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Distribution

Binomial Distribution Calculator

Free binomial probability calculator. Compute P(X = k), P(X ≤ k), and P(X ≥ k) for n trials with success probability p — with the exact binomial PMF and CDF.

📊 n = 20, p = .30, k = 6 → P(X = k) = .1916
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P(X = k)
P(X ≤ k)
P(X ≥ k)
Mean = np
Variance = np(1−p)
SD
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Formula, reference table & FAQ

The binomial distribution models the number of successes X in n independent trials, each with the same probability of success p — a coin flipped n times, a fixed number of patients each responding or not to treatment, a fixed number of manufactured items each defective or not. It is fully specified by two parameters: n (number of trials) and p (probability of success on each trial). Unlike the normal, t, chi-square, and F distributions above, the binomial is discrete — X can only take whole-number values 0, 1, 2, …, n — so its distribution function is a probability mass function (PMF), not a density.

Binomial PMF
P(X=k)=(nk)pk(1p)nk,k=0,1,,n

Frequently Asked Questions

When can I approximate the binomial with the normal distribution?
A common rule of thumb requires both np ≥ 5 and n(1−p) ≥ 10 (or the stricter np ≥ 10 and n(1−p) ≥ 10) — when both hold, the binomial is well-approximated by a normal distribution with mean np and variance np(1−p), usually with a continuity correction of ±0.5. This calculator instead computes the exact binomial probability directly, so no approximation or rule-of-thumb check is needed.
What is the difference between P(X = k) and P(X ≤ k)?
P(X = k), the PMF, is the probability of exactly k successes. P(X ≤ k), the CDF, is the cumulative probability of k or fewer successes — the sum of the PMF over 0, 1, …, k. Hypothesis tests based on the binomial (e.g. the sign test, or an exact binomial test of a proportion) typically need the cumulative (or tail) probability, not the single-point PMF.
Does order matter in the binomial formula?
No — the binomial coefficient C(n,k) = n!/(k!(n−k)!) in the formula exists precisely to count every possible ORDER in which k successes could occur among n trials, so the final probability already accounts for all orderings. You do not need to (and should not) multiply the result by anything else to account for order.