Chapter VIII · 28 min

Discrete Families

A field guide to named stories

“To conjecture about some affair is to measure its probability.”
— Jacob Bernoulli, Ars Conjectandi (1713)

A few discrete families appear so often that they have names. Each one is a counting story that hardened into a formula. If you remember the story, you can rebuild the formula; if you remember only the formula, you will apply it to the wrong story.

Before the names, four quiet words. A distribution is the whole assignment of chance to values. A pmf (probability mass function) is that assignment written as a list $p_{X} (k) = P (X = k)$ . A family is a shape with a knob or two — $n$ and $p$ for binomial, $λ$ for Poisson. A parameter is one of those knobs. The family is the story; the parameter is how the story is tuned.

A field guide

Keep this table nearby. The question is never “which formula is famous?” It is “which story did the experiment actually tell?”

Family	Story	Lives on	Mean
Bernoulli	one yes / no	{0, 1}	p
Binomial	n independent yes/no, same p	0, 1, …, n	np
Geometric	wait for the first yes	1, 2, 3, …	1/p
Poisson	rare events, rate λ	0, 1, 2, …	λ (also the variance)
Hypergeometric	draws without replacement	0 … min(K, n)	nK/N

Bernoulli and binomial

A Bernoulli random variable is a single yes/no: $P (X = 1) = p$ , $P (X = 0) = 1 - p$ . Then $E [X] = p$ , $Var (X) = p (1 - p)$ . It is an indicator in costume.

A binomial $X \sim Bin (n, p)$ is the number of yeses in $n$ independent Bernoulli trials with the same $p$ . The counting is the classical one: there are $(k n)$ sequences with $k$ yeses, each with probability $p^{k} (1 - p)^{n - k}$ , so

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

Linearity gives $E [X] = n p$ without touching the pmf. Independence of the trials gives $Var (X) = n p (1 - p)$ . If the trials are not independent, you still have a count, but you no longer have a binomial.

The binomial pmf for n = 8, p = 1/2. The shape is the row of Pascal’s triangle, normalised. Change p and the mass slides toward 0 or n.

The binomial workshop is the same picture with sliders.

Geometric waiting times

Flip until the first yes. Let $X$ be the trial number of that first yes (supported on ${1, 2, 3, \dots}$ ). Then

P (X = k) = (1 - p)^{k - 1} p, E [X] = \frac{1}{p} .

The geometric law is memoryless: given that you have already failed $m$ times, the remaining wait has the original distribution. A coin does not get embarrassed. (Some texts put the number of failures, supported on ${0, 1, 2, \dots}$ , under the same name. Check the support before you copy a mean.)

A geometric pmf with p = 0.3. Most mass sits on the early integers; the tail is a persistent (1−p) factor.

Poisson rare events

The Poisson law with rate $λ > 0$ is

P (X = k) = e^{- λ} \frac{λ ^{k}}{k !}, k = 0, 1, 2, \dots

with mean and variance both equal to $λ$ . It is the limit of $Bin (n, p_{n})$ when $n \to \infty$ and $n p_{n} \to λ$ : many trials, each unlikely, a moderate expected count. Typographical errors, arrivals at a switchboard, mutations on a strand of DNA — the same counting limit.

Poisson with λ = 3. A unimodal mass on the non-negative integers, centred near its mean, with a right tail.

Without replacement: hypergeometric

An urn holds $N$ balls, $K$ of them gold. You draw $n$ without putting them back. Let $X$ be the number of gold balls in the hand. Then

P (X = k) = \frac{( k K ) ( n - k N - K )}{( n N )},

whenever the binomial coefficients make sense, and $E [X] = n K / N$ by linearity (each position in the hand is equally likely to be gold). This is the binomial’s honest cousin: same “how many gold?”, different story, because each draw changes the urn. If $N$ is huge and $n$ is small, the two families hug — which is why people sometimes get away with a binomial on a deck of cards, and why they should not on a deck of ten.