For a fixed number of independent trials, each with the same chance of success, the binomial distribution gives the probability of any particular number of successes. This computes exactly, at least, and at most k successes at once.
How it works
Binomial probability
P(X = k) = C(n, k) × p^k × (1−p)^(n−k)
n is the number of trials, k is the number of successes, p is the success probability per trial, and C(n, k) is the number of ways to choose which k of the n trials succeed.
Three conditions this needs, and where they quietly fail
A fixed number of trials, the same success probability on every trial, and independence between trials. A coin flipped ten times satisfies all three cleanly. Ten cards drawn from a deck without replacement do not — each draw changes what remains for the next one, breaking both the “same probability” and “independence” conditions at once. Drawing without replacement is one of the most common ways this assumption silently fails in a problem that otherwise looks exactly like a textbook binomial case.
“Exactly” is not the same question as “at least” or “at most”
Most real questions are not actually “exactly k” — they are “k or more” (at least five heads in ten flips) or “k or fewer.” Answering those needs the exact probabilities for every relevant outcome summed together, which this calculator does automatically across all three framings rather than leaving several separate lookups to be added by hand.
How to use this calculator
- Enter the number of trials and the success probability per trial.
- Enter the number of successes you are interested in.
- Read exactly, at least, and at most that many successes.
Frequently asked questions
Why do coin flip probabilities work so cleanly with this formula?
Because a fair coin flipped repeatedly satisfies all three conditions exactly — fixed number of flips, identical 50% probability each time, and each flip is genuinely independent of the others. It is the textbook example precisely because nothing about it breaks the assumptions.
What if my trials are not actually independent?
This formula gives a wrong answer. Drawing cards without replacement, or any situation where an earlier outcome changes a later probability, needs a different distribution (the hypergeometric distribution, for card-drawing specifically) rather than the binomial one.
Why does P(exactly k) plus every other possible k sum to 100%?
Because those outcomes are exhaustive and mutually exclusive — exactly one of “0 successes,” “1 success,” … “n successes” must occur, so their probabilities must sum to certainty.
How does this relate to the “at least” and “at most” figures?
“At least k” sums the exact probabilities from k up to n. “At most k” sums from 0 up to k. Both include the “exactly k” case once, which is why “at least k” plus “at most (k−1)” together sum to exactly 100%.
Can k be larger than n?
No — you cannot have more successes than there are trials, so this calculator declines that input rather than returning a meaningless negative or undefined probability.
