For a Poisson count averaging 3 support tickets per hour, what is the probability of zero tickets in an hour?

## The distribution A Poisson random variable counts how many events occur in a fixed window — an hour, a day, a page of text — when events happen independently at a stable average rate. It has a single parameter λ (lambda), the **expected number of events in that window**, and its probability mass function is `P(X = k) = e^(-λ) · λ^k / k!` for k = 0, 1, 2, ... Here `k!` is the factorial (0! = 1) and e ≈ 2.71828. The support is all non-negative integers with no upper bound, which distinguishes it from a binomial count capped at n. ## Answering the question With λ = 3 tickets per hour and k = 0: `P(X = 0) = e^(-3) · 3^0 / 0! = e^(-3) · 1 / 1 = e^(-3) ≈ 0.0498`. So about a 5 percent chance of a silent hour, and P(at least one ticket) = 1 - e^(-3) ≈ 0.9502. The whole low end of the distribution is worth having at your fingertips: - P(0) ≈ 0.0498 - P(1) = 3e^(-3) ≈ 0.1494 - P(2) = e^(-3)·9/2 ≈ 0.2240 - P(3) = e^(-3)·27/6 ≈ 0.2240 - P(X ≤ 2) ≈ 0.4232 Notice P(2) = P(3): whenever λ is a whole number the distribution is flat-topped, with modes at λ - 1 and λ. ## λ scales with the window The single most-tested mechanic here is that λ is tied to an interval, and it scales **linearly** with that interval's length when the rate is constant. At 3 per hour: - 30 minutes → λ = 1.5, P(0) = e^(-1.5) ≈ 0.223 - 20 minutes → λ = 1, P(at least one) = 1 - e^(-1) ≈ 0.632 - 8 hours → λ = 24, P(0) = e^(-24), effectively zero A candidate who plugs λ = 3 into a half-hour question has made the classic error. Always restate the rate in the units of the window you are asked about. ## Additivity If two independent Poisson counts have rates λ₁ and λ₂, their sum is Poisson with rate λ₁ + λ₂. Email tickets at 3 per hour plus chat tickets at 5 per hour, if independent, give a combined Poisson(8) count. This makes Poisson models convenient to aggregate across channels, regions or time blocks — but only under independence. ## The assumptions The Poisson model is the right description when, in the window of interest: 1. events occur **independently** — one ticket neither invites nor suppresses the next; 2. the average rate is **constant** across the window; 3. events do not occur **simultaneously** — in a short enough slice the chance of two is negligible. Real support traffic often breaks all three. An outage produces a burst of correlated tickets. Rates vary sharply by hour of day and by weekday. A batch integration can file several tickets at the same instant. Each violation produces counts that are more dispersed than the model expects, so Poisson tail probabilities computed from a single λ will be too optimistic: a "1-in-500 hour" under the model may occur weekly in reality. ## Practical framing When you use this for capacity planning, the useful outputs are not point probabilities but tail questions: P(X > c) for a staffing threshold c. With λ = 3, P(X ≥ 8) is well under 1 percent, so a desk sized for eight concurrent tickets is comfortable in a *typical* hour. The judgment call is that a single λ fitted to an all-day average will badly understate the peak hour; fitting separate rates by time block, or planning against the busiest block's λ, is the honest approach. ## Common errors Treating λ as a probability (it is a count, and can exceed 1), forgetting to rescale λ to the asked-about window, saying P(0) must be zero because "some tickets always arrive", or writing P(0) as λe^(-λ), which is actually P(1). Also remember λ need not be an integer — 1.5 tickets per half hour is a perfectly valid rate even though you cannot observe 1.5 tickets.