Binomial Distribution Calculator

What the binomial distribution answers

A binomial count answers one kind of question: out of a fixed number of tries, how many will come out one way? Ten free throws, 20 parts off a line, 50 orders that might arrive late. Each try has two possible results, which the math calls success and failure, and the count of successes is the random variable.

Four conditions have to hold before the binomial formula applies, and the OpenStax statistics text lists the same four: the number of trials n is fixed in advance, each trial is independent of the others, each trial has exactly two outcomes, and the chance of success p is the same on every trial.

When those four hold, the whole distribution is pinned down by n and p. You do not need the raw data, a histogram or a fitted curve. Two numbers describe the setup, and the formula below gives the chance of any count from 0 to n.

The results on this page are theoretical probabilities, not forecasts. A 23.35 percent chance of exactly 8 successes means that if you repeated the whole set of 10 trials many times, about 23 of every 100 sets would land on 8. It does not mean your next set will.

The formula, one piece at a time

The probability of exactly k successes in n trials is:

P(X = k) = C(n, k) x p^k x (1 - p)^(n - k)

Each factor does a separate job. C(n, k), read as n choose k, counts the number of different orders in which k successes and n minus k failures can appear. The factor p^k is the chance of the successes happening in one particular order. The factor (1 - p)^(n - k) is the chance of all the failures in that same order. Multiply the number of orders by the chance of one order, and you have the total.

The binomial coefficient, which MathWorld writes as C(n, k) = n! / (k! x (n - k)!), uses the exclamation mark for factorial, the product of every whole number down to 1. Order does not matter inside it: choosing which 3 of 20 parts are defective gives the same count as choosing which 17 are fine.

One warning about notation: some books write q for 1 - p, and some write the formula with the trials listed in the other order. The value is identical. The version above matches the one used by the NIST engineering statistics handbook.

Example: 8 free throws out of 10 at 70 percent

A player who makes 70 percent of free throws takes 10 shots. What is the chance of exactly 8 makes? Start with the number of orders: C(10, 8) = 10! / (8! x 2!) = 45. There are 45 different sequences of 10 shots that contain exactly 8 makes and 2 misses.

Next, the chance of one such sequence. The makes contribute 0.7^8 = 0.057648, and the misses contribute 0.3^2 = 0.09. Multiply those with the 45 orders: 45 x 0.057648 x 0.09 = 0.2335, which is 23.35 percent.

The chance of 8 or more makes is the sum of those three rows: 0.2335 + 0.1211 + 0.0282 = 0.3828, or 38.28 percent. The chance of 8 or fewer is 85.07 percent, and the two answers overlap on exactly 8, which is the 23.35 percent counted in both: 38.28 + 85.07 - 23.35 = 100.

The average count is n x p = 10 x 0.7 = 7 makes, which is also the single most likely count at 26.68 percent. A player who hits 8 is one make above average, and the standard deviation of 1.45 makes that gap look ordinary rather than remarkable.

At most, at least, and between

Most real questions are cumulative. A coach rarely asks for the chance of exactly 8 makes. The useful question is the chance of at least 8. That is the sum of every count from 8 to 10, and the calculator returns it in one step.

The complement rule saves work, but it is the place where people drop a count. The chance of 8 or more is 1 minus the chance of 7 or fewer, not 1 minus the chance of 8 or fewer. Here, P(X <= 7) = 61.72 percent, so P(X >= 8) = 1 - 0.6172 = 0.3828, or 38.28 percent. Subtract the wrong tail and you lose the 23.35 percent that belongs to exactly 8.

Between questions work the same way: subtract the chance of everything below the lower number. For 20 parts with a 25 percent defect rate, the chance of 4 to 8 defects is P(X <= 8) minus P(X <= 3), which is 0.9591 - 0.2252 = 0.7339, or 73.39 percent. For that same setup, 3 or fewer defects happens 22.52 percent of the time, and 3 or more happens 90.87 percent of the time, because the middle count is counted in both tails.

The two tails always add to 100 percent plus the chance of the count you split on. That is not a quirk of the arithmetic, it is a useful check: if 3 or fewer is 22.52 percent and 3 or more is 90.87 percent, the two numbers should overlap by exactly P(X = 3) = 13.39 percent, and 22.52 + 90.87 - 13.39 = 100.

Mean, variance and standard deviation

The mean of a binomial count is n x p. Ten shots at 70 percent average 7 makes, 20 parts at a 25 percent defect rate average 5 defects, and 50 orders at a 6 percent late rate average 3 late orders. The mean does not have to be a whole number, and for a single set of trials it does not have to be a count you can observe.

The variance is n x p x (1 - p), and the standard deviation is its square root. For the free throws that gives 10 x 0.7 x 0.3 = 2.1 for the variance and 1.45 makes for the standard deviation. For the 50 orders the variance is 50 x 0.06 x 0.94 = 2.82 and the standard deviation is 1.68 late orders.

Both come from adding n independent yes or no trials. Those two numbers set the scale for what counts as surprising. A count within one standard deviation of the mean is routine. Two standard deviations out is worth a second look at the process, and three is a signal that something changed. With a mean of 3 and a standard deviation of 1.68, six late orders out of 50 is under two standard deviations away, so a single bad week is not proof of a worse supplier.

Note how flat the peak is. Even at the most likely count, the chance is only about one in four. That is normal for a count variable, and it is why single results tell you so little about a process.

The most likely count and the shape of the distribution

The most likely count is (n + 1) x p, rounded down to a whole number. For 20 parts at p = 0.25 that is 21 x 0.25 = 5.25, so 5 defects is the peak. For 50 orders at p = 0.06 it is 51 x 0.06 = 3.06, so 3 late orders is the peak.

When p is 0.5 the distribution is symmetric, and the peak sits at the middle count. Twelve coin flips give P(X = 6) = 22.56 percent, and the chance of landing between 4 and 8 heads is 85.40 percent. Push p away from 0.5 and the shape leans: with p = 0.7 the long tail runs toward zero, and with p = 0.06 it runs toward the high counts.

That lean is why the normal curve is a rough stand-in at best for small samples. A common rule of thumb allows a normal approximation only when n x p is at least 10 and n x (1 - p) is at least 10. Ten shots at 70 percent give n x p = 7, which fails the rule, so the exact binomial is the right tool here. The normal distribution calculator takes over when both counts clear 10 and you want a quick z score instead of an exact sum.

When the binomial model does not fit

The formula is only as good as its four conditions, and everyday data breaks them in predictable ways. The first to check is independence. If one defective part makes the next one more likely, because a worn tool keeps drifting, the trials are not independent and the true spread is wider than n x p x (1 - p) suggests.

The second is a constant p. A supplier who starts the week strong and fades by Friday does not have one success chance across 50 orders. Split the data into groups that share a condition, such as shift or machine, and run the binomial inside each group.

The third is sampling without replacement. Drawing 20 parts from a bin of 25 changes the chance after each draw, so the count follows a hypergeometric distribution, not a binomial one. A common guard is to keep the population at least 10 times the sample size, and the binomial stays a close approximation.

Finally, count the outcomes. A trial with three or more possible results, such as a delivery that arrives early, on time or late, needs a multinomial model. A trial that keeps going until the first success needs a geometric model. Using the binomial on either one gives a confident answer to the wrong question.

Checking a result by hand

It helps to know what a hand calculation looks like, because a wrong entry in a calculator is easy to miss. Take 50 orders with a 6 percent late rate and ask for the chance of exactly 2 late orders. The orders term is C(50, 2) = 1,225. The success term is 0.06^2 = 0.0036. The failure term is 0.94^48 = 0.051302. Multiply the three: 1,225 x 0.0036 x 0.051302 = 0.2262, or 22.62 percent.

The chance of 2 or more late orders is 1 minus the chance of 1 or fewer: 1 - 0.19 = 0.81, or 81.00 percent. That is the number worth quoting to a supplier, since it describes the risk of a bad week rather than the chance of one exact count.

Three habits keep hand checks honest. Keep at least four decimal places until the final step, because 0.94^48 rounds to 0.05 and that error carries into the final answer. Compute the failure term as (1 - p) raised to n - k, not to n. And sanity check the answer against the mean: a count far from n x p should have a small probability, and if it does not, an exponent or a factorial is likely wrong.

The calculator above handles trial counts up to 2,000, which covers most classroom and quality control work. For counts beyond that, the normal approximation with a continuity correction becomes the practical route, and the exact formula starts to lose digits in ordinary floating point arithmetic.