Binomial Probability Calculator
The probability of getting exactly k successes in n independent trials, each with the same success probability — the standard model behind coin flips, pass/fail testing, and quality-control sampling.
Inputs
Result
20.5078%
P = 0.205078
How the binomial probability calculator works
P(exactly k successes) = C(n,k) × p^k × (1−p)^(n−k), where C(n,k) is the number of ways to choose which k of the n trials succeed.
This assumes every trial is independent and has the identical probability p of success — changing conditions between trials breaks the model.
Worked example: 10 coin flips, exactly 4 heads, p = 0.5
- C(10,4) = 10! ÷ (4!×6!) = 210.
- p^k = 0.5^4 = 0.0625, (1−p)^(n−k) = 0.5^6 = 0.015625.
- P = 210 × 0.0625 × 0.015625 ≈ 0.2051, or about 20.5%.
Common mistakes to avoid
Using this for 'at least k' instead of 'exactly k' questions
This formula gives the probability of precisely k successes. For 'at least k' or 'at most k', you need to sum this formula across every relevant value of k, not just look up one.
Applying it when trials aren't truly independent or identical
Drawing cards without replacement, for example, changes the probability with each draw — the binomial model assumes p stays constant, which sampling without replacement violates.
Frequently asked questions
What counts as a 'trial' in this formula?
Any repeatable event with exactly two outcomes (success/failure) and a fixed probability each time — a coin flip, a pass/fail test, a yes/no survey response.
How would I find the probability of at least 4 successes instead of exactly 4?
Sum this formula's result for k = 4, 5, 6... up to n, or equivalently subtract the sum for k = 0 through 3 from 1.
Why does C(n,k) appear in the formula at all?
Because there are multiple different orders in which exactly k successes among n trials could occur, and each order is equally likely — C(n,k) counts how many such orders exist.
What happens if p = 0.5 exactly, like a fair coin?
The distribution becomes symmetric around n/2 successes, so k successes and (n−k) successes have identical probability.
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