Module 04 · Probability distributions
Use a binomial model for successes in a fixed number of trials and a Poisson model for events in an interval. Translate the question into exactly, at most, or at least.
A random variable assigns a number to an uncertain outcome. Let X be the number of customers who accept an offer among the next ten contacted customers. Before observing the responses, X can take several values. Afterward, x = 3 might be the single value actually observed.
A discrete probability mass function gives P(X = x) at each possible count. Its probabilities are nonnegative and sum to one. The cumulative distribution function F(k) = P(X ≤ k) adds the probabilities through k. “Exactly two,” “at most two,” and “at least two” describe different events.
The expected value E[X] is a probability-weighted average over the possible outcomes, not a promised result for one trial. An expected count of 2.5 is meaningful even though no single observed count is 2.5. Variance describes dispersion around that expectation.
Choose a model, set its parameters, and shade the event you want to evaluate.
Green bars belong to the selected event. Binomial support is finite. The Poisson display ends in the upper tail; its model has no finite maximum.
The event is X ≤ 2. Its probability is a sum of masses at the selected integer counts. The expected count is a long-run average, not a guaranteed outcome.
For the binomial model, P(X ≤ 2) = 0.677800. Expected count 2.000, variance 1.600, standard deviation 1.265. Green bars identify the selected event.
The binomial model counts successes in n independent trials, each with two outcomes and the same success probability p. “Success” is simply the event being counted; it can be an unwanted outcome such as a defective item. The count ranges from 0 through n.
The probability of exactly k successes is P(X = k) = C(n,k)pᵏ(1 − p)ⁿ⁻ᵏ. The combination C(n,k) counts the different placements of the k successes. The expected count is np, and the variance is np(1 − p).
For an illustrative campaign with n = 10 independent contacts and p = 0.20, the expected number accepting is 2. The probability of exactly two acceptances is 45 × 0.20² × 0.80⁸ ≈ 0.3020. The probability of at most two is P(0) + P(1) + P(2) ≈ 0.6778. A count equal to its expectation is not certain.
A Poisson model counts events in a specified exposure interval, such as arrivals during ten minutes or defects along a fixed length of material. Its parameter λ is the expected count for that interval. The possible counts are 0, 1, 2, and so on, with no finite upper bound.
The mass function is P(X = k) = e⁻λ λᵏ / k!. The mean and variance both equal λ. A homogeneous Poisson-process interpretation assumes a constant event rate, independent counts in disjoint intervals, and no meaningful simultaneous batches.
If a service desk receives an illustrative average of four arrivals per ten minutes, then λ = 4 for a ten-minute interval. The probability of no arrivals is e⁻⁴ ≈ 0.0183, and the probability of at most two is e⁻⁴(1 + 4 + 4²/2) ≈ 0.2381. Under a constant rate, a twenty-minute interval has λ = 8; the time unit cannot be omitted.
Ten fixed calls with an accept-or-decline result naturally suggest a binomial model. The number of calls arriving during an hour may suggest a Poisson model. These choices still require checks: customer responses can be dependent, success probabilities may differ, and arrival rates may vary during the day.
Sampling a substantial fraction of a finite batch without replacement creates dependence; the binomial model may then be inappropriate. Arrivals in groups or a changing rate can produce more variability than a Poisson model allows. A plausible-looking plot is not a substitute for checking the data-generating process.
For integer k, P(X ≥ k) = 1 − P(X ≤ k − 1). The shift by one matters because the count k belongs in the event “at least k.” The explorer sums the selected probabilities directly and highlights those counts. Poisson support extends beyond the displayed bars; the probability calculation includes the tail beyond the display.
Try each question before revealing the solution.
The complement is no successes. P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.80¹⁰ ≈ 0.8926. Subtracting P(X ≤ 1) would incorrectly remove the outcome X = 1, which belongs in “at least one.”
Twenty minutes is one third of an hour, so λ = 3 × 1/3 = 1. The no-arrival probability is e⁻¹ ≈ 0.3679. Using λ = 3 for this shorter interval would mix time units and answer a different question.
No. Removing an item changes the remaining defect proportion, so the trials are dependent and their conditional success probabilities change. A hypergeometric model is exact for a simple random sample from this fixed batch. The binomial model is not exact merely because the initial defect proportion is 0.20.
Further reading