Binomial Distribution Calculator

P(X = k) = C(n, k) · p^k · (1 − p)^(n − k) where C(n, k) = n! / (k! (n − k)!)

n

The number of trials, up to 1000 — independent repetitions of the same yes-or-no event. The limit is a performance bound rather than a mathematical one: the binomial is perfectly well defined for far larger n, but the counts along this page grow with n and a page that hangs is worse than one that declines

k

The number of successes you are asking about. It cannot exceed n, since more successes than trials is not an unlikely outcome but an impossible request

p

The success probability of a single trial, entered as a percentage from 0 to 100. It is the same for every trial, which is what makes the trials exchangeable and the count binomial

C(n, k)

The binomial coefficient: how many ways k successes can be arranged among n trials. It is the reason the middle counts are the likely ones — a count near the middle can be arranged in far more ways than an extreme one

p^k · (1 − p)^(n − k)

The chance of one particular arrangement of k successes and n − k failures. Multiplying by the number of arrangements turns "one arrangement" into "any arrangement"

np

The mean number of successes. The variance is np(1 − p) and the standard deviation is its square root, which is what the last three rows of the panel report

Use it when trials are counted rather than measured: how many of 20 components pass inspection, how many of 12 free throws go in, how likely a run of at least 8 heads is in 10 flips. The two cumulative rows are the ones to reach for in practice, because a target is nearly always a threshold — at least k for passing a quality gate, at most k for staying inside a budget — and reading a threshold off the exact-count row means adding up counts by hand. Pick a different tool when the number of trials is not fixed in advance (that is a different distribution), when the success probability changes from trial to trial, or when the trials influence one another; the third case is the one that most often breaks the assumption in practice.

1. Ten trials, three successes, a 50% chance each time

   1. There are C(10, 3) = 120 ways to place three successes among ten trials
   2. One arrangement has probability 0.5³ × 0.5⁷ = 1/1024
   3. 120 / 1024 = 0.1171875, so exactly three successes happens 11.72% of the time
   4. Adding up the counts from 0 to 3 gives 17.19% at most; the mean is np = 10 × 0.5 = 5

   The gap between the two cumulative rows is the point of this example: 17.19% at most and 94.53% at least leave 11.72% between them, and that missing slice is exactly the probability of three — the count you asked about. It is the arithmetic that catches the common mistake of reading at least as one minus at most, which would say 82.81% here. The mean of 5 with a standard deviation of 1.58 also tells you that 3 is unremarkable for this setup: it sits a little over one standard deviation below the average.

2. Twenty trials with a 25% chance each time

   1. The mean is np = 20 × 0.25 = 5 successes
   2. The variance is np(1 − p) = 20 × 0.25 × 0.75 = 3.75, and its square root is about 1.94
   3. Exactly four successes has probability 18.97%, the single most likely count here
   4. At most four is 41.48%, and at least four is 77.48%

   Four successes is both the mean-adjacent count and the single most likely one, which is why the two cumulative figures sit either side of it so asymmetrically: 41.48% of the distribution is at or below the most likely count. That asymmetry is general and worth remembering — for any count at or below the mean, the chance of being at most that count is larger than intuition suggests, because the distribution has a long upper tail when p is small. The summary rows are what let you see this without adding up terms: a standard deviation of 1.94 around a mean of 5 puts 4 well inside the ordinary range.

3. One hundred trials, no successes, a 2% chance each time

   1. With k = 0 there is exactly one arrangement — no successes at all — so C(100, 0) = 1
   2. Its probability is 0.98¹⁰⁰ ≈ 0.1326, so getting none is 13.26%
   3. At most zero successes is the same event, so that row matches it exactly
   4. "At least zero successes" includes every possible outcome, so it is 100%

   This is the case that pins down the at-least row. At least zero successes is not an uninformative edge but a certainty — every outcome has zero or more successes — so 100% is the correct answer and not a placeholder. Read the three rows together and the off-by-one becomes visible: at most 13.26% plus at least 100% exceeds 100% because the count of exactly zero is inside both, and the amount they overlap is that 13.26%. The other half of the example is the mean: two expected successes out of a hundred sounds safe, yet getting none at all is still about one run in eight.

4. A thousand trials with a 50% chance each time

   1. The mean is 1000 × 0.5 = 500 successes and the variance is 250, so the standard deviation is about 15.81
   2. Exactly 500 successes — the most likely single count — has probability 2.52%
   3. At most 500 is 51.26%, and at least 500 is 51.26% as well, because 500 is the centre
   4. The two cumulative figures are equal here, which only happens at the mean

   Two things are worth taking from this one. The first is that the single most likely count is not a likely event: 500 is the peak of the distribution and it still only happens about one run in forty, because there are a thousand different counts to spread the probability over. The second is that the two cumulative rows are equal at exactly 51.26% — a useful self-check, since at the mean the distribution is symmetric and any deviation from equality means the count is off-centre. Note also that a 2.52% probability here is being computed from binomial coefficients with hundreds of digits, which is why the tool caps the trial count rather than trying to be brave about it.

The binomial model rests on three assumptions, and when one of them fails the numbers stay plausible while being wrong. The number of trials must be fixed in advance rather than decided as you go, the success probability must be the same for every trial, and the trials must be independent of one another — and the third is the one that breaks most often in practice, because trials drawn from the same batch, the same person or the same day tend to move together. When successes cluster, the real spread is wider than this page reports. Two further limits are worth stating. The trial count is capped at 1000, which is a performance bound rather than a mathematical one: the model is defined far beyond it, but the counts grow with the trial count and a page that hangs helps nobody. And there is no table of probabilities by count on this page: the table people want is one row per possible number of successes, which depends on the trials and the probability you enter, and a reference table here cannot see those inputs — the three rows on the panel are the shape of that table for the numbers you chose.

Why is at least k not simply 100% minus at most k?

Because the two events are not complements. At most k means zero through k successes; at least k means k through n. The count of exactly k sits in both, so subtracting one from 100% leaves out that shared slice — which is the very probability you asked about. Use the at most k − 1 row if you want the complement of at least k, or read the three rows together: exactly k is always the gap between the other two, and that gap is a useful check on all three.

Can the success probability be 0% or 100%?

Yes, and both are meaningful rather than edge cases. A 0% chance means the event never happens, so the count of successes is 0 with certainty; a 100% chance means it always does, so the count is n with certainty. The panel reports the resulting 0% and 100% rows plainly. If you are looking at a flat 0% for a count you expected to be small but possible, check the probability field — a 0 entered where 10 was meant produces exactly that.

Why is there no table showing the probability of every number of successes?

Because such a table depends on the trials and the success probability you enter, and a reference table on this page cannot see those inputs — it is computed from constants, so it would show some other setup's numbers under a heading that looks like yours. Instead the three rows on the result panel give the shape of that table at the count you asked about: the exact count, everything below it, everything above it. To see the whole distribution, change the count of successes and read the exact row again — those values are the table, one row at a time.

What does the mean tell me that the probability does not?

It tells you where the distribution sits, which is what makes a count interpretable. A mean of 5 successes with a standard deviation of 1.58 makes 3 an ordinary result, while the same count of 3 against a mean of 12 would be a long way out. The mean is np, the variance is np(1 − p), and the standard deviation is the square root of that — the three summary rows are the distribution described by its centre and its spread rather than by one probability at a time.

What happens if I ask for more successes than trials?

It is rejected rather than answered with 0%. More successes than trials is not an unlikely outcome — it is not an outcome at all, and reporting 0% would suggest the model had considered it and ruled it out. The panel reports an error, and the same applies to a trial count above 1000, which is refused as a performance limit rather than because the mathematics stops working there.

Do the trials really have to be independent?

They do, and this is the assumption that most often fails in real data. If successes cluster — components from the same production batch, shots by the same player on the same day, patients from the same clinic — then the trials carry information about each other, and the true spread of the count is wider than the binomial says. The mean stays roughly right in the presence of clustering, while the standard deviation and the tail probabilities do not, so the three summary rows are the ones to distrust first. A model with extra variation built in is the right tool when you know the trials cluster.

• 1.3.6.6.18. Binomial Distribution — e-Handbook of Statistical Methods (the binomial probability function, its mean np and variance np(1 − p), and the assumptions the model rests on) — National Institute of Standards and Technology (NIST)
• 4.3 Binomial Distribution — Introductory Statistics 2e (the fixed number of independent trials with a constant success probability, and the cumulative form used by the at most and at least rows) — OpenStax, Rice University
• 1.3.6.1. What is a Probability Distribution — e-Handbook of Statistical Methods (what it means for a count to have a distribution, and how probabilities are read off the outcomes it assigns them to) — National Institute of Standards and Technology (NIST)