Binomial Distribution Table

The binomial distribution table gives cumulative probabilities P(X ≤ x) — the chance of at most x successes — for n = 5, 10, 15 and 20 trials across the standard success probabilities, with the subtraction tricks that turn cumulative values into any answer.

n = 10 trials

n = 20 trials

n = 5 trials

n = 15 trials

How to use a cumulative table

Read the cell at row x, column p: it is P(X ≤ x) for n trials with success probability p. Three moves extract everything else. Exactly x: subtract the cell above: P(X = 3) for n = 10, p = 0.5 is P(X ≤ 3) − P(X ≤ 2) = 0.1719 − 0.0547 = 0.1172. At least x: flip to the complement: P(X ≥ 8) for n = 10, p = 0.5 is 1 − P(X ≤ 7) = 1 − 0.9453 = 0.0547 — equal to P(X ≤ 2) by the symmetry of p = 0.5, a coincidence worth noticing because it checks your work. Between a and b: subtract two cumulatives: P(3 ≤ X ≤ 6) = P(X ≤ 6) − P(X ≤ 2) = 0.8281 − 0.0547 = 0.7734. Every textbook exercise is one of these three in costume.

When the binomial model applies

Four conditions, all checkable: a fixed number of trials n; each trial binary (success/failure); a constant success probability p; and independence between trials. Quality sampling (defective or not, from a large stable process), conversion counts, multiple-choice guessing and A/B test sign-ups all fit. Sampling without replacement from a small population breaks independence — that territory belongs to the hypergeometric distribution — and varying p breaks the model entirely. Note the family resemblance to the rest of this site: for large n, the binomial is approximated by a normal curve with mean np and standard deviation √(np(1−p)), which is why the z table can answer approximate binomial questions (with a continuity correction) when tables run out — the usual rule asks for np and n(1−p) both at least 10.

The "at least one" problem, worked

The commonest binomial question in applied work is whether something happens at all. A component fails with probability 0.05 per unit; a batch holds 20 units. P(at least one failure) = 1 − P(none) = 1 − P(X = 0). The n = 20 table's p = 0.05 column gives P(X ≤ 0) = 0.3585 directly — that cell is P(X = 0) — so the answer is 1 − 0.3585 = 0.6415: nearly a two-in-three chance that a batch contains a failure, even though each unit is 95% reliable. The complement move (1 − the zero cell) is the same subtraction as the right-tail move on the z table, and it is why cumulative tables are built the way they are: one cell answers the question everyone actually asks.

Values computed from the binomial distribution's cumulative function · Verified November 2026

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