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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

nn
number of independent trials - n∈{1,2,… }n \in \{1,2,\dots\}
pp
probability of success on one trial - 0≤p≤10 \le p \le 1
Support
x∈{0,1,…,n}x \in \{0, 1, \dots, n\}

The formulas

p(x)
p(x)=(nx)px(1−p)n−xp(x) = \binom{n}{x} p^{x} (1-p)^{n-x}
F(x)
F(x)=∑k=0⌊x⌋(nk)pk(1−p)n−kF(x) = \displaystyle\sum_{k=0}^{\lfloor x \rfloor} \binom{n}{k} 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]=npE[X] = np
Variance
Var⁡(X)=np(1−p)\operatorname{Var}(X) = np(1-p)
MGF
M(t)=(1−p+pet)nM(t) = \left(1 - p + p e^{t}\right)^{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

  1. Write X=∑k=1nBkX = \sum_{k=1}^{n} B_k where each BkB_k is Bernoulli(pp) and the BkB_k are independent.
  2. E[Bk]=pE[B_k] = p and Var⁡(Bk)=E[Bk2]−p2=p−p2=p(1−p)\operatorname{Var}(B_k) = E[B_k^2] - p^2 = p - p^2 = p(1-p), since Bk2=BkB_k^2 = B_k.
  3. Expectation is always additive: E[X]=npE[X] = np.
  4. Variance is additive because the trials are INDEPENDENT: Var⁡(X)=np(1−p)\operatorname{Var}(X) = np(1-p).
  5. The MGF is the product of nn identical Bernoulli MGFs 1−p+pet1-p+pe^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.

  1. This is X∼Binomial(n=10, p=0.15)X \sim \text{Binomial}(n = 10,\, p = 0.15) - a fixed number of independent trials with a common success probability.
  2. P(X=3)=(103)(0.15)3(0.85)7P(X = 3) = \binom{10}{3}(0.15)^3(0.85)^7.
  3. (103)=120\binom{10}{3} = 120, (0.15)3=0.003375(0.15)^3 = 0.003375, and (0.85)7=0.320577(0.85)^7 = 0.320577.
  4. P(X=3)=120×0.003375×0.320577=0.1298P(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 that test run.

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).

Related

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.