Binomial Distribution Calculator

The probability of exactly k successes in n independent trials is P(X = k) = C(n, k) p^k (1 − p)^{n − k}, where p is the chance of success on each trial and C(n, k) = n! / (k!(n − k)!). For 5 heads in 10 fair flips, it is 252 × 0.5¹⁰ = 0.2461.

Choose n, p and k, then count the ways to place k successes among n trials with C(n, k), multiply by p to the k, then by (1 − p) to the n − k. The calculator does this for exact, cumulative and between-range questions and shows the answer to six decimals.

A binomial distribution describes the number of successes in a fixed number of independent trials, each with the same probability of success. Flipping a coin 10 times and counting heads, or checking 50 items for defects when each has a 2% chance, are examples.

There must be a fixed number of trials, two possible outcomes per trial (success or failure), independent trials and the same probability of success each time. If any of these fails, the binomial model does not fit. The checklist table above lists a failing example for each.

In the binomial formula, n is the number of trials, p is the probability of success on a single trial (a number between 0 and 1) and k is the number of successes you are asking about. Enter p as a decimal: 0.25 for 25%.

Add the probabilities of k, k + 1, up to n successes, or take 1 minus the probability of k - 1 or fewer. With 20 trials at p = 0.05, P(X ≥ 2) = 1 - P(X ≤ 1) = 0.2642. Choose the 'at least' option in the calculator.

Add the probabilities of 0, 1, up to k successes. For 10 fair coin flips, P(X ≤ 3) = 0.0010 + 0.0098 + 0.0439 + 0.1172 = 0.1719. This is the cumulative binomial probability, available as the 'at most' option.

Add the exact probabilities from the lower to the upper number of successes, inclusive. With n = 10 and p = 0.8, P(7 ≤ X ≤ 9) = 0.2013 + 0.3020 + 0.2684 = 0.7718. Choose 'between' and fill in both numbers.

Cumulative probability is the chance of getting k or fewer successes, P(X ≤ k), found by adding the exact probabilities from 0 up to k. The calculator shows it beside the exact probability, along with the 'at least' tail, so you can read all three answers for the same k at once.

The mean, or expected number of successes, is np. For 100 trials at p = 0.02 it is 2. It does not need to be a whole number: 20 trials at p = 0.05 has a mean of 1, and 30 trials at p = 0.05 has a mean of 1.5. It is a long-run average, not a prediction.

The variance is np(1 − p) and the standard deviation is its square root. For n = 10 and p = 0.5, the variance is 2.5 and the standard deviation is 1.581. The calculator reports both with every result, along with the mean, so you can see how spread out the counts will be.

It is the square root of np(1 − p). With n = 100 and p = 0.02 it is the square root of 1.96, which is 1.4. About two thirds of outcomes fall within one standard deviation of the mean for moderate n.

The binomial coefficient C(n, k) counts the ways to choose which k of the n trials are successes. It equals n! / (k!(n − k)!), so C(10, 5) = 252. These numbers form Pascal's triangle, and the coefficient table above lists them up to n = 12.

Use p = 0.5. For exactly 5 heads in 10 flips, the probability is C(10, 5) × 0.5¹⁰ = 252 / 1024 = 0.2461. For at least one head in 10 flips, it is 1 − 0.5¹⁰ = 0.9990. Both are in the tables above.

Treat rolling the target number as a success with p = 1/6 and the number of rolls as n. For exactly 2 sixes in 6 rolls the probability is C(6, 2) × (1/6)² × (5/6)⁴ = 0.2009. The dice table lists the common values.

With 4 options per question, p = 0.25. For 10 questions the chance of at least 6 correct is 0.0197. The quiz table lists the exact and 'at least' values for 0 to 10 correct. Use the calculator for other test lengths and option counts.

It is 1 minus the probability of no successes: 1 − (1 − p)^n. With p = 0.05 and 20 trials, it is 1 − 0.95²⁰ = 0.6415. The table above shows it for many combinations of n and p.

A common rule of thumb is when both np and n(1 − p) are at least 10 (some books use 5). The binomial is then roughly bell-shaped with mean np and standard deviation √(np(1 − p)). The calculator gives exact values, so the approximation is not needed.

Poisson is a good approximation of the binomial when n is large and p is small, with the mean np held moderate. It also stands alone for counting events in a fixed interval. See the Poisson distribution calculator linked below for that model.

The binomial is discrete: it counts whole successes in a fixed number of trials. The normal is continuous. For large n the binomial resembles a normal curve, but small n or p near 0 or 1 gives a skewed shape, so exact binomial values are better.

A Bernoulli trial is a single trial with two outcomes. A binomial random variable counts the number of successes in n independent Bernoulli trials with the same probability. With n = 1 the binomial reduces to a Bernoulli trial, so the Bernoulli distribution is the simplest special case of the binomial.

Exact probabilities spread over every possible count, so each is a fraction of the total. Even the most likely count rarely exceeds 0.25 for large n. A cumulative probability, such as at least or at most, covers many counts and is larger.

The mode is the whole number at or just below (n + 1)p. For n = 10 and p = 0.5 the mode is 5, with probability 0.2461. For n = 20 and p = 0.05 it is 1, with probability 0.3774. The distribution table shows where the peak falls.

Yes, when each item independently has the same chance of being defective. With 100 items and a 2% defect rate, the chance of 5 or more defects is 0.0508. The calculator gives the cumulative tail directly. For samples drawn without replacement from a small lot, the hypergeometric model is better.

Roughly, if people are chosen independently and each has the same chance of a given answer. For a small population sampled without replacement, the binomial is only an approximation. The calculator treats each trial as independent with a fixed p.

Up to 100,000 trials. Large values are computed with logarithms of factorials to avoid overflow, so results stay accurate. For counts that big, the normal approximation is also very close, but the calculator still returns the exact binomial figure rather than an estimate.

If p is 0, there are never any successes and P(X = 0) = 1. If p is 1, every trial succeeds and P(X = n) = 1. The calculator accepts p from 0 to 1 and returns these degenerate cases correctly.

Use C(n, 3) x p³ × (1 − p)^{n − 3}. With n = 10 and p = 0.5, C(10, 3) = 120, so the probability is 120 / 1024 = 0.1172. Choose 'exactly' and enter k = 3 in the calculator.

It is (1 − p)^n. With 10 trials at p = 0.5, the chance of no successes is 0.5¹⁰ = 0.000977, and with 20 trials at p = 0.05 it is 0.3585. Enter k = 0 and the 'exactly' option in the calculator.

Subtract the chance of none from 1: 1 − (1 − p)^n. For 20 trials at p = 0.05, this is 1 − 0.3585 = 0.6415. Choose 'at least' with k = 1 in the calculator for the same answer.

The cumulative distribution function gives P(X ≤ k), the sum of the exact probabilities from 0 to k. For n = 10 and p = 0.5, F(5) = 0.6230. The calculator shows this as 'at most k' next to the exact probability.

The probability mass function gives the probability of each exact count: P(X = k) = C(n, k) p^k (1 − p)^{n − k}. For n = 10 and p = 0.5 it ranges from 0.0010 at 0 or 10 to 0.2461 at 5. The chart above plots it.

Work out C(n, k) from factorials, raise p to the power k, raise 1 - p to the power n - k and multiply. For 4 heads in 6 flips, C(6, 4) = 15 and 0.5⁶ = 0.015625, so the probability is 15 × 0.015625 = 0.2344.