Discrete Probability Distributions

Match the model to how the count is generated

A discrete distribution assigns probability mass to possible numerical outcomes. Different count processes require different models; “my data are integers” is not enough to choose Poisson or binomial.

Model	What is counted	Essential condition
Binomial(n,p)	Successes in n trials	Independent trials, common p
Poisson(λ)	Events in a fixed exposure	Constant-rate independent-increment model
Geometric(p)	Trials until first success	Independent trials, common p
Negative binomial(r,p)	Failures before r successes	State the counting convention
Hypergeometric(N,K,n)	Successes in n draws without replacement	Uniform selection from a finite population

All model probabilities must be nonnegative and sum to 1 over their support. Their parameters are not interchangeable: a probability p lies between 0 and 1, while a Poisson expected count λ can exceed 1.

Worked example: events per hour

Under a hypothetical Poisson model with expected count λ = 3 arrivals per hour:

$P(X=k)=e^{-\lambda}\frac{\lambda^k}{k!}.$

The probability of exactly two arrivals is e⁻³ × 3²/2 = 0.2240. The probability of no arrivals is 0.0498. The mean and variance both equal 3, so the standard deviation is √3 ≈ 1.7321 arrivals.

The equality of mean and variance is a model property, not an exact requirement for sample summaries. Substantial extra variability can suggest changing rates, clusters or another model. For a half-hour at the same rate, λ is 1.5; the exposure window must match the parameter.

Waiting for success uses a different variable

If each independent attempt succeeds with probability p = 0.25 and T counts trials including the first success, then:

$P(T=k)=(1-p)^{k-1}p,\quad k=1,2,\ldots$

P(T = 3) = 0.75² × 0.25 = 0.140625, and E[T] = 1/p = 4 trials. Some software instead counts failures before the first success, starting at 0. Translate the convention before comparing results.

Without replacement changes dependence

A box contains N = 10 distinct items, K = 3 defective. Select n = 2 uniformly without replacement. The probability of exactly one defect is:

$\frac{\binom31\binom71}{\binom{10}2}=\frac{21}{45}=\mathbf{0.4667}\text{ (rounded)}.$

Using a binomial with p = 0.3 would give 2(0.3)(0.7) = 0.42. After a draw, the box composition changes, so the exact sampling model is hypergeometric. A binomial approximation may work when the sampled fraction is small, but here it is visibly different.

Practice

Which model fits the number of attempts up to and including the first success, if independent attempts have a constant success probability?

Show the explanation

The geometric distribution under the trials-until-success convention. Binomial instead fixes the number of trials and counts successes. Always state which quantity is random.

Continue

Read the binomial lesson, or the Poisson table. NIST describes the Poisson functions; SciPy documents the hypergeometric convention.