A binomial distribution with many trials and a small chance on each, such as defects in a day's production, is tedious to compute exactly. The Poisson distribution is its limit, and a very good approximation.
Key idea
If $n$ is large, $p$ is small and $\lambda = np$ is moderate, then $$\binom{n}{k}p^{k}(1 - p)^{n - k} \approx e^{-\lambda}\frac{\lambda^{k}}{k!}.$$
Why it works
Put $p = \frac{\lambda}{n}$ and let $n$ grow with $k$ fixed: $$\binom{n}{k}\left(\frac{\lambda}{n}\right)^{k}\left(1 - \frac{\lambda}{n}\right)^{n - k} = \frac{\lambda^{k}}{k!} \cdot \frac{n(n - 1)\cdots(n - k + 1)}{n^{k}} \cdot \left(1 - \frac{\lambda}{n}\right)^{n}\left(1 - \frac{\lambda}{n}\right)^{-k}.$$ The second factor tends to 1, the third to $e^{-\lambda}$ (the limit from calculus), and the last to 1. What is left is the Poisson pmf. This is also why Poisson counts arise: split time into many tiny slots, each with a tiny chance of an event.
The slot picture is worth keeping. A window holds a huge number of tiny moments, each with a tiny, independent chance of an event; the binomial count over those moments is essentially Poisson. The same picture, run through time, becomes the Poisson process later in this course.
Worked examples
Worked example: Defective chips
A factory makes 500 chips a day, each defective with probability $0.004$, independently. Approximate the probability of no defective chips.
- $\lambda = np = 500 \times 0.004 = 2$.
- $P(X = 0) \approx e^{-2} \approx 0.1353$.
- The exact binomial value is $0.996^{500} \approx 0.1348$.
Answer: about $0.135$
Worked example: At least two
With the same chips, approximate $P(X \ge 2)$.
- $$P(X \ge 2) = 1 - P(X = 0) - P(X = 1) \approx 1 - e^{-2}(1 + 2)$$
- $\approx 1 - 0.406 = 0.594$ (exact: $0.5945$).
Answer: about $0.594$
Worked example: When it fails
Compare binomial $n = 20$, $p = 0.5$ with Poisson $\lambda = 10$ at $k = 10$.
- Binomial: $\binom{20}{10}(0.5)^{20} \approx 0.176$.
- Poisson: $e^{-10}\frac{10^{10}}{10!} \approx 0.125$.
- The binomial's variance is $np(1 - p) = 5$, while the Poisson's is 10: the shapes differ.
Answer: Far apart: $p = 0.5$ is not small.
A common rule of thumb: the approximation is good when $n \ge 100$ and $np \le 10$, or more loosely when $p \le 0.05$. The error is roughly of size $p$, so the smaller $p$ is, the better. For quick work, $P(\text{no events}) = (1 - p)^{n} \approx e^{-np}$ is the most used case.
Common mistake
Using the approximation when $p$ is not small. The Poisson has variance equal to its mean; a binomial has variance $np(1 - p)$, which is close to $np$ only when $p$ is near 0. Check: is $p$ below about $0.05$ with $n$ in the hundreds? If not, compute the binomial exactly or use a normal approximation.