Uniform, Bernoulli & Binomial

The last lesson promised a few shapes of randomness that come up so often you never re-derive their two numbers — you recognise the setup and read them off. Here are the three discrete ones, and almost every discrete question on the exam is one of them in costume. A fair die, where every face is equally likely. A single coin flip, heads or tails. And many coin flips at once, counting how many come up heads.

Three recipes, built one on the next

Start with the simplest and stack upward. A discrete uniform is the fair die: every value equally likely. A Bernoulli is a single yes-or-no trial — one flip. A Binomial is what you get when you line up many such flips and count the yeses. The third is built from the second, which is why we meet them in this order.

The discrete uniform spreads its chance evenly over the integers a to b — that is n = b − a + 1 values, each with chance 1/n. By symmetry the mean sits dead centre, (a+b)/2, and the variance works out to (n² − 1)/12. A fair die (n = 6): mean (1+6)/2 = 3.5, variance (36−1)/12 = 35/12 ≈ 2.92 — the very numbers we ground out by hand last lesson, now had for free.

The Bernoulli(p) is the atom: one trial, success (X = 1) with chance p, failure (X = 0) with chance 1 − p. Its mean is 1·p + 0·(1−p) = p, and its variance is p(1 − p) — largest at p = 0.5, where the outcome is most uncertain, and zero at p = 0 or p = 1, where nothing is in doubt.

The Binomial(n,p) stacks n of those atoms and counts the successes. To get exactly k successes, you first choose which k of the n trials succeed — C(n,k) ways, the same combination from the counting lesson — and each such pattern has chance p^k (1−p)^(n−k). Together:

P(X = k) = C(n,k) · p^k · (1 − p)^(n−k),   k = 0, 1, …, n

And because a Binomial is just n independent Bernoulli atoms added up, its two numbers are n copies of the Bernoulli ones: mean np and variance np(1−p). That stacking works only when the trials are genuinely repeatable, which is the three binomial conditions — a fixed number of trials n, independent trials, and a constant chance p on every one.

Watching it on a small case

Let X be Binomial(n = 5, p = 0.5). Find P(X = 3), the mean, and the variance.

Put k = 3 into the PMF, and use that p = 1 − p = 0.5 so the two powers fold together:

P(X = 3) = C(5,3) · (0.5)³ · (0.5)²
         = C(5,3) · (0.5)⁵
         = 10 · (1/32)
         = 0.3125

mean     = np        = 5 · 0.5        = 2.5
variance = np(1 − p) = 5 · 0.5 · 0.5  = 1.25

So P(X = 3) = 0.3125, mean 2.5, variance 1.25 — the C(5,3) = 10 patterns of three heads, each of chance 1/32.

A question to carry forward

Every variable so far has landed on a separated list — 0, 1, 2, 3 heads, a face 1 to 6. But a great many real quantities are not counts at all: a waiting time, a height, a measurement that can fall anywhere in a range, with no next value to step to. Here is the thread onward: when the values form a continuum, what takes the place of the PMF, and where does the chance live then?

In one breath

• Discrete uniform on n values: each 1/n; mean (a+b)/2, variance (n²−1)/12 (fair die → 3.5, 35/12).
• Bernoulli(p) — one trial (the atom): mean p, variance p(1−p) (largest at p=0.5).
• Binomial(n,p) = n stacked Bernoulli atoms: P(X=k) = C(n,k)·p^k·(1−p)^(n−k), mean np, variance np(1−p).
• The C(n,k) is the counting lesson; the three conditions are fixed n, independent trials, constant p.
• Variance is np(1−p), not np — and sampling without replacement breaks constant-p (hypergeometric, not binomial).

Practice