Back to courseLesson 8 of 14

The binomial distribution

What you'll learn

Count successes across repeated trials — coin flips, free throws, or defective parts.

Flip a coin 10 times — how many heads? Shoot 20 free throws — how many go in? Whenever you count successes in a fixed number of independent yes/no trials, you're looking at the binomial distribution.

When it applies

A situation is binomial when all four hold:

  • a fixed number of trials, n
  • each trial is independent of the others
  • each trial has only two outcomes (success / failure)
  • the probability of success, p, is the same every trial

The formula

The probability of exactly k successes in n trials is:

P(X=k)=(nk)pk(1p)nkP(X = k) = \binom{n}{k} \, p^k \, (1-p)^{n-k}

It has three parts: pᵏ is the chance of k successes, (1 − p)ⁿ⁻ᵏ the chance of the other n − k failures, and the binomial coefficient — read "n choose k" — counts how many different orders those k successes could come in.

n = 10 trials, p = 0.5012345678910k (number of successes)
Each bar is P(X = k) = C(n, k) · pᵏ · (1 − p)ⁿ⁻ᵏ — the chance of exactly k successes in n independent trials. With p = 0.5 the shape is symmetric, peaking at the most likely count.

Its center and spread

You don't have to sum the whole distribution to know where it sits:

  • Expected number of successes: μ = n·p
  • Standard deviation: σ = √(n·p·(1−p))

For n = 10, p = 0.5, that's μ = 5 (the peak) and σ ≈ 1.58 — matching the bars above.

Why this matters

The binomial is the workhorse for counts: defect rates in manufacturing, conversion rates in marketing, votes in a poll, genetics. And when n is large it starts to look like a bell — the doorway to the normal distribution and the reason it shows up everywhere next.

Check your understanding

Question 1 of 2

Which of these would DISQUALIFY a situation from being binomial?

Next lesson