Binomial Distribution Calculator

When can you use the binomial distribution?

A situation is binomial when it meets four conditions, often remembered as BINS:

• Binary: each trial has two outcomes – success or failure.
• Independent: one trial does not affect another.
• Number fixed: you know in advance how many trials, n.
• Same probability: p is the same on every trial.

Rolling a die 12 times and counting sixes is binomial. Drawing cards without replacement is not, because the probability changes after each draw – that situation uses the hypergeometric distribution. Counting how many trials until the first success is a geometric (or negative binomial) problem, not binomial.

The binomial formula, step by step

P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ. C(n, k) counts how many different orders the k successes could occur in; pᵏ(1 − p)ⁿ⁻ᵏ is the probability of any one such order.

Example: the probability of exactly 5 heads in 10 fair coin flips. C(10, 5) = 252, and 0.5⁵ × 0.5⁵ = 1/1,024, so P = 252 ÷ 1,024 = 0.2461 – about 24.6%. Even the single most likely result happens less than a quarter of the time.

Exactly, at most, at least and between

• Exactly k: P(X = k). Exactly 5 heads = 0.2461.
• At most k: add P(X = 0) through P(X = k). At most 3 heads = 0.1719.
• At least k: 1 − P(X ≤ k − 1). At least 7 heads = 1 − 0.8281 = 0.1719.
• Fewer than k = at most k − 1; more than k = at least k + 1.
• Between a and b inclusive: P(X ≤ b) − P(X ≤ a − 1). Between 4 and 6 heads = 0.6562.

Watch the wording: 'at least 3' includes 3, 'more than 3' does not. Many exam mistakes come from that one word.

Mean, variance and shape

The mean (expected number of successes) is np, the variance is np(1 − p) and the standard deviation is √(np(1 − p)). With n = 100 and p = 0.3, you expect 30 successes with a standard deviation of about 4.58. The distribution is symmetric when p = 0.5, skewed right when p is small and skewed left when p is large; the most likely value (mode) is the integer part of (n + 1)p.

Worked examples

Quality control

A machine makes 5% defective parts. In a sample of 20, the chance of no defects is 0.95²⁰ = 0.3585 (about 36%), and the chance of at least two defects is 1 − P(0) − P(1) = 0.2642 (about 26%). If samples regularly show two or more defects, the true defect rate is probably higher than 5%.

Free throws

A player makes 80% of free throws. In 10 attempts, the probability of making at least 8 is P(8) + P(9) + P(10) = 0.3020 + 0.2684 + 0.1074 = 0.6778 – about two times in three. Making all ten happens only about 11% of the time.

Surveys and polls

If 40% of a population supports a policy and you survey 50 random people, the number who say yes is approximately binomial with mean 20 and SD √(50 × 0.4 × 0.6) ≈ 3.46. Results between about 13 and 27 are well within normal variation – a reminder that small samples move around a lot.

Approximations for large n

Normal approximation

When np and n(1 − p) are both at least about 10, the binomial is close to a normal distribution with mean np and SD √(np(1 − p)). Use a continuity correction: P(X ≤ 55) ≈ P(Y < 55.5). For 100 fair coin flips, P(at most 55 heads) ≈ Φ((55.5 − 50) ÷ 5) = Φ(1.1) ≈ 0.864; the exact binomial answer is about 0.864 too.

Poisson approximation

When n is large and p is small (rare events), the binomial is close to a Poisson distribution with mean λ = np. For 1,000 trials with p = 0.002, λ = 2, and P(no events) ≈ e⁻² ≈ 0.135.

Binomial probability in Excel, Google Sheets and Python

Common mistakes

• Using the binomial when trials are not independent or p changes (sampling without replacement from a small population).
• Forgetting the C(n, k) factor and computing the probability of only one order.
• Mixing up 'at least' and 'more than'.
• Entering p as a percentage (30) instead of a decimal (0.30).
• Using the normal approximation when np or n(1 − p) is small.

Computing C(n, k) by hand

C(n, k) – read 'n choose k' – is the number of ways to pick which k of the n trials are successes. For small numbers, cancel before multiplying: C(10, 3) = (10 × 9 × 8) ÷ (3 × 2 × 1) = 120. Pascal's triangle gives the same values row by row; row 5 is 1, 5, 10, 10, 5, 1, so with 5 trials there are 10 ways to get exactly 2 successes. The values are symmetric because choosing k successes is the same as choosing n − k failures.

The probability of at least one success

One of the most useful binomial results is P(at least one) = 1 − (1 − p)ⁿ. If an item drops with a 1% chance per attempt, the chance of seeing it at least once in 50 attempts is 1 − 0.99⁵⁰ ≈ 39.5% – far from certain. To have a 50% chance you need about 69 attempts, because ln 0.5 ÷ ln 0.99 ≈ 69. Our Drop Chance Calculator is built around exactly this formula.

Using the binomial to test a claim

Suppose a coin lands heads 16 times in 20 flips. If the coin were fair, P(X ≥ 16) = 0.0059 – about 6 in 1,000. That is the one-sided p-value for the test 'the coin is fair' against 'the coin favours heads'. Because it is well below the common 0.05 threshold, the result is strong evidence that the coin is biased (or the flips were not fair). This 'exact binomial test' is used in quality control, A/B tests with small samples and medical trials.

Binomial in genetics

Two parents who both carry one copy of a recessive gene have a 1 in 4 chance (p = 0.25) of each child inheriting the condition, independently for every child. With four children, the probability that exactly one is affected is C(4, 1) × 0.25 × 0.75³ ≈ 0.422, the probability that none is affected is 0.75⁴ ≈ 0.316, and the probability that at least one is affected is about 0.684. Genetic counsellors use these numbers to explain risks for each pregnancy.

Guessing on a multiple-choice test

Guessing every answer on a 10-question test with four options gives p = 0.25 per question. The expected score is 2.5, and the chance of getting 6 or more right by luck is only about 2.0%. On a 20-question test, the chance of guessing 12 or more correctly is less than 0.1%. This is why multiple-choice exams are hard to pass by chance – and why some exams subtract marks for wrong answers.

Binomial vs hypergeometric

When you sample without replacement from a small population, the probability changes after each draw, and the hypergeometric distribution gives the exact answer. Drawing 5 cards from a 52-card deck, the probability of exactly one ace is 0.299 using the hypergeometric distribution; treating each card as an independent 4/52 chance (binomial) gives 0.279. When the population is much larger than the sample – a common rule is at least 20 times larger – the binomial is a good approximation.

Confidence interval for a proportion

If 20 of 50 people in a survey say yes, the sample proportion is 0.40. A quick 95% confidence interval uses the normal approximation: 0.40 ± 1.96 × √(0.40 × 0.60 ÷ 50) = 0.40 ± 0.136, or about 26% to 54%. For small samples or proportions near 0 or 1, statisticians prefer the Wilson or Clopper–Pearson (exact binomial) intervals, which behave better.

Shape of the distribution

With n = 20 and p = 0.1, the distribution is strongly right-skewed: P(0) = 0.122, P(1) = 0.270, P(2) = 0.285, P(3) = 0.190, P(4) = 0.090 and P(5) = 0.032, with a long thin tail beyond. As n grows with p fixed, the skew fades and the bars approach the bell curve – the reason the normal approximation works for large samples. The calculator's chart lets you see this as you change n and p.

A little history

The binomial distribution is named after the binomial theorem, (a + b)ⁿ, whose expansion gives exactly the C(n, k) coefficients. Jacob Bernoulli studied repeated success/failure trials in Ars Conjectandi, published in 1713, and proved an early version of the law of large numbers: as n grows, the proportion of successes gets closer to p. Abraham de Moivre later showed that the binomial approaches the normal curve, an early form of the central limit theorem.

Choosing n and p from a real problem

Most binomial errors happen before any calculation, when n and p are set up. Ask three questions. What is one trial? (One flip, one customer, one part.) What counts as a success? (It does not have to be a good thing – a defect or a missed shot can be the 'success'.) Is p the probability for one trial, in decimal form? For example, if 3 in every 200 emails bounce and you send 1,000, then n = 1,000 and p = 3 ÷ 200 = 0.015, so the expected number of bounces is 15 with a standard deviation of about 3.84. Seeing 25 bounces – more than 2.5 standard deviations above the mean – would suggest the bounce rate has gone up.

Reading the calculator's chart and table

The chart shows a bar for every possible number of successes from 0 to n, with the bars for your selected range highlighted. The table lists P(X = k) and the running total P(X ≤ k), so you can read any cumulative probability directly. For large n, the calculator hides values that round to zero and marks the mean and one standard deviation either side, which is where about two-thirds of the probability lies when n is large.

Sources

1. NIST/SEMATECH e-Handbook of Statistical Methods – Binomial distribution — itl.nist.gov/div898/handbook
2. Microsoft Support – BINOM.DIST function — support.microsoft.com