The binomial distribution and its formula

Write X for the number of successes obtained over the n repetitions. The distribution of X is called the binomial distribution.

The formula

                     n-k
P(X = k) = C(n,k) · p^k · (1-p)

for k from 0 to n. We write X ~ B(n , p).

Where it comes from

The three factors can be read straight off the tree:

p^k          the probability of the path's k successes
(1-p)^(n-k)  that of its n-k failures
C(n,k)       the NUMBER of paths with exactly k successes
             (choosing in which k of the n repetitions the success falls)

This is the binomial coefficient of Pascal's triangle in its most concrete role: counting the possible positions of the successes in the sequence.

A worked example

A machine produces 20 % defective parts. Five are taken. Let X be the number of defective ones.

X ~ B(5 ; 0.2)

P(X = 2) = C(5,2) × 0.2² × 0.8³
         =   10   × 0.04 × 0.512
         = 0.2048

About a 20 % chance of getting exactly two defective parts out of five.

The full distribution:

  k          0        1        2        3        4        5
  P(X=k)  0.3277   0.4096   0.2048   0.0512   0.0064   0.0003

  total = 1  ✔

 P(X=k)
  0.41 |     █
  0.33 |  █  █
  0.20 |  █  █  █
  0.05 |  █  █  █  █
  0.00 |  █  █  █  █  █  █
       +--0--1--2--3--4--5--> k

The recurring computations

"At least one" — always via the complementary event:

P(X ≥ 1) = 1 - P(X = 0) = 1 - (1-p)^n

Probability of getting at least one 6 in 4 rolls of a die:

  1 - (5/6)⁴ = 1 - 0.4823 = 0.5177        about 52 %

This is the computation that, it is said, founded probability theory: the Chevalier de Méré had noticed the bet was slightly favourable, and asked Pascal why.

"At most" or "at least k" — add the relevant terms, or use the complement if shorter:

P(X ≤ 2) = P(X=0) + P(X=1) + P(X=2)
P(X ≥ 3) = 1 - P(X ≤ 2)

The shape of the distribution

p = 0.5 (symmetric)         p = 0.2 (right-skewed)      p = 0.9 (left)

      ___                     _                                  _
     /   \                   / \_                            ___/ \
  __/     \__               /    \___                    ___/
  0   n/2   n              0                n            0        n

• p = 0.5: symmetric distribution;
• small p: the values cluster near 0;
• large n: the curve takes a bell shape — the announcement of the normal distribution.

Summary

• X ~ B(n ; p) counts the successes over n independent repetitions.
• P(X = k) = C(n,k) p^k (1-p)^(n-k).
• The three factors: probability of the successes, of the failures, and the number of paths.
• "At least one" is computed as 1 - (1-p)^n.
• The distribution is symmetric for p = 0.5, skewed otherwise, and tends to a bell as n grows.