Binomial Distribution

A factory produces LED bulbs. Quality control tests show that 90% of bulbs work perfectly and 10% are defective. You pick 5 bulbs at random from a large production batch. How many defective bulbs should you expect to find?

You could enumerate every possibility. Zero defective, one defective, two defective, all the way up to five defective. For each case, you could compute the probability. Then you could find the expected value.

But this particular setup — repeating the same yes/no experiment multiple times and counting the "yes" outcomes — appears everywhere. Coin flips, medical trials, election polls, quality inspections, free-throw shooting. It appears so often that it has its own name and its own formula, and once you see where the formula comes from, you'll never need to enumerate cases by hand again.

One trial: the Bernoulli experiment

Start with the smallest possible version of the problem. You pick one bulb. It is either defective (probability p = 0.1) or not defective (probability 1 - p = 0.9). That's it — two outcomes, one trial.

This single yes/no experiment is called a Bernoulli trial. You label one outcome "success" (which doesn't have to be a good thing — here, "success" means "defective," because that's what you're counting) and the other "failure." The random variable X takes value 1 for success and 0 for failure, with

The expected value is E(X) = 0 \cdot (1 - p) + 1 \cdot p = p. The variance is E(X^2) - [E(X)]^2 = p - p^2 = p(1 - p).

A single Bernoulli trial is simple. The interesting question is what happens when you repeat it.

From one trial to n trials

Go back to the 5 bulbs. Each bulb is an independent Bernoulli trial with p = 0.1. You want the probability of getting exactly k defective bulbs out of 5.

Take a specific case first: exactly 2 defective bulbs. One way this could happen is if the first two bulbs are defective and the last three are good:

The probability of this specific sequence is 0.1 \times 0.1 \times 0.9 \times 0.9 \times 0.9 = (0.1)^2 (0.9)^3.

But this is not the only arrangement that gives exactly 2 defectives. The defective bulbs could be in positions 1 and 3, or positions 2 and 5, or any other pair of positions. Each such arrangement has exactly the same probability (0.1)^2 (0.9)^3 — because the trials are independent, the probability depends only on how many are defective, not on which ones.

How many such arrangements are there? You are choosing 2 positions (for the defective bulbs) out of 5. That is \binom{5}{2} = 10.

So the total probability of exactly 2 defectives is

The logic generalises immediately.

The binomial probability formula

Derivation. Each trial is independent. A specific sequence with k successes and n - k failures has probability p^k (1 - p)^{n-k}, because you multiply the probabilities of each trial and get p for each success and (1-p) for each failure. The number of distinct sequences with exactly k successes is \binom{n}{k} — you choose which k of the n positions are successes. Since these sequences are mutually exclusive (they represent different outcomes), you add their probabilities, giving \binom{n}{k} p^k (1-p)^{n-k}. \square

Three assumptions must hold for the binomial model to apply:

1. Fixed number of trials n.
2. Each trial has exactly two outcomes (success or failure).
3. Trials are independent, and the probability p is the same for every trial.

If any of these fail — if p changes from trial to trial, or the outcomes are dependent, or the number of trials is not fixed — then you do not have a binomial distribution.

Notice how the probabilities verify: 0.590 + 0.328 + 0.073 + 0.008 + 0.0004 + 0.00001 = 1. This must happen, by the binomial theorem: \sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k} = (p + (1-p))^n = 1^n = 1. The binomial probability formula and the binomial theorem from algebra are the same identity, just viewed through different lenses.

Mean of the binomial distribution

You could compute the mean directly from the definition: E(X) = \sum_{k=0}^{n} k \binom{n}{k} p^k (1-p)^{n-k}. This sum is not obviously easy to evaluate. There is a cleaner path.

Proof using linearity. Think of X as a sum. Each trial i produces a Bernoulli random variable X_i that is 1 if trial i is a success and 0 otherwise. Then

By linearity of expectation (which holds whether or not the variables are independent):

That's it. No complicated sums, no binomial coefficient manipulations. The entire proof rests on writing X as a sum of indicator variables and using the fact that expectation distributes over addition.

For the bulb example, E(X) = 5 \times 0.1 = 0.5. On average, half a defective bulb per batch.

Variance of the binomial distribution

Proof using independence. The X_i are independent Bernoulli variables, each with variance p(1 - p). For independent variables, variances add:

The standard deviation is \sigma = \sqrt{np(1-p)}.

For the bulb example, \text{Var}(X) = 5 \times 0.1 \times 0.9 = 0.45 and \sigma = \sqrt{0.45} \approx 0.671.

Alternative proof via the direct sum. For completeness, you can also compute E(X^2) directly. Since X = \sum X_i:

Since X_i is 0 or 1, X_i^2 = X_i, so E(X_i^2) = p. Since X_i and X_j are independent for i \neq j, E(X_i X_j) = E(X_i) E(X_j) = p^2. There are \binom{n}{2} pairs, so:

Both proofs arrive at the same answer. The first is elegant; the second shows you the algebra that's hiding behind the elegance.

This symmetry is not a coincidence. If "success" has probability p, then "failure" has probability 1 - p, and the number of failures in n trials follows B(n, 1-p). The two distributions are reflections.

Two worked examples

Common confusions

• "The binomial formula works for sampling without replacement." Only approximately, and only when the population is much larger than the sample. If you pick 5 bulbs from a box of 20 without replacing them, the probability changes after each draw. The exact model is the hypergeometric distribution, not the binomial. But if you pick 5 from a box of 10,000, the change in probability after each draw is negligible, and the binomial is an excellent approximation.

• "\binom{n}{k} counts the probability." No — \binom{n}{k} counts the number of arrangements, not the probability. Each arrangement has its own probability p^k(1-p)^{n-k}, and you multiply the count by this probability to get the final answer. The two factors play very different roles.

• "If p = 0.5, the distribution is always symmetric." Correct, but the converse is false — a binomial distribution with p \neq 0.5 is always skewed. The skewness increases as p moves further from 0.5.

• "More trials means more variance." True in absolute terms (\text{Var} = np(1-p) grows with n), but the variance per trial is p(1-p)/n, which shrinks. The relative spread decreases as you do more trials — this is why polling becomes more reliable with larger samples.

Going deeper

Connection to the binomial theorem

The name "binomial distribution" is not a coincidence. The binomial theorem says

Set a = p and b = 1 - p. The right side becomes \sum_{k=0}^{n} \binom{n}{k} p^k (1-p)^{n-k}, which is the sum of all binomial probabilities. The left side becomes (p + 1 - p)^n = 1. So the binomial theorem is the algebraic statement that probabilities sum to 1 — the two ideas are literally the same identity.

Mode of the binomial

The most probable value of k — the mode — is the integer part of (n+1)p. More precisely, P(X = k) increases as k goes up from 0, peaks at or near k = (n+1)p - 1, and then decreases. You can show this by examining the ratio P(X = k+1) / P(X = k):

This ratio exceeds 1 (meaning probabilities are still increasing) when k < (n+1)p - 1, and drops below 1 after that.

The normal approximation

When n is large and p is not too close to 0 or 1, the binomial distribution looks increasingly like a bell curve. Specifically, B(n, p) is well approximated by a normal distribution with the same mean and variance:

A common rule of thumb is that the approximation is good when np \geq 5 and n(1-p) \geq 5. This is the bridge between discrete and continuous probability, and it is the reason the normal distribution is so central to statistics.

Where this leads next

• Other Discrete Distributions — the geometric distribution (how many trials until the first success) and the Poisson distribution (rare events in a large population), with derivations.
• Normal Distribution — the bell curve that the binomial approaches for large n, and the most important distribution in statistics.
• Binomial Theorem for Positive Integer — the algebraic identity that powers the binomial distribution formula.
• Combinations — Basics — a closer look at \binom{n}{k} and why it counts what it counts.
• Conditional Probability — Advanced — when the probability of success changes based on what you know. Extends the fixed-p world of the binomial.