if adequate; Fixed n independent Bernoulli trials with common p? — No → Stable rate and independent increments over exposure?; Stable rate and independent increments over exposure? — Yes → Poisson count, λ = rate × exposure; Stable rate and independent increments over exposure? — No → Independent constant-p trials continue until r successes?; Independent constant-p trials continue until r successes? — Yes → Is r = 1?; Is r = 1? — r = 1 → Geometric: trials to first success; Is r = 1? — r > 1 → Negative Binomial: r > 1 successes; Independent constant-p trials continue until r successes? — No → Hold: resolve the process or model assumptions

Define the count, sampling plan, and exposure → Finite population sampled without replacement?; Finite population sampled without replacement? — Yes → Hypergeometric count; identify N, K, n; Finite population sampled without replacement? — No → Fixed n independent Bernoulli trials with common p?; Fixed n independent Bernoulli trials with common p? — Yes → Binomial count; Poisson approximation only if adequate; Fixed n independent Bernoulli trials with common p? — No → Stable rate and independent increments over exposure?; Stable rate and independent increments over exposure? — Yes → Poisson count, λ = rate × exposure; Stable rate and independent increments over exposure? — No → Independent constant-p trials continue until r successes?; Independent constant-p trials continue until r successes? — Yes → Is r = 1?; Is r = 1? — r = 1 → Geometric: trials to first success; Is r = 1? — r > 1 → Negative Binomial: r > 1 successes; Independent constant-p trials continue until r successes? — No → Hold: resolve the process or model assumptions

  • Define the count, sampling plan, and exposure: terminator
  • Finite population sampled without replacement?: decision
  • Hypergeometric count; identify N, K, n: document
  • Fixed n independent Bernoulli trials with common p?: decision
  • Binomial count; Poisson approximation only if adequate: document
  • Stable rate and independent increments over exposure?: decision
  • Poisson count, λ = rate × exposure: document
  • Independent constant-p trials continue until r successes?: decision
  • Is r = 1?: decision
  • Geometric: trials to first success: document
  • Negative Binomial: r > 1 successes: document
  • Hold: resolve the process or model assumptions: terminator
Key Takeaways
  • Random Variables: Numerical values assigned to experimental outcomes.
  • Expected Value (E[X]E[X]): The long-run average of a discrete distribution.
  • Binomial: Used for independent trials with exactly two outcomes (success/failure) and constant probability pp.
  • Poisson: Models a count over a stated exposure when the Poisson count assumptions are adequate; for a homogeneous process, the expected count is rate times exposure (λ=rE\lambda=rE). A fixed-trial Poisson approximation is a separate modeling choice.
  • Geometric/Negative Binomial: Focuses on the number of trials needed to achieve a specified number of successes.
  • Hypergeometric: Used for finite populations when sampling without replacement (probability changes trial-to-trial).

References