Named Discrete Distributions

You'll learn to recognize which named discrete distribution — Bernoulli, binomial, geometric, negative binomial, hypergeometric, or Poisson — fits a counting scenario, and pull its mean and variance straight from the reference table.

Predict: for the Binomial family shown below, if you drag \(p\) toward 0.5 (keeping \(n\) fixed), will the pmf's spread (variance) grow or shrink? Then drag the slider and check the variance readout. Pick a family and drag its sliders — the bars redraw live as the pmf shape changes. The dashed line marks the mean; hover or focus a bar for the exact probability at that value.

Intuitive

Ask yourself "what am I counting, and under what rule?" One trial, success or fail → Bernoulli. Successes across \(n\) independent tries → binomial. Failures before the first (or \(r\)th) success → geometric (or negative binomial). Draws without replacement from a finite pool → hypergeometric. Rare events over a continuous stretch of time or space → Poisson.

Formal

Use the reference table below for the pmf, mean, and variance of each family. A few relationships worth memorizing: Bernoulli = Binomial\((1,p)\); Binomial is a sum of \(n\) iid Bernoullis; Negative Binomial is a sum of \(r\) iid Geometrics; Poisson\((\lambda)\) is the \(n\to\infty\), \(p\to0\), \(np=\lambda\) limit of Binomial\((n,p)\); Hypergeometric \(\to\) Binomial as \(N\to\infty\) with \(K/N\) fixed. On the interactive above, the dashed mean line is exactly the mean row of this table.

Applied

Say you're an insurer tracking the number of claims filed per policyholder per year — often modeled as Poisson\((\lambda)\). If you're underwriting a batch of \(n\) applicants where each independently qualifies with probability \(p\), use Binomial\((n,p)\). If you're auditing a sample of \(k\) policies drawn without replacement from a book of \(N\) (with \(K\) known defects), use Hypergeometric.

Named discrete distributions — pmf, mean, variance
Distribution pmf p(x) Mean Variance
Bernoulli(p) p if x=1, 1−p if x=0 p p(1−p)
Binomial(n,p) C(n,x) px(1−p)n−x, x=0..n np np(1−p)
Geometric(p) (1−p)x p, x=0,1,2,… (1−p)/p (1−p)/p²
Negative Binomial(r,p) C(x+r−1,x)(1−p)xpr, x=0,1,2,… r(1−p)/p r(1−p)/p²
Hypergeometric(N,K,n) C(K,x)C(N−K,n−x)/C(N,n) nK/N n(K/N)(1−K/N)(N−n)/(N−1)
Poisson(λ) e−λλx/x!, x=0,1,2,… λ λ
Worked example
\(N=20\) items, \(K=5\) defective, draw \(n=4\) without replacement. \(X \sim\) Hypergeometric\((20,5,4)\). \(E[X]\) \(= 4(5)/20 = 1\). \(\text{Var}(X)\) \(= 4(0.25)(0.75)(16/19) \approx\) 0.632.
Now you try — faded example

A batch of \(N=50\) parts contains \(K=8\) defective. You inspect \(n=6\) without replacement and let \(X\) be the number of defectives found.

Step 1 — which family applies? Hypergeometric, since you're sampling without replacement from a finite, fixed population.

Step 2 — set up the mean: \(E[X] = nK/N = 6(8)/50\).

Step 3 — compute the value:

Reveal the answer
\(E[X] = 48/50 = \) 0.96.

More info — why Binomial(n,p) becomes Poisson(λ) as n grows

Here's another angle on the Binomial → Poisson limit mentioned above: hold \(np=\lambda\) fixed while \(n\to\infty\) and \(p\to0\) — many trials, each with a tiny success chance. Try it on the interactive: switch to Binomial, push \(n\) up toward 40 while dragging \(p\) down so \(np\) stays around 4, then switch to Poisson with \(\lambda=4\) — the two bar shapes converge. This is why claim counts (rare events across many "trials") are usually modeled directly as Poisson rather than Binomial. See the StatQuest link under Dive deeper below for a worked derivation.

Check your understanding

Question 1 of 4

A shipment has 20 items, 5 defective. You draw 4 without replacement and count defectives X. Which distribution and mean apply?

Question 2 of 4

Which single property, if you observed it in claims-count data, would make Poisson(λ) a poor fit even though the mean looks right?

Question 3 of 4

In the Binomial pmf p(x) = C(n,x)·pˣ·(1−p)ⁿ⁻ˣ, what does the C(n,x) factor count, and which rule computes it?

Question 4 of 4

X ~ Negative Binomial(r, p) is built as a sum of r independent Geometric(p) variables (failures before each success). If one Geometric(p) has Var = (1−p)/p², what's Var(X)?

Recap

  • Match the scenario to the family first: fixed \(n\) independent trials → Binomial; failures before a success → Geometric / Negative Binomial; sampling without replacement from a finite pool → Hypergeometric; rare events over time/space → Poisson.
  • Bernoulli = Binomial\((1,p)\); Binomial sums \(n\) iid Bernoullis; Negative Binomial sums \(r\) iid Geometrics.
  • Binomial\((n,p)\to\) Poisson\((\lambda)\) as \(n\to\infty,\ p\to0,\ np=\lambda\); Hypergeometric \(\to\) Binomial as \(N\to\infty\) with \(K/N\) fixed.
  • Every family's pmf, mean, and variance is in the reference table above — memorize the table, not just the names.

Dive deeper

Sources

  • Named Discrete Distributions