Khayyam Math

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

The probability of getting exactly k successes in n independent trials — drawn as a bar chart for n = 10 fair-coin flips.

Binomial(n = 10, p = 0.5) — P(X = k) = C(10, k) · 0.5¹⁰01230.11740.20550.24660.20570.11789100.10.2k (number of successes)

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What this shows

A binomial random variable X counts the number of "successes" in n independent Bernoulli trials, each of which succeeds with probability p. Its probability mass function is

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

for k = 0, 1, 2, …, n. The C(n, k) factor (the binomial coefficient, "n choose k") counts the number of ways the k successes can be arranged among the n trials; the pᵏ and (1−p)ⁿ⁻ᵏ factors are the probabilities of any one specific arrangement.

For n = 10 and p = 0.5 (ten fair coin flips, counting heads), the figure plots P(X = k) for k = 0 through 10. Values:

    k = 0,  10:  1/1024 ≈ 0.001
    k = 1,   9: 10/1024 ≈ 0.010
    k = 2,   8: 45/1024 ≈ 0.044
    k = 3,   7: 120/1024 ≈ 0.117
    k = 4,   6: 210/1024 ≈ 0.205
    k = 5    :  252/1024 ≈ 0.246

The distribution is symmetric (because p = 0.5), peaking at k = 5. The probabilities sum to exactly 1.

Where it shows up

The binomial distribution is the right model for any "count of successes in n independent yes/no trials" — quality-control sampling (defects in a batch), A/B testing (clicks out of impressions), survey responses (yes-votes out of n people), polling. Its mean is n·p and its variance is n·p·(1 − p).

For large n the binomial is well approximated by a normal distribution with the same mean and variance — that's the de Moivre-Laplace theorem and a special case of the Central Limit Theorem. For large n and small p (with n·p moderate) it's better approximated by a Poisson distribution with rate n·p.

Frequently asked questions

What's C(n, k)?

C(n, k) — also written ⁿCₖ or 'n choose k' — counts the number of ways to choose k items from n without regard to order. Computed as n! / (k!·(n−k)!). For example C(10, 3) = 10·9·8 / (3·2·1) = 120.

Does the distribution have to be symmetric?

Only when p = 0.5. For p < 0.5 the distribution is right-skewed (the mode and most of the mass sit at low k); for p > 0.5 it's left-skewed. The skewness disappears as n grows, by the Central Limit Theorem.

What if the trials aren't independent?

Then the binomial isn't the right model. For sampling without replacement from a finite population, use the hypergeometric distribution. For correlated trials, you need a more general model (e.g. a beta-binomial when the success probability itself is uncertain).

Is there a closed-form for P(X ≤ k)?

Not a simple one — the cumulative distribution function is a regularised incomplete beta function. For practical work you either sum the PMF directly (cheap for small n) or use the normal approximation (cheap for large n).

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