Most word problems in this unit are solved the moment the right model is named. This lesson collects the hypergeometric, negative binomial and Poisson models with the binomial and geometric, and gives a short set of questions for telling them apart.
Idea clave
Ask in order: Is there a fixed number of draws without replacement from a finite group? Hypergeometric. A fixed number $n$ of independent trials with the same $p$, counting successes? Binomial. Trials until the first success? Geometric. Until the $r$th? Negative binomial. Events at random at a steady rate in time or space, with no fixed number of trials? Poisson.
Por qué funciona
Each model is the exact answer for its story, and each formula uses the story's features. The binomial's $\binom{n}{k}p^{k}(1 - p)^{n - k}$ needs independence and a constant $p$; without replacement those fail, and counting subsets replaces them. The waiting-time models fix the number of successes and let the number of trials vary; the binomial does the opposite. The Poisson is the limit of many tiny independent chances. Using a model whose story does not fit gives a precise-looking wrong answer.
| Model | Story | $P(X = k)$ | Mean | Variance |
|---|---|---|---|---|
| Binomial | $n$ independent trials, count successes | $\binom{n}{k}p^{k}(1 - p)^{n - k}$ | $np$ | $np(1 - p)$ |
| Geometric | trials until the first success | $(1 - p)^{k - 1}p$ | $\frac{1}{p}$ | $\frac{1 - p}{p^{2}}$ |
| Negative binomial | trials until the $r$th success | $\binom{k - 1}{r - 1}p^{r}(1 - p)^{k - r}$ | $\frac{r}{p}$ | $\frac{r(1 - p)}{p^{2}}$ |
| Hypergeometric | $n$ draws without replacement, $K$ of $N$ special | $\frac{\binom{K}{k}\binom{N - K}{n - k}}{\binom{N}{n}}$ | $\frac{nK}{N}$ | $n\frac{K}{N}\frac{N - K}{N}\frac{N - n}{N - 1}$ |
| Poisson | random events at a steady rate | $e^{-\lambda}\frac{\lambda^{k}}{k!}$ | $\lambda$ | $\lambda$ |
Worked examples
Ejemplo resuelto: A card hand
Five cards are dealt from a shuffled deck. $X$ is the number of hearts. Which model, and what is $E[X]$?
- A fixed number of draws without replacement from 52 cards, 13 of them hearts: hypergeometric.
- $E[X] = 5 \cdot \frac{13}{52} = \frac{5}{4}$.
Respuesta: Hypergeometric, mean $\frac{5}{4}$.
Ejemplo resuelto: Waiting for sixes
Roll a die until the third six. $X$ is the number of rolls. Which model, and what is $E[X]$?
- Independent trials, same $p = \frac{1}{6}$, until the 3rd success: negative binomial with $r = 3$.
- $E[X] = \frac{3}{1/6} = 18$.
Respuesta: Negative binomial, mean 18.
Ejemplo resuelto: Two counts that look alike
(a) The number of earthquakes above magnitude 5 in a region in a year. (b) The number of defective items in a sample of 10 from a shipment of 100,000. Which models?
- (a) Events at random times at a steady average rate, no fixed number of trials: Poisson.
- (b) Strictly hypergeometric, but 10 draws from 100,000 barely change the mix, so binomial with $n = 10$ is an excellent approximation.
Respuesta: Poisson; binomial (as an approximation to the hypergeometric).
Error común
Reaching for the binomial whenever the word "success" appears. Before using it, check all four conditions: a fixed number of trials, two outcomes, independence, and the same $p$ every time. A question that says "until" has a random number of trials; one that says "without replacement" from a small group breaks independence.