The binomial distribution counts successes in a fixed number of independent yes-or-no trials, each with the same chance of success. It answers questions like “If 30% of customers redeem a coupon, what is the chance that at most 5 of the next 20 do?” Enter the number of trials, the success probability and the count you care about; the calculator returns the exact probability for any event (exactly, at most, fewer than, at least, more than or between), the mean and standard deviation, a highlighted bar chart and a distribution table.
How to use the binomial probability calculator
- Enter the number of trials n and the probability of success p as a decimal (0.3 for 30%).
- Choose the event: exactly k, at most k, fewer than k, at least k, more than k, or between k and k₂.
- Enter k (and k₂ for a range).
- Read the probability on the tape. The chart highlights the bars included in your event, and the table lists P(X = k), P(X ≤ k) and P(X ≥ k) for every relevant k.
Binomial formula
The probability of exactly k successes is
where C(n, k) = n! ÷ (k!(n − k)!) counts the ways to place k successes among n trials. Cumulative probabilities add these terms; for example P(X ≤ k) sums from 0 to k. The distribution’s center and spread are
Worked example
A store finds that 30% of customers use a mailed coupon. Of the next 20 customers, what is the chance that at most 5 use it?
- Exactly 5: C(20, 5) × 0.3⁵ × 0.7¹⁵ = 15,504 × 0.00243 × 0.0047476 = 0.178863.
- At most 5 adds the terms for 0 through 5: 0.000798 + 0.006839 + 0.027846 + 0.071604 + 0.130421 + 0.178863 = 0.416371.
- So there is about a 41.6% chance that 5 or fewer use the coupon, and a 58.4% chance that 6 or more do.
- The mean is 20 × 0.3 = 6 customers, with a standard deviation of √4.2 = 2.049.
Between 4 and 8 customers inclusive covers 0.779582 of the probability, which is why a planner might stock enough coupons for 8 redemptions and expect to be right roughly 89% of the time (P(X ≤ 8) = 0.886669).
Getting the event right
Wording matters, and off-by-one mistakes are the most common error:
| Phrase | Event | Equivalent |
|---|---|---|
| at most 5 | X ≤ 5 | 1 − P(X ≥ 6) |
| fewer than 5 | X < 5, i.e. X ≤ 4 | 1 − P(X ≥ 5) |
| at least 5 | X ≥ 5 | 1 − P(X ≤ 4) |
| more than 5 | X > 5, i.e. X ≥ 6 | 1 − P(X ≤ 5) |
| no more than 5 | X ≤ 5 | same as at most |
When the binomial does not apply
If you sample without replacement from a small population, the trials are not independent, and the hypergeometric distribution is correct. A rough rule is that the binomial is fine when the sample is under 10% of the population. If successes are counted over continuous time or space with no fixed n, such as calls per hour, use the Poisson distribution calculator. The Poisson is also the limit of the binomial when n is large and p is small, with λ = np.
Connection to counting
The C(n, k) factor is the binomial coefficient; the combinations calculator computes it exactly for any size. For a probability with two events rather than repeated trials, see the probability calculator.
Frequently asked questions
When does a situation follow a binomial distribution?
When four conditions hold: a fixed number of trials n, each trial has only two outcomes (success or failure), the probability of success p is the same every time, and the trials are independent. Flipping a coin 10 times, or checking 50 parts from a stable process, fits; drawing cards without replacement from a small deck does not.
What is the difference between binomial PDF and CDF?
The probability mass function (often called binompdf on calculators) gives the chance of exactly k successes. The cumulative distribution function (binomcdf) adds those up from 0 to k, giving the chance of at most k successes. This calculator shows both, plus the at-least and more-than tails.
How do I calculate the probability of at least one success?
Use the complement: P(X ≥ 1) = 1 − P(X = 0) = 1 − (1 − p)^n. With p = 0.3 and n = 20 that is 1 − 0.7^20 ≈ 0.9992. Choose At least k with k = 1 to get the same result.
How large can n be?
Up to one billion trials. Probabilities are computed with log-gamma functions and the regularized incomplete beta function rather than by multiplying huge numbers, so they stay accurate even when the binomial coefficient has millions of digits.
When can I use the normal approximation instead?
A common rule is np ≥ 10 and n(1 − p) ≥ 10. Then the binomial is close to a normal curve with mean np and SD √(np(1 − p)), using a continuity correction of 0.5. When the rule holds, the calculator prints the approximation next to the exact value so you can compare.