Binomial probability calculator
Enter trials, successes and the chance of success, get the probabilities.
What is the binomial probability formula?
P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ: the number of ways to get k successes times the chance of any one such way. Exactly 5 heads in 10 coin flips is 252 × 0.5¹⁰ = 24.6%.
How do I find "at least" probabilities?
Add up the probabilities from k to n, or use 1 − P(fewer than k). At least one six in 4 rolls of a die is 1 − (5/6)⁴ = 51.8%. Mind the wording: “at least 5” includes 5, while “more than 5” starts at 6, so it equals P(X ≥ 6).
When does the binomial distribution apply?
When there is a fixed number of independent trials, each with the same chance of success — coin flips, free throws, defective parts in a batch. OpenStax lists the conditions: a fixed number of trials n, only two outcomes (success and failure), and independent trials run under identical conditions, so p stays the same on every trial.
What are the mean and standard deviation?
Mean = n × p and SD = √(n × p × (1 − p)): for 10 coin flips, 5 heads on average with an SD of 1.58.
Worked example: workers with a high school diploma
OpenStax (Introductory Statistics 2e, section 4.3) notes that about 41% of adult workers have a high school diploma but no further education. Pick 20 workers at random, so X is binomial with n = 20 and p = 0.41; enter 20, 12 and 41%. The expected count is 20 × 0.41 = 8.2 workers, with an SD of √(20 × 0.41 × 0.59) = 2.20. The chance that at most 12 of them fit is P(X ≤ 12) = 0.9738, so more than 12 has probability 0.0262. Exactly 12 has 0.0417, and at least 12 has 0.0679.
How is it related to the Poisson distribution?
For a large number of trials with a small chance of success, the Poisson distribution with mean np is a good substitute (OpenStax section 4.6). With n = 200 and p = 0.0102, the chance of exactly 2 successes is 0.2719 by the binomial formula and 0.2706 by the Poisson with mean 2.04.
Worked example: defects in a batch
A process makes 2% defective parts, and you inspect a sample of 25 (n = 25, p = 2%). The chance that none is defective is 0.98²⁵ = 60.3%, so the chance of at least one defect is 1 − 0.6035 = 39.7%. The expected number of defects is 0.5, with an SD of 0.70, and the chance of 3 or more is 1.3%. A sample can look clean even when the defect rate is not zero: with only 25 parts, finding no defects is the most likely single outcome.
What is the most likely number of successes?
The most probable value of X, the mode, is the whole number part of (n + 1) × p, or two neighboring values when that product is itself a whole number. For the 20 workers with p = 0.41 it is ⌊21 × 0.41⌋ = 8, which has a probability of 0.1790; computing every P(X = k) from the formula confirms that the largest is at 8. The mode and the mean (here 8.2) are close but not equal, so quote the one you mean.
Common mistakes with binomial probabilities
First, using the binomial when the trials are not independent. Drawing 5 cards from a deck, or 10 items from a small batch without putting them back, changes the odds after each draw; the hypergeometric distribution is the right model there. Second, typing the probability as a decimal: this calculator takes it as a percentage, so 0.41 would mean 0.41%, and 41 means 41%. Third, mixing up “exactly”, “at most” and “at least”; the table lists all five versions so you can pick the right one. Fourth, forgetting that p must be the same for every trial; a shooter who tires or a machine that drifts is not binomial.
Sources: OpenStax, Introductory Statistics 2e, section 4.3 (Binomial Distribution) and section 4.6 (Poisson Distribution, binomial approximation). Related: hypergeometric calculator, Poisson distribution calculator, geometric distribution calculator, negative binomial calculator, permutation and combination calculator and normal distribution calculator.