Poisson Distribution

The binomial distribution needs a fixed trial count n and a success probability p. The Poisson distribution is its limiting case for rare events: n very large, p very small, product λ = n·p moderate; as a rule of thumb n ≥ 50 and p ≤ 0.1. Then P(X=k) ≈ λᵏ·e^(−λ)/k!, and only ONE parameter λ is needed instead of two. Poisson is moreover the natural model when there is no fixed n at all, but events arrive randomly in time or space: calls per minute, decays per second, misprints per page. Example: n = 1000 tickets, p = 0.002: exact binomial P(X=2) ≈ 27.07 %, with λ = 2 the Poisson approximation 2²·e⁻²/2 ≈ 27.07 %, practically identical.

λ is the mean number of events per considered interval, i.e. the expected value: E(X) = λ. With "on average 2 calls per minute", λ = 2 if you look at one minute. The special feature of the Poisson distribution: the variance is also λ, so Var(X) = λ and σ = √λ. You can even use this as a diagnostic: if mean and variance of real count data lie far apart, the Poisson model does not fit. λ need not be an integer (λ = 1.5 is perfectly normal), because it is an average, not a count. The most probable value (mode) lies at ⌊λ⌋; for λ = 2 the values 1 and 2 are equally most likely (each ≈ 27.1 %).

Via the complement, since summing "at least one" directly would mean adding infinitely many terms. The opposite of "at least one event" is "no event", and for that there is the simplest Poisson formula of all: P(X=0) = λ⁰·e^(−λ)/0! = e^(−λ). Hence P(X ≥ 1) = 1 − e^(−λ). Example: if on average 1.5 customers arrive per minute, P(X ≥ 1) = 1 − e^(−1.5) = 1 − 0.2231 ≈ 77.7 %. The same pattern works for "at least two": P(X ≥ 2) = 1 − P(0) − P(1) = 1 − e^(−λ)·(1 + λ), here 1 − 0.2231·2.5 ≈ 44.2 %. Complement thinking is almost always the fastest route with Poisson.

λ scales linearly with the interval length: if λ = 2 per minute, then λ = 10 for 5 minutes and λ = 1 for 30 seconds. Only after this adjustment may you use the formula. Example: on average 2 calls per minute; probability of NO call in 5 minutes: not e⁻², but P(X=0) = e^(−10) ≈ 0.0000454, i.e. practically impossible. The difference is enormous and precisely why it is a popular exam mistake. The same scaling holds spatially: 0.3 errors per page mean λ = 3 over 10 pages. The prerequisite is that the rate stays constant and events arrive independently; at peak times with changing rate the simple model breaks down.

Signal phrases are "on average ... per ..." plus a question about a specific count: so many calls per hour on average, accidents per month, errors per page. Recipe: first read off λ and, if necessary, scale it to the requested interval. Second clarify what is asked: exactly k, at most k or at least k. Third substitute: P(X=k) = λᵏ·e^(−λ)/k!, working with the complement for "at least". Complete example: hotline with on average 2 calls per minute, wanted P(exactly 3 in one minute): P(X=3) = 2³·e⁻²/3! = 8·0.1353/6 ≈ 0.180, i.e. 18 %. Finally check plausibility: k near λ should be relatively likely, k far above it rare.