Coin flips add up: the binomial counts yes–no successes
Flip a biased coin six times and ask for exactly four heads: the binomial distribution answers. It models how many successes appear in n independent Bernoulli trials that each succeed with probability p—and underpins the binomial significance test.
A binomial random variable with parameters n and p counts successes in n independent yes–no experiments, each with success chance p and failure chance q = 1 − p. One trial alone is a Bernoulli distribution; a string of trials is a Bernoulli process. Sampling with replacement from a finite population fits this picture; sampling without replacement yields a hypergeometric law instead, though when the population N dwarfs the sample n the binomial remains a practical approximation.
If X ~ B(n, p), the chance of exactly k successes is the probability mass C(n,k) p^k (1−p)^{n−k}, where the binomial coefficient counts which of the n slots hold the successes. Every specific success–failure pattern with those counts shares the same probability because trials are independent and identically rated; multiplying by the count of patterns finishes the formula. For a coin with heads probability 0.3, the chance of exactly four heads in six tosses is about 0.0595.
The mode—the most probable k—sits where the ratio of consecutive masses crosses one. There is always an integer M maximizing the mass; when (n+1)p is itself an integer, two adjacent modes share the peak. Expectation is np by linearity, since X is the sum of n Bernoulli variables each with mean p. Independence likewise adds variances to give Var(X) = np(1−p). Higher central moments follow closed polynomials in n and p.
The cumulative distribution sums the mass up to floor(k) and also equals a regularized incomplete-beta expression, linking it to beta and F distribution cdfs—handy for computation and for bounding tails when tables only list half the range and exploit symmetry.
Source: Binomial distribution
