Independent trials

Binomial Probability Calculator

Find the probability of a chosen number of successes when every trial has the same independent chance of success.

Calculate a binomial probability

A whole number from 0 to 1,000.

Enter 50 for 50%, not 0.5. Range: 0–100%.

Success counts must be whole numbers from 0 to n. Every event includes its stated boundary.

Example: exactly 3 heads in 10 fair coin flips.

Event probability

Choose your trials, success chance, and event, then select Calculate.

≈ marks a rounded display. Calculations run in your browser. Press Enter to calculate. Successful results appear in this tab's Recent calculations.

Coin flips, event ranges, and complements

Exactly

Exactly 3 heads in 10 fair flips

P(X = 3) = C(10, 3) × 0.5³ × 0.5⁷

There are 120 ways to place three heads among ten flips. Each complete sequence has probability 1/1,024. The event probability is 120/1,024 = 11.71875%. The expected number of heads is 10 × 0.5 = 5.

At least

At least 1 success in 5 independent trials

P(X ≥ 1) = 1 − P(X = 0)

With a 20% chance each time, all five trials fail with probability 0.8⁵ = 0.32768. At least one success therefore has probability 1 − 0.32768 = 67.232%. Five trials do not guarantee a success.

Inclusive range

3 through 7 heads in 10 fair flips

P(3 ≤ X ≤ 7) = P(3) + … + P(7)

Add the five counts' sequence totals: 120 + 210 + 252 + 210 + 120 = 912. Dividing by 1,024 gives 89.0625%. Its complement is 10.9375%: fewer than 3 or more than 7 heads.

Formula, input limits, and precision

For X, the number of successes in n independent trials with a constant success probability p, the binomial formula is P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ. Here C(n, k) = n! ÷ [k! × (n − k)!]. Cumulative events add the probabilities for all included whole-number counts. The complement is 1 minus the selected event probability, and the expected count is n × p.

The probability field uses percentages: 0.5 means 0.5%, while 50 means 50%. Enter up to 10 decimal places with a period; a percent sign, expressions, and scientific notation are not accepted. Probability input is limited to 64 characters and 15 integer digits before its 0–100 range is checked. Trial and success counts accept digits and correctly grouped thousands commas, up to 16 characters; they must be whole numbers from 0 to 1,000, with success counts no greater than n.

With zero trials, X equals 0 with probability 1. With a 0% success chance, every trial fails; with a 100% chance, every trial succeeds. These cases are evaluated directly, including the usual empty-product convention behind zero-trial probabilities.

Finite decimal inputs are converted to exact fractions. Integer binomial weights are summed before division, so very small probabilities are retained without subtracting rounded decimal values. The 1,000-trial limit keeps this exact calculation bounded in your browser. This calculator does not use a normal approximation or simulation.

Only the final display is rounded, with halfway values rounded up and trailing zeros omitted. ≈ indicates rounding. A value displayed as ≈ 0% can still be positive, and ≈ 100% can still be below certainty. In these cases, additional decimal scientific notation is shown to six significant figures; its own ≈ marker indicates rounding. Changing decimal places does not add a recent-calculation entry.

A mathematically exact calculation still depends on your assumed probability and independence. Drawing items without replacement from a small collection usually changes the chance after each draw and calls for a different model. This calculator does not estimate p from observations, produce confidence intervals, or model dependent outcomes.

Reference: NIST/SEMATECH e-Handbook — Binomial Distribution.

Common questions

Does “at least” include the number I enter?

Yes. At least k means k or more successes, and at most k means k or fewer. The range option includes both endpoints. To calculate more than 3 successes, choose at least 4.

Why is an expected count allowed to be a decimal?

Expected successes describes an average over many repetitions of the whole experiment. For example, three trials at a 50% chance have an expected count of 1.5, even though each actual experiment has 0, 1, 2, or 3 successes.

Can I use this for card draws without replacement?

Generally no. Removing a card changes the next draw's probability, so the trials do not meet this model's independence and constant-probability conditions. Drawing with replacement and reshuffling can meet those conditions if every draw has the same chance.

Why is a tiny probability shown as ≈ 0?

The chosen number of decimal places may not be enough to show a positive value. The calculation retains its exact integer fraction and also shows scientific notation for extreme displayed probabilities. A rounded zero does not mean the event is impossible.