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MasterMath

Binomial Distribution Calculator

The probability of getting exactly k successes in n trials, plus the cumulative ones, when every trial is independent and always has the same probability.

Exactly k successes

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Exactly k successes—
At most k—
At least k—
Expected successes—
Standard deviation—
Variance—
kExactly kUp to k

How this was worked out

    The formula

    P(X = k) = C(n, k) · pᵏ · (1 − p)ⁿ⁻ᵏ

    What it means

    The binomial counts successes across a fixed number of independent trials with the same probability each: heads in ten tosses, defective parts in a batch, correct answers on a multiple-choice test. The formula has two parts: the powers give the probability of one specific sequence with k successes, and the combination counts how many different ways those k successes can sit among the n trials.

    How to work it out by hand

    1. Count how many trials there are in total and how likely each one is to succeed
    2. Work out C(n, k): how many ways the k successes can be arranged
    3. Multiply by p to the power k and by (1 − p) to the power n − k
    4. For the cumulative figures, add up the probabilities of all the cases you care about

    What is worth knowing

    The three conditions — fixed number of trials, independence and constant probability — are not a formality: draw cards without replacing them and the probability changes at every draw, which calls for the hypergeometric, not the binomial. When n is large and p small, the binomial closely resembles a Poisson with mean n·p, and when n is large with p in the middle, a normal; both approximations exist because the exact calculation was unworkable by hand, which is no longer the case here.

    Frequently asked questions

    What does the binomial count?

    Successes across a fixed number of independent trials that all share the same probability.

    Does it work for drawing without replacement?

    No. If you do not put back what you drew, the probability changes each time and the right model is the hypergeometric.

    What is the difference between "exactly k" and "at least k"?

    The first is a single probability; the second adds the one for k and for every value above it.

    Why are the expected successes not a whole number?

    Because n · p is a long-run average. Averaging 2.5 heads does not mean 2.5 heads ever come up.

    Does it work with very large n?

    Yes. The calculation is done in logarithms, so nothing overflows even when the combination is astronomical.