Week 9 — Common discrete models

Bernoulli, binomial, geometric, and Poisson

The week question

When you face a new chance situation, you do not have to invent a probability model from scratch every time. A handful of named models show up again and again, and most discrete problems you will meet are one of them in disguise. So the week’s question is a recognition question: given a described scenario, which standard discrete model fits, and what makes you sure? Once you can name the model, the pmf, the mean, and the variance come along for free.

We have already met the pieces. In Week 6 we counted outcomes; in Week 7 we packaged those counts into a random variable and its pmf; in Week 8 we summarized a random variable with \(E[X]\) and \(\mathrm{Var}(X)\). This week we collect the four discrete models that cover most of what you will see — Bernoulli, binomial, geometric, and Poisson — and we practice matching a story to a model.

Why this matters

Naming the model is the move that turns a wordy scenario into something you can compute. If you can say “this is \(\text{Binomial}(10, 0.5)\),” then you immediately know the pmf, that the mean is \(5\), and that the variance is \(2.5\) — no fresh derivation required. The hard part is almost never the arithmetic; it is reading the assumptions in the story carefully enough to choose correctly.

The skill also travels. The same Poisson model that describes shuttle arrivals describes typos on a page, calls to a help desk in an hour, or radioactive decays in a second. The same binomial that describes a guessed quiz describes defective items in a batch or voters favoring a measure in a small sample. Learning the assumptions behind each model — not just its formula — is what lets you reuse it.

Learning goals

By the end of this week you should be able to:

State, in words and in symbols, the Bernoulli, binomial, geometric, and Poisson models, including each pmf and (as stated results from Week 8) each mean and variance.
Read a scenario and identify which standard discrete model its assumptions select — and explain the assumption that rules the others out.
Compute a probability from the binomial pmf and from the Poisson pmf for our recurring commuter case.
Recognize a new scenario (defective items in a batch) as binomial and set up its pmf.
Know the boundary of each model: what happens when its assumptions are only approximately true.

Core vocabulary

Bernoulli trial — a single experiment with exactly two outcomes, conventionally “success” (1) and “failure” (0), with \(P(\text{success}) = p\). The atom out of which the other models are built.
Binomial — the count of successes in a fixed number \(n\) of independent Bernoulli trials, each with the same \(p\). Written \(X \sim \text{Binomial}(n, p)\).
Geometric — the number of trials up to and including the first success in a sequence of independent Bernoulli\((p)\) trials. Written \(X \sim \text{Geometric}(p)\), support \(\{1, 2, 3, \dots\}\).
Poisson — the count of events in a fixed interval of time or space when events occur at a constant average rate, independently of one another. Written \(N \sim \text{Poisson}(\lambda)\).
Rate / parameter \(\lambda\) — for Poisson, the expected number of events per interval; it is both the mean and the variance.
Support — the set of values the random variable can actually take (binomial: \(0\) to \(n\); geometric: \(1, 2, 3, \dots\); Poisson: \(0, 1, 2, \dots\)).

Concept development

Bernoulli and binomial: a fixed number of yes/no trials

Start with the smallest model. A Bernoulli\((p)\) random variable is one trial with two outcomes: \(X = 1\) with probability \(p\) and \(X = 0\) with probability \(1 - p\). Its pmf is just \(p(1) = p\), \(p(0) = 1 - p\). From Week 8, \(E[X] = p\) and \(\mathrm{Var}(X) = p(1-p)\). By itself it is almost too simple to be interesting; its value is as a building block.

Now run \(n\) independent Bernoulli\((p)\) trials and count the successes. That count is \(X \sim \text{Binomial}(n, p)\). To take a particular value \(x\), you need \(x\) successes and \(n - x\) failures in some order; each such ordering has probability \(p^x (1-p)^{\,n-x}\), and there are \(\binom{n}{x}\) orderings (the Week 6 counting reappears here). Adding them gives the binomial pmf:

\[ p(x) = \binom{n}{x}\, p^{x} (1-p)^{\,n-x}, \qquad x = 0, 1, 2, \dots, n. \]

The mean and variance (stated results from Week 8, where \(X\) is a sum of \(n\) Bernoulli pieces) are

\[ E[X] = np, \qquad \mathrm{Var}(X) = np(1-p). \]

The assumptions that define binomial are worth memorizing, because they are exactly what you check when recognizing it: (1) a fixed number of trials \(n\); (2) each trial has two outcomes; (3) the same \(p\) on every trial; (4) trials are independent. Break any one and you are likely in a different model.

Geometric: how long until the first success

Keep the independent Bernoulli\((p)\) trials, but change the question. Instead of fixing \(n\) and counting successes, keep going until the first success and count the trials. That count is \(X \sim \text{Geometric}(p)\). To have the first success on trial \(x\), the first \(x - 1\) trials must all be failures and trial \(x\) a success:

\[ p(x) = (1-p)^{\,x-1}\, p, \qquad x = 1, 2, 3, \dots . \]

The support is unbounded — in principle you could wait arbitrarily long — but the probabilities shrink geometrically. The mean is the clean result

\[ E[X] = \frac{1}{p}. \]

The intuition is friendly: if a success happens one trial in five (\(p = 0.2\)), you wait about five trials on average. The difference between binomial and geometric is what is fixed: binomial fixes the number of trials and lets the success count vary; geometric fixes “first success” and lets the number of trials vary.

Poisson: counts of events at a steady rate

The last model drops the “trials” framing entirely. A Poisson\((\lambda)\) random variable counts events in a fixed window — an hour, a page, a square meter — when events arrive at a constant average rate \(\lambda\) and do not influence one another. Its pmf is

\[ p(k) = \frac{e^{-\lambda}\, \lambda^{k}}{k!}, \qquad k = 0, 1, 2, \dots . \]

Its defining feature is that the mean equals the variance:

\[ E[N] = \lambda, \qquad \mathrm{Var}(N) = \lambda. \]

One useful way to see where Poisson comes from: imagine chopping the hour into many tiny slots, each either holding an event or not — a binomial with huge \(n\) and tiny \(p\), but with \(np = \lambda\) held fixed. In that limit the binomial pmf becomes the Poisson pmf. That is why Poisson is the natural model for “rare events over many opportunities”: arrivals, accidents, typos, decays. You do not need a fixed \(n\); you need a rate.

Recognizing the model: a short checklist

When a scenario lands in front of you, ask in order:

Is there a fixed number of yes/no trials, and do you count the successes? → binomial (with \(n = 1\), a single Bernoulli).
Are you waiting for the first success and counting the trials it takes? → geometric.
Are you counting events over a span of time or space at a steady rate, with no fixed number of trials? → Poisson.

The same questions also tell you when no standard model fits cleanly: if \(p\) changes from trial to trial, or trials are dependent (drawing without replacement from a small population), or the rate is not constant, you are outside these models and should say so rather than force a fit. Recognizing the limit of a model is part of using it honestly.

Worked examples

Worked example — the guessing quiz, as a binomial (recurring slice)

Maya takes the 10-question true/false quiz from Weeks 6–8 by pure guessing. Let \(X\) be the number she gets correct. Each question is an independent yes/no trial with the same success probability \(p = 0.5\), and there are a fixed \(n = 10\) of them — the four binomial assumptions hold exactly — so

\[ X \sim \text{Binomial}(10,\ 0.5), \qquad p(x) = \binom{10}{x} (0.5)^{x} (0.5)^{10 - x} = \binom{10}{x} (0.5)^{10}. \]

Because \(p = 0.5\), every term shares the factor \((0.5)^{10} = 1/1024\), so a probability is just a count of favorable orderings over \(1024\). Suppose we want \(P(X \ge 8)\) — guessing eight or more correctly. Add the top three terms:

\[ P(X \ge 8) = \big[\tbinom{10}{8} + \tbinom{10}{9} + \tbinom{10}{10}\big]\,(0.5)^{10} = \frac{45 + 10 + 1}{1024} = \frac{56}{1024} \approx 0.0547. \]

So a pure guesser scores 8-or-better only about 5.5% of the time. As a check, \(E[X] = np = 5\) and \(\mathrm{Var}(X) = np(1-p) = 2.5\) from Week 8, so a score of \(8\) sits about two standard deviations (\(\sqrt{2.5} \approx 1.58\)) above the mean — consistent with it being uncommon. (Synthetic scenario; numbers fixed for the course.)

Worked example — shuttle arrivals, as a Poisson (recurring slice)

Shuttles arrive at Maya’s stop at an average rate of \(\lambda = 4\) per hour — one every fifteen minutes on average — and arrivals do not coordinate with one another. Let \(N\) be the number that arrive in a given hour. There is no fixed number of trials here; we are counting events over a fixed span at a steady rate, so this is Poisson:

\[ N \sim \text{Poisson}(4), \qquad p(k) = \frac{e^{-4}\, 4^{k}}{k!}. \]

What is the probability that exactly four arrive in the hour — the rate’s “typical” value?

\[ P(N = 4) = \frac{e^{-4}\, 4^{4}}{4!} = \frac{e^{-4}\,(256)}{24} \approx 0.195. \]

About a 20% chance of exactly four. Note this is well below \(1\) even though \(4\) is the mean: the Poisson spreads its probability across \(0, 1, 2, \dots\), and “exactly the average” is just one of many plausible counts. Here \(E[N] = \mathrm{Var}(N) = 4\), the Poisson signature. (Synthetic scenario; numbers fixed for the course.)

Worked example — defective items in a batch, as a binomial (transfer)

Now a new context. A supplier ships components in batches, and each component is independently defective with probability \(p = 0.05\). An inspector pulls a fixed sample of \(n = 20\) from a batch and counts the defectives, \(D\). Two outcomes per item (defective / not), a fixed sample size, the same \(p\) each time, and independence — this is binomial:

\[ D \sim \text{Binomial}(20,\ 0.05), \qquad p(x) = \binom{20}{x} (0.05)^{x} (0.95)^{20 - x}. \]

The probability the sample is clean (no defectives) is the \(x = 0\) term:

\[ P(D = 0) = \binom{20}{0} (0.05)^{0} (0.95)^{20} = (0.95)^{20} \approx 0.358. \]

So even at a 5% defect rate, a clean sample of 20 happens only about a third of the time; defects are common enough that you would expect to see one. The expected number is \(E[D] = np = 20(0.05) = 1\), which matches the intuition that a clean draw should be the minority outcome. (Synthetic scenario; numbers fixed for the course.) The same setup — independent yes/no items at a fixed sample size — is why “defectives in a batch” and “correct answers on a guessed quiz” are the same model wearing different clothes.

A common mistake

The most frequent error this week is forcing the binomial onto a sampling-without-replacement problem. If you draw 5 cards from a 52-card deck and count hearts, the trials are not independent and \(p\) changes after each draw (one fewer heart, one fewer card), so \(\text{Binomial}\) is only an approximation — exactly, it is a different (hypergeometric) model. The binomial assumes the same \(p\) on every trial, which requires replacement or an effectively infinite pool. When the population is large relative to the sample, the binomial is a fine approximation; when it is small, it is not. Always check the “same \(p\), independent trials” assumption before reaching for \(\binom{n}{x} p^{x}(1-p)^{n-x}\).

A related slip: confusing binomial and geometric. Both use independent Bernoulli\((p)\) trials, but binomial fixes the number of trials and counts successes, while geometric fixes the first success and counts trials. If the question is “how many out of 10?” it is binomial; if it is “how many until the first?” it is geometric.

Low-stakes self-checks (ungraded)

These are for your own practice — ungraded, no submission, just a way to test whether the recognition skill is sticking. Try them before peeking at the reasoning.

A basketball player makes free throws independently with probability \(0.8\). You watch her shoot until she misses for the first time and count the shots. Which model is the count of shots, and what is its mean? (Geometric of the miss, \(p = 0.2\); mean \(1/0.2 = 5\).)
In our shuttle-arrivals model \(N \sim \text{Poisson}(4)\), is \(P(N = 0)\) bigger or smaller than \(P(N = 4)\), without a calculator? (Smaller — \(P(N=0) = e^{-4} \approx 0.018\), far below the near-the-mean count.)
A 12-item multiple-choice quiz has 4 options each; you guess every one. What model and parameters describe the number correct, and what is its mean? (Binomial\((12,\ 0.25)\); mean \(12 \times 0.25 = 3\).)
You inspect items one at a time off a line where each is defective with probability \(0.05\), stopping at the first defective. Binomial or geometric — and why? (Geometric: you are counting trials until the first success, not successes in a fixed number.)

You can check all of these by simulation in the companion lab, Lab 9 — Simulating discrete models, where you draw many samples and compare the empirical frequencies to these pmfs.

Reading and source pointer

This week tracks Grinstead & Snell, Chapter 5 — Important Distributions, specifically the discrete distributions (Bernoulli, binomial, geometric, and Poisson), for the catalogue of named models and the parameterizations we use. The recognition checklist and the “which assumption picks which model” framing are supported by the review of discrete distributions in MIT OCW 18.05. Our parameterizations follow the course ledger: \(\text{Binomial}(n, p)\); \(\text{Geometric}(p)\) as trials-to-first-success with support \(\{1, 2, \dots\}\) and mean \(1/p\); \(\text{Poisson}(\lambda)\) with mean and variance both \(\lambda\). These notes are the course’s own synthesis, grounded in but not copied from the sources. All scenario data are synthetic, with seeds set in any simulation.

Public vs. graded

These notes, the examples, and the practice here are public and ungraded — study material only. No graded prompts, answer keys, rubrics, point values, or due dates appear on this site. Graded checkpoints, quizzes, homework, labs, the midterm, the project, and the final live in Blackboard (the LMS), which is authoritative for due dates, submissions, and grades. If this page and Blackboard ever disagree, follow Blackboard.

Looking ahead

Every model this week was discrete — the random variable took separated values you could list and add. Next week we cross to continuous random variables, where the variable can take any value in a range and a single point has probability zero. The bridge is our shuttle: this week we counted how many shuttles arrive in an hour (Poisson); next week we ask how long until the next one arrives (a continuous wait), and probability becomes the area under a density rather than a sum of pmf bars. In Week 11 we return to “common models,” now continuous — the exponential wait and the normal commute time — mirroring this week’s catalogue.