Binomial Distribution: Formula, Worked Examples, and the Normal Approximation

Flip a coin ten times — how many heads should you expect, and how likely is exactly seven? Any situation with a fixed number of independent trials, each with only two possible outcomes and the same probability of "success" every time, follows the binomial distribution. It's one of the most widely used discrete distributions in statistics, underlying everything from quality control sampling to clinical trial design.

The Four Conditions (BINS)

The binomial distribution is one of a family of discrete probability distributions built on the same basic rules covered in basic probability — it's simply the multiplication and addition rules applied systematically across every possible sequence of successes and failures in a fixed number of trials.

A situation follows a binomial distribution only when all four hold:

• Binary outcomes — each trial has exactly two possible results (success/failure, heads/tails, defective/not defective).
• Independent trials — the outcome of one trial doesn't affect any other.
• Number of trials is fixed in advance (n is set before data collection, not a stopping rule based on results).
• Same probability of success on every trial (p stays constant throughout).

Sampling without replacement from a small population violates independence (each draw changes the composition of what remains, exactly as in the non-binomial sampling examples elsewhere on this site) — but when the population is large relative to the sample, this violation is small enough to safely ignore in practice.

The Formula

Here n is the number of trials, k is the number of successes you're finding the probability for, p is the probability of success on a single trial, and C(n,k) — "n choose k" — counts how many different ways k successes can be arranged among n trials: C(n,k) = n! / [k!(n−k)!].

Worked Example

Problem: A multiple-choice quiz has 10 questions, each with 4 options (1 correct). A student guesses randomly on every question. What's the probability of getting exactly 3 correct?

n = 10, k = 3, p = 0.25 (probability of guessing correctly on any one question)

C(10,3) = 10! / (3!×7!) = 120

P(X=3) = 120 × (0.25)³ × (0.75)⁷ = 120 × 0.015625 × 0.13348 ≈ 0.2503

There's roughly a 25% chance of getting exactly 3 correct by pure guessing — a useful benchmark for judging whether a student's actual score suggests real knowledge beyond chance.

Cumulative Probabilities: At Least, At Most

Often the real question isn't "exactly k" but "at least k" or "at most k" — found by summing the exact-k formula across the relevant range:

For the quiz example, P(passing with at least 6 correct by pure guessing) = P(X=6)+P(X=7)+...+P(X=10) — summing these individual terms gives a small number (around 0.02), showing that scoring 6+ out of 10 by pure random guessing is genuinely unlikely, which is exactly the kind of benchmark used to argue a student's score reflects real knowledge, not luck.

Mean and Standard Deviation

For the quiz example: μ = 10 × 0.25 = 2.5 (on average, a random guesser gets 2.5 questions right out of 10), and σ = √[10 × 0.25 × 0.75] = √1.875 ≈ 1.37. These formulas require no summation across the whole distribution — they follow directly from n and p.

When Does the Binomial Distribution Look Like a Bell Curve?

As n grows large (and p isn't too close to 0 or 1), the binomial distribution's shape increasingly resembles a normal distribution — a specific case of the Central Limit Theorem, since a binomial count is really a sum of n independent 0/1 trials. A common rule of thumb: the normal approximation is reasonable when both np ≥ 5 and n(1−p) ≥ 5. When this holds, probabilities can be estimated using the normal distribution calculator with μ = np and σ = √[np(1−p)], which is much faster than summing many individual binomial terms by hand for large n.

Real-World Uses

Quality control: if a manufacturing process has a known defect rate, the binomial distribution predicts the probability of finding a given number of defects in a sample batch — directly informing acceptance sampling plans.

Clinical trials: the number of patients responding to a treatment, out of a fixed trial size, with a hypothesized response rate, is a textbook binomial scenario used to plan trial sizes and interpret results.

Election polling: the number of respondents in a fixed-size random sample who support a candidate follows a binomial distribution, the basis for the same margin-of-error calculations covered in the confidence intervals guide.

A/B testing: the number of conversions out of a fixed number of visitors, at a hypothesized conversion rate, is binomial — directly underlying the two-proportion z-test used in A/B testing.

A Second Worked Example: At-Least Probability

Problem: A factory's parts have a 2% defect rate. In a random sample of 20 parts, what's the probability of finding at least 1 defective part?

Working directly with "at least 1" would require summing P(X=1) through P(X=20) — nineteen separate calculations. The complement rule shortcuts this dramatically: P(at least 1) = 1 − P(exactly 0).

P(X=0) = C(20,0) × (0.02)⁰ × (0.98)²⁰ = 1 × 1 × 0.98²⁰ ≈ 0.6676

P(at least 1 defective) = 1 − 0.6676 ≈ 0.3324

Despite a low 2% individual defect rate, there's still roughly a 1-in-3 chance of finding at least one defective part in a sample of 20 — a useful, sometimes counterintuitive illustration of how individually rare events become common when enough independent trials are involved.

Common Mistakes

Using the binomial formula when trials aren't independent. Sampling without replacement from a small, finite population (where removing one item meaningfully changes the probabilities for the rest) technically follows the hypergeometric distribution, not the binomial. The binomial remains a good approximation when the population is large relative to the sample size — a common rule of thumb is that the sample should be less than about 10% of the population — but breaks down for small populations.

Forgetting C(n,k) and only calculating pᵏ(1−p)ⁿ⁻ᵏ. This gives the probability of ONE specific ordering of successes and failures, not the total probability of exactly k successes in any order — the combinations term C(n,k) accounts for all the different orderings that produce the same count of successes, and omitting it is one of the most common calculation errors with this distribution.

Applying the normal approximation when np or n(1−p) is too small. With a small n or an extreme p, the binomial distribution can be noticeably skewed, and the normal approximation becomes unreliable — always check the np≥5 and n(1−p)≥5 rule of thumb before substituting a normal-distribution calculation for the exact binomial one.

Using Software for Larger Problems

Calculating binomial probabilities by hand becomes unwieldy for large n or when summing many terms for a cumulative probability. In practice, statistical software (R's dbinom/pbinom functions, Python's scipy.stats.binom module, or a spreadsheet's BINOM.DIST function) computes exact binomial probabilities instantly for any n, k, and p — worth knowing the underlying formula well enough to sanity-check software output, even once you're relying on a tool for the actual calculation.