Binomial distribution

If \(n\) independent Bernoulli trials are repeated, each with success probability \(p\), the total number of successes \(X\) is binomial; we write \(X\sim\mathrm{Bin}(n,p)\). To get exactly \(k\) successes there are \(\binom{n}{k}\) ways to choose which trials succeed, and each has probability \(p^k(1-p)^{n-k}\); that gives the mass function.

The four conditions are checked one by one: \(n\) fixed in advance, two outcomes per trial, the same \(p\) in every trial, and independent trials. If one fails, the model does not apply: drawing cards without replacement breaks independence, and counting tosses until the first success leaves \(n\) unfixed.

Since \(X\) is a sum of \(n\) independent Bernoulli variables, linearity gives \(E[X]=np\) and independence gives \(\operatorname{Var}(X)=np(1-p)\). With \(n=20\) and \(p=0.3\) we expect 6 successes with variance 4.2.

The distribution is symmetric when \(p=\tfrac12\) and skewed otherwise. For “at least one” the complement is convenient, \(1-(1-p)^n\). When \(n\) is large and \(p\) small, the binomial is approximated by the Poisson with \(\lambda=np\).