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Binomial distribution

The number of successes in a fixed number of independent trials that each succeed with the same probability.

Parameters and support

$n$: number of independent trials — $n \in {1, 2, \dots}$
$p$: probability of success on one trial — $0 \leq p \leq 1$
Support: $x \in {0, 1, \dots, n}$

The formulas

p(x): $p (x) = (x n) p^{x} (1 - p)^{n - x}$
F(x): $F (x) = k = 0 \sum ⌊ x ⌋ (k n) p^{k} (1 - p)^{n - k}$
No shorter closed form. For large n the normal approximation with a continuity correction is the intended exam route.
Mean: $E [X] = n p$
Variance: $Var (X) = n p (1 - p)$
MGF: $M (t) = (1 - p + p e^{t})^{n}$

Memory hook. A binomial is n independent Bernoullis added together, so both the mean and the variance are n times the Bernoulli's: n·p and n·p(1−p). Variance is always SMALLER than the mean.

Where the moments come from

Write $X = \sum_{k = 1}^{n} B_{k}$ where each $B_{k}$ is Bernoulli( $p$ ) and the $B_{k}$ are independent.
$E [B_{k}] = p$ and $Var (B_{k}) = E [B_{k}^{2}] - p^{2} = p - p^{2} = p (1 - p)$ , since $B_{k}^{2} = B_{k}$ .
Expectation is always additive: $E [X] = n p$ .
Variance is additive because the trials are INDEPENDENT: $Var (X) = n p (1 - p)$ .
The MGF is the product of $n$ identical Bernoulli MGFs $1 - p + p e^{t}$ .

Worked example

An insurer issues 10 independent one-year policies, each with a 0.15 probability of producing a claim. Find the probability that exactly 3 policies produce a claim.

This is $X \sim Binomial (n = 10, p = 0.15)$ — a fixed number of independent trials with a common success probability.
$P (X = 3) = (3 10) (0.15)^{3} (0.85)^{7}$ .
$(3 10) = 120$ , $(0.15)^{3} = 0.003375$ , and $(0.85)^{7} = 0.320577$ .
$P (X = 3) = 120 \times 0.003375 \times 0.320577 = 0.1298$ .

Answer: 0.1298

The mean, variance, CDF and moment generating function above are re-derived numerically from this distribution’s own density on every test run — summed over the support for a discrete distribution, integrated by quadrature for a continuous one — and compared with the closed forms printed here. A typo on this page fails the build.

Traps

Using the binomial when sampling is WITHOUT replacement from a small population — that is hypergeometric.
Forgetting the continuity correction when the normal approximation is used on a discrete count.
Reading 'at least 3' as P(X = 3) rather than 1 − P(X ≤ 2).

Drill this: the Exam P question bank has original questions on this distribution, and the recall trainer builds its prompts from exactly the formulas above.

Binomial distribution

Parameters and support

The formulas

Where the moments come from

Worked example

Traps

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