Binomial Distribution Calculator
Exact in log space, so large n still works.
Work out Binomial Distribution. Exact in log space, so large n still works. Refuses out-of-range input instead of guessing.
Probability of exactly 5 successes
24.6094%
62.305% of getting that many or fewer · 62.305% that many or more
The binomial distribution counts successes across a fixed number of independent trials with a constant probability. Independence and constancy are the assumptions that fail first in practice — drawing without replacement changes p as you go, which is a hypergeometric problem rather than a binomial one. The probabilities are computed in log space, through the log-gamma function. Computing n-choose-k directly overflows a double at around n = 1030, which is well inside the range people actually ask about. The normal approximation flag follows the usual rule that np and n(1−p) should both be at least 5. Below that the distribution is too skewed for a symmetric bell to stand in for it, and the exact figures above are the ones to use.
How the Binomial Distribution Calculator works
Exact binomial probabilities — exactly k, at most k, at least k — with the mean, variance and a check on whether the normal approximation applies. Computed through log-gamma, so n in the thousands still returns a real answer.
Also known as: probability of exactly k successes · coin flip probability calculator · n choose k probability · at least 3 out of 10 chance
Frequently asked questions
When does the binomial distribution apply?
A fixed number of independent trials, each with the same probability of success, and you are counting successes. Independence and a constant probability are the assumptions that fail first in practice.
Why is drawing cards not binomial?
Because drawing without replacement changes the probability as you go. That is a hypergeometric problem. The binomial requires the probability to stay constant across every trial.
Why compute in log space?
Because n-choose-k overflows a double at around n = 1030, which is well inside the range people ask about. Working through log-gamma keeps n in the thousands perfectly workable.
When can I use the normal approximation?
The usual rule is that np and n(1−p) should both be at least 5. Below that the distribution is too skewed for a symmetric bell to stand in for it, and the exact figures are the ones to use.
What is the most likely number of successes?
The mode, which is the floor of (n+1)p. It is usually but not always the same as the rounded mean, and the two can differ by one.
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