The Binomial Distribution

§ 01What Is a Binomial Experiment?

Before we write a single formula, we need to recognise the type of situation the binomial distribution models, and understand why four very specific conditions must hold.

Many real-world problems involve repeating the same action multiple times and counting how often a particular outcome occurs: flipping a coin ten times and counting heads, testing fifty manufactured parts and counting defects, guessing on a twenty-question multiple-choice exam. All of these are binomial experiments.

The binomial distribution gives us the probability of getting exactly k successes in n independent trials when each trial has the same probability p of success. The word "binomial" refers to the two possible outcomes on every single trial: success or failure, yes or no, head or tail.

• 1
  Fixed number of trials, nThe experiment is performed a predetermined number of times. You do not stop early based on results.
• 2
  Each trial has exactly two outcomesWe label one outcome "success" and the other "failure". Which is which depends on the question, not on which outcome is desirable.
• 3
  Constant probability of success, pThe probability of success is the same on every single trial. This is the condition most often violated in practice (sampling without replacement from a small population violates it).
• 4
  Trials are independentThe outcome of one trial does not affect the probability on any other trial. Independence and constant p go hand in hand.

When all four conditions hold, we write X ∼ B(n, p) (or sometimes Bin(n, p)), read "X follows a binomial distribution with parameters n and p."

§ 02The Probability Mass Function

The PMF tells us the probability that X takes the value k, that is, exactly k successes in n trials.

There are three multiplicative pieces, each with a clear meaning:

C(n, k), the combinatorial piece. This counts the number of distinct orderings in which exactly k successes and n−k failures can be arranged across n trials. For example, if n = 4 and k = 2, the sequences SSFF, SFSF, SFFS, FSSF, FSFS, FFSS are all different arrangements of 2 successes in 4 trials, and C(4, 2) = 6 counts them correctly.

pᵏ, the probability of any specific arrangement that contains exactly k successes, since each success occurs with probability p and the trials are independent, these multiply to pᵏ.

(1−p)ⁿ⁻ᵏ, the probability of the remaining n−k failures in that same specific arrangement, each occurring with probability q = 1−p.

Multiplying the three pieces gives the probability of any specific k-success sequence and then counts how many such sequences exist, yielding the correct total probability.

This follows directly from the binomial theorem: (p + q)ⁿ = Σ C(n,k) pᵏ qⁿ⁻ᵏ = 1ⁿ = 1. In other words, the probabilities of all possible outcomes sum to exactly 1, a useful internal check on calculations.

§ 03Mean, Variance, and Standard Deviation

You could compute μ = Σ k · P(X = k) from scratch, but the binomial's elegant structure yields closed-form expressions directly.

Intuition for the Mean

If you flip a fair coin (p = 0.5) 100 times, you expect about 50 heads: μ = 100 × 0.5 = 50. This is simply the number of trials multiplied by the probability of success on each trial, exactly what common sense suggests.

Intuition for the Variance

The variance np(1−p) is maximised when p = 0.5 (most uncertainty) and approaches zero as p → 0 or p → 1 (near-certain outcome). It can never exceed n/4. This captures the idea that outcomes are most spread out when success and failure are equally likely, and most concentrated when one outcome is almost certain.

Proof of the Mean (Outline)

Write X = X₁ + X₂ + ⋯ + Xₙ where each Xᵢ is a Bernoulli(p) random variable equal to 1 for success and 0 for failure. By linearity of expectation, E[X] = Σ E[Xᵢ] = n · p. The variance follows similarly from Var(Xᵢ) = p(1−p) and the independence of the trials.

§ 04Shape of the Distribution — Visual Guide

The binomial distribution changes shape dramatically depending on p. For p = 0.5 it is perfectly symmetric; for p < 0.5 it is right-skewed; for p > 0.5 it is left-skewed.

The chart above shows X ∼ B(10, 0.3). The distribution is right-skewed because p = 0.3 < 0.5: the bulk of the probability sits on lower values of k, with a long tail trailing to the right. The mean is μ = 10 × 0.3 = 3, and the mode (most probable outcome) is k = 3 (highlighted in gold).

Compare this mental picture to a symmetric B(10, 0.5) where the distribution would form a perfect bell shape centred at k = 5, or a left-skewed B(10, 0.8) where most probability sits near k = 8.

§ 05Worked Examples

Ten fully worked problems covering single-value probabilities, cumulative probabilities, mean and variance, complement method, and real-world contexts.

§ 06The Normal Approximation

When n is large, computing dozens of binomial terms becomes tedious. The Central Limit Theorem rescues us: the binomial distribution is well approximated by a normal distribution.

For X ∼ B(n, p) with np ≥ 5 and n(1−p) ≥ 5, we approximate X ≈ N(np, np(1−p)). The rule of thumb checks that both the expected number of successes and expected number of failures are at least 5, ensuring neither outcome is too rare for the bell curve to apply.

The continuity correction (adding or subtracting 0.5) compensates for the fact that we are approximating a discrete distribution with a continuous one. Without it, answers can be off by several percentage points, especially for smaller n.

§ 07Common Mistakes

§ 08Related Topics

The binomial distribution connects naturally to several other topics. Explore them to deepen your understanding.

§ 0910-Question Quiz

Test your understanding. Each question gives immediate feedback. Click "See Explanation" after answering to read a full solution.