The binomial distribution

Four conditions, the pmf with ₙCₓ, and μ = np, σ² = np(1 − p) — on an audit of twelve expense claims.

Counting successes in a fixed number of trials

A Bernoulli trial has two outcomes — call them success and failure, whatever they are in context. Repeat it and count the successes, and under four conditions the count has a binomial distribution:

a fixed number $n$ of trials;
each trial has two outcomes (success / failure);
the success probability $p$ is the same on every trial;
the trials are independent.

Write $X \sim Bin (n, p)$ . The pmf is

$P (X = x) = (x n) p^{x} (1 - p)^{n - x}, x = 0, 1, \dots, n$

Read it as a product of three things: the number of arrangements of $x$ successes among $n$ trials (L5.6's combination), the probability of the $x$ successes, and the probability of the $n - x$ failures. The mean and variance need no table:

$μ = n p, σ^{2} = n p (1 - p), σ = n p (1 - p)$

Worked example — twelve expense claims

Maple Ledger audits 12 expense claims. Historically 15 % of claims contain an error, and the claims come from different employees. Let $X$ = the number of the 12 that contain an error. Fixed $n = 12$ ; each claim has or lacks an error; $p = 0.15$ throughout; independent claims — $X \sim Bin (12, 0.15)$ .

Exactly two with an error:

$P (X = 2) = (2 12) (0.15)^{2} (0.85)^{10} = 66 \times 0.0225 \times 0.19687 = 0.2924$

None: $P (X = 0) = (0.85)^{12} = 0.1422$ — the combination is $(0 12) = 1$ and $p^{0} = 1$ , so only the failures remain.

Mean and SD: $μ = 12 \times 0.15 = 1.8$ errors; $σ^{2} = 12 \times 0.15 \times 0.85 = 1.53$ ; $σ = 1.237$ . Expect about two erroneous claims, give or take one.

The scaffold the paper rewards

Every binomial question is answered in the same three lines: define $X$ in words (" $X$ = number of the 12 claims with an error"), state the distribution ( $X \sim Bin (12, 0.15)$ ), translate the phrase into an inequality. Then compute. The first two lines earn marks on their own and prevent the third from going wrong.

When it is not binomial

The condition that fails most often is the fixed $n$ . "The number of claims audited until the first error is found" has no fixed number of trials — the count is open-ended and the distribution is not binomial. And condition 4 is strained by sampling without replacement: drawing 12 claims from a file of 40 changes $p$ with each draw. The convention is that the binomial is acceptable when the sample is less than 10 % of the population — 12 of 400 claims, yes; 12 of 40, no.