The Binomial Distribution

Flip a coin ten times and count the heads. Inspect twenty light bulbs and count the faulty ones. Ask fifty people a yes/no question and count the "yes"es. In every case you repeat the same little experiment a fixed number of times and count how often one particular result — call it a success — happens. The binomial distribution is the distribution of that count.

We write

X \sim B(n, p)

to say that X — the number of successes — follows a binomial distribution with n trials, each having success probability p. The symbol \sim is read "is distributed as".

a coin

A single coin flip is the cleanest example of a binomial trial. It has exactly two outcomes (heads or tails), the probability never changes from flip to flip, and one flip tells you nothing about the next. Call "heads" a success with p = 0.5, flip n times, and the number of heads is B(n, 0.5). Almost every binomial story is really a coin in disguise — a coin that may be bent so that p \ne 0.5.

The four conditions

A count is binomial only when all four of these hold. They are worth learning by heart, because spotting a broken one is how you decide whether the model applies:

A quick memory hook: fixed n, two outcomes, same p, independent.

an apple

A packing plant knows that 3\% of its apples are bruised. An inspector pulls a box of 20 apples at random and counts the bruised ones. Each apple is a trial (bruised = success, sadly), the count is fixed at n = 20, the rate is a constant p = 0.03, and one apple's state does not change the next. So the number of bruised apples is B(20, 0.03) — and the plant can work out how often a box will contain 0, 1, or more duds.

Building the formula

Where does the probability of exactly r successes come from? Build it in two pieces.

One particular sequence. Suppose n = 5 and we want exactly r = 2 successes in the specific order S S F F F. Because the trials are independent we multiply their probabilities:

p \cdot p \cdot (1-p)\cdot(1-p)\cdot(1-p) = p^{2}(1-p)^{3}.

Any other order with two successes — say F S F S F — has exactly the same probability p^{2}(1-p)^{3}, because multiplication does not care about order. So every arrangement of two S's and three F's is equally likely.

How many arrangements? The number of ways to place r successes among n trials is the binomial coefficient \binom{n}{r} ("n choose r"). Add up the equally-likely arrangements and you get the binomial probability formula:

P(X = r) = \binom{n}{r}\, p^{\,r}\,(1-p)^{\,n-r}, \qquad r = 0, 1, 2, \ldots, n.

Read it left to right: \binom{n}{r} counts the arrangements, p^{r} is the chance the r successes all happen, and (1-p)^{n-r} is the chance the remaining n-r trials all fail.

Worked examples

Three heads in ten flips. A fair coin, n = 10, p = 0.5. The chance of exactly 3 heads is

P(X = 3) = \binom{10}{3}(0.5)^{3}(0.5)^{7} = 120 \cdot (0.5)^{10} \approx 0.117.

Two bruised apples in a box. With B(20, 0.03),

P(X = 2) = \binom{20}{2}(0.03)^{2}(0.97)^{18} \approx 190 \cdot 0.0009 \cdot 0.578 \approx 0.099.

A small hand check. With n = 5 and p = 0.3, the chance of exactly 2 successes is

P(X = 2) = \binom{5}{2}(0.3)^{2}(0.7)^{3} = 10 \cdot 0.09 \cdot 0.343 = 0.3087.

See it: the shape of the distribution

The whole distribution is a bar chart: one bar for each possible count r = 0, 1, \ldots, n, whose height is P(X = r). Because X lands on some value, the bar heights add up to 1.

Press Refresh for a fresh n and p. Watch the shape: when p = 0.5 the bars are symmetric; when p is small the peak slides left (few successes are likely), and when p is large it slides right. The tallest bar sits near np, the average number of successes.