Binomial Distribution Calculator

What this calculator answers

Given n independent trials that each succeed with the same probability p, it gives the chance of exactly k successes and of every “at most”, “fewer than”, “at least” and “more than” version of that question at once, or the chance of a range from k₁ to k₂. It also shows the mean, standard deviation and most likely count, the whole distribution as a table and chart, and the matching TI-84 and Excel calls. Two more questions use the same model: how many trials you need to reach a target chance, and the smallest k whose cumulative probability reaches a given level.

How to find binomial probabilities: exactly, at most, at least

• Number of trials (n): how many times the trial is repeated, a whole number up to 100,000.
• Probability of success (p): the chance of success on one trial. Type it as a decimal (0.85), a percent (85%) or a fraction (17/20, 1/6). A line under the field shows how it was read. On a phone, the / and % keys sit under the field.
• Probability of: the question in words. Pick exactly, at most, fewer than, at least, more than, or between k₁ and k₂.
• Number of successes (k): the count the question is about. A k above n is allowed; its probability is simply 0.

Results update as you type. The headline answers the question you picked, and the table below it answers the other four versions for the same k and lists the counts each one includes, so you can check that you picked the right one. The Try buttons load the examples on this page: two 6s in 12 dice rolls, fewer than 10 survey replies, a range of defects, the rolls needed for a 6, and a defect limit. Switch Find to “Trials needed to reach a chance” or “k at a cumulative probability” for the reverse questions.

Binomial probability formula

The chance of exactly k successes multiplies the number of ways to place k successes among n trials by the chance of any one such arrangement:

$$P(X = k) = \binom{n}{k}\, p^{k} (1 - p)^{n-k}, \qquad \binom{n}{k} = \frac{n!}{k!\,(n-k)!}$$

The cumulative probability adds these terms from 0 up to k:

$$P(X \le k) = \sum_{i=0}^{k} \binom{n}{i}\, p^{i} (1 - p)^{n-i}$$

• $n$ is the number of trials and $k$ the number of successes.
• $p$ is the chance of success on each trial, and $1 - p$ the chance of failure.
• $\binom{n}{k}$, read “n choose k”, counts the ways to choose which k trials succeed.

Every other version follows from these two: $P(X < k) = P(X \le k - 1)$, $P(X \ge k) = 1 - P(X \le k - 1)$ and $P(X > k) = 1 - P(X \le k)$.

Worked example: at least 20 of 24 seeds sprout

A seed packet says 85% of its seeds germinate. You plant 24. What is the chance that at least 20 sprout?

1. Here n = 24 and p = 0.85, so 1 − p = 0.15. “At least 20” means 20, 21, 22, 23 or 24 sprout.
2. Exactly 20: P(X = 20) = C(24, 20) × 0.85²⁰ × 0.15⁴ = 10,626 × 0.0387595 × 0.00050625 = 0.208504.
3. The same formula gives P(X = 21) = 0.225051, P(X = 22) = 0.173903, P(X = 23) = 0.0856915 and P(X = 24) = 0.0202327.
4. Add them: P(X ≥ 20) = 0.713382, about 71.34%. The shortcut gives the same: 1 − P(X ≤ 19) = 1 − 0.286618 = 0.713382.

The mean is 24 × 0.85 = 20.4 seedlings, with a standard deviation of 1.7493, and the most likely count is 21. The chance that between 18 and 22 sprout, inclusive, is 0.836884.

“At least” vs. “more than”: picking the right inequality

“At least 20” includes 20; “more than 20” starts at 21. In the seed example, at least 20 has probability 0.713382 while more than 20 has only 0.504879. The gap, 0.208504, is exactly P(X = 20). The same trap applies to “at most” (includes k) and “fewer than” (stops at k − 1).

“Between” is ambiguous in everyday speech, so the range option shows both readings: including k₁ and k₂, and strictly between them.

Survey example. If each of 40 invited people replies with probability 30%, the chance of fewer than 10 replies is P(X ≤ 9) = 0.195925, not P(X ≤ 10) = 0.308743.

Dice example. The chance of exactly two 6s in 12 rolls is C(12, 2) × (1/6)² × (5/6)¹⁰ = 66 × 0.0277778 × 0.161506 = 0.296094.

Range example. If 1.5% of parts are defective, the chance that a sample of 200 holds 2 to 5 defective parts is P(X ≤ 5) − P(X ≤ 1) = 0.917608 − 0.196897 = 0.720712.

Binomial distribution table and chart

For up to 200 trials the table lists every k from 0 to n with P(X = k), P(X ≤ k) and P(X ≥ k), and the bar chart draws P(X = k), highlighting the bars your question counts. Bars too small to see are left off the chart, and its Chart data table lists every value drawn. With more than 200 trials the table shows the 201 values of k nearest the mean. Download CSV saves either table with full precision.

Mean, variance and standard deviation of a binomial distribution

The mean is $\mu = np$, the variance $\sigma^2 = np(1 - p)$ and the standard deviation $\sigma = \sqrt{np(1 - p)}$. The mean is the long-run average count, not necessarily a count that can happen: 24 seeds at 85% average 20.4 seedlings. The most likely count, the mode, is the whole number between (n + 1)p − 1 and (n + 1)p: for the seeds, 25 × 0.85 = 21.25, so the mode is 21. When (n + 1)p is a whole number, that count and the one just below it tie: 5 rolls of a die give (5 + 1) × 1/6 = 1, and 0 sixes and 1 six are equally likely.

When the binomial model applies (and when it doesn’t)

The count of successes is binomial only when all four hold:

1. The number of trials is fixed in advance.
2. Each trial has two outcomes, success or failure.
3. The trials are independent.
4. The chance of success is the same on every trial.

Drawing without replacement breaks the last two: after a red card leaves the deck, the next card is less likely to be red. That setting is the hypergeometric distribution. When the sample is a small share of the group, under about 5%, the binomial is a close stand-in.

Matching binompdf, binomcdf and BINOM.DIST

binompdf gives the chance of exactly k; binomcdf gives the cumulative chance of k or fewer. The calculator lists the matching calls for your inputs:

Excel’s BINOM.DIST puts the successes first and returns #NUM! when k is more than n. For the seeds, =1-BINOM.DIST(19, 24, 0.85, TRUE) returns 0.713382.

How many trials do you need?

You need the smallest n for which P(X ≥ k), the chance of at least k successes, reaches the chance you want. Choose Trials needed to reach a chance and enter p, the number of successes you need and that chance; the calculator finds n and shows the chance one trial earlier. To be 90% sure of at least 20 seedlings from the 85% packet, plant 26 seeds: the chance is 0.916673 with 26 and 0.838485 with 25.

For at least one success there is a shortcut, $n \ge \ln(1 - \text{target}) \div \ln(1 - p)$. To have a 95% chance of rolling at least one 6, ln(0.05) ÷ ln(5/6) = 16.431, so you need 17 rolls, with a chance of 0.954927 (16 rolls give 0.945912). A table repeats the search for targets from 50% to 99.9%.

Finding k for a cumulative probability (BINOM.INV)

Choose k at a cumulative probability to find the smallest k with P(X ≤ k) at or above a level, which is what Excel’s BINOM.INV returns. Quality checks use it to set a limit: if 2% of parts are defective, a sample of 500 has 15 or fewer defective parts with probability 0.953003, while 14 or fewer has only 0.918643, so 15 is the 95th percentile, and more than 15 happens with probability 0.0469971. Because the count jumps in whole numbers, P(X ≤ k) usually passes the level rather than landing on it.

Normal approximation to the binomial

When np and n(1 − p) are both large, a normal distribution with mean np and standard deviation √(np(1 − p)) is close to the binomial. Treat each whole number k as the interval from k − 0.5 to k + 0.5 (the continuity correction). The usual rule asks for np and n(1 − p) both above 5, and the match is better when both are 10 or more. The calculator shows this estimate next to the exact value so you can see the difference:

• Survey: np = 12 and n(1 − p) = 28, so fewer than 10 replies is P(Y < 9.5) = 0.194184, close to the exact 0.195925.
• Seeds: n(1 − p) = 3.6, below 5, and P(Y > 19.5) = 0.696547 misses the exact 0.713382 by 0.0168.

With the exact value available, use the approximation only to check your intuition or to follow a textbook method. Continue in the Normal Distribution and Z-Score Calculator carries np, √(np(1 − p)) and the corrected cut-off over, so you can see the same approximation on a bell curve.

Very large n and very small probabilities

For up to 1,000 trials the calculator normally computes each probability as an exact fraction and rounds it once. Above that, it switches to a log-space saddle-point method that stays accurate to about 12 significant digits for up to 100,000 trials. The accuracy note under the results says which method was used. Probabilities below 1E-300 are reported as ”< 1E-300” rather than as 0: with n = 200 and p = 0.01, the chance that all 200 trials succeed is 0.01²⁰⁰ = 10⁻⁴⁰⁰, tiny but not impossible. A probability shown as ”> 0.999999” is below 1 but rounds to 1 at six digits; its complement in the table shows how much is left.

Common mistakes

• Reading “at least” as “more than”. At least 20 includes 20. In the seed example the two differ by 0.208504.
• Using binomcdf for “at least”. binomcdf(n, p, k) is P(X ≤ k). At least k is 1 − binomcdf(n, p, k − 1), with k − 1, not k.
• Typing a percent as a whole number. p = 85 is not a probability. Type 85% or 0.85.
• Swapping n and k in BINOM.DIST. Excel wants the successes first: BINOM.DIST(k, n, p, …).
• Using the binomial for draws without replacement from a small group. Five cards from one deck are not independent trials; the hypergeometric distribution fits.
• Rounding p early. Using 0.17 for 1/6 turns the chance of two 6s in 12 rolls from 0.296094 into 0.295953. Type 1/6.