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Binomial probability for a fixed number of trials

Enter fixed trials, a success percentage and a success count. Compare exactly, at least and at most without confusing overlapping events or guaranteed outcomes.

Start with your own numbers

Editable example

An example is filled in. Replace it with your own data to calculate.

First check that trials are independent and have the same probability. Example: ten fair coin tosses, a 50% chance of heads each time and five heads of interest. Drawing without replacement or changing chances does not fit this model.

Enter a fixed total from 0 to 1000, such as 10 for ten coin tosses. This is not a rule to keep trying until success. With zero trials the only allowed success count is zero.
Enter a percentage: 50 means 50%, while 0.5 means 0.5%, not 50%. Use the same reliable probability for each trial; do not guess unknown chances. Both 0 and 100 are allowed, with up to six decimal places.
Enter the success count you want to compare, from zero to the total trials. At least 5 includes 5; at most 5 also includes 5. Success simply names the event you choose, not necessarily a good outcome.

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Example result

Probability of exactly the chosen count

24.609375%

The main answer is exactly the chosen count. At least and at most both include that count; do not add the three probabilities. These are model probabilities, not a promise about the next batch. Tiny positive chances are not shown as zero.

Probability of at least the chosen count (inclusive)62.304688 %
Probability of at most the chosen count (inclusive)62.304688 %
Theoretical mean success count (not guaranteed)5 times

Worked steps

  1. n = 10, k = 5, p = 50 ÷ 100
  2. P(X = 5); P(X ≥ 5); P(X ≤ 5)

For k successes in n trials, P(X=k) = C(n,k) × p^k × (1−p)^(n−k), with p equal to the entered percentage divided by 100. At most adds counts 0 through k; at least adds k through n. Each tail is summed directly using logarithms for numerical stability. Exactly refers only to the chosen count; at least and at most include it and overlap. Do not add the three answers. The theoretical mean n×p is a long-run average, not the count guaranteed in each batch. Zero trials can only have zero successes. The model does not fit dependent trials, draws without replacement, guaranteed outcomes after retries or changing chances. Values are floating-point approximations: a positive chance below 0.000001% is shown as <0.000001, and a non-certain chance close to 100% as >99.999999. Save the event definition and probability source with your plan. Arrangement counts and observed-data statistics belong to the related tools. No actual outcome, investment gain or medical conclusion is guaranteed.

Method and scope

The main answer is exactly the chosen count. At least and at most both include that count; do not add the three probabilities. These are model probabilities, not a promise about the next batch. Tiny positive chances are not shown as zero.

The model does not fit dependent trials, draws without replacement, guaranteed outcomes after retries or changing chances. Values are floating-point approximations: a positive chance below 0.000001% is shown as <0.000001, and a non-certain chance close to 100% as >99.999999. Save the event definition and probability source with your plan. Arrangement counts and observed-data statistics belong to the related tools. No actual outcome, investment gain or medical conclusion is guaranteed.

References:NIST: binomial probability and cumulative formulas、NTU Mathematics: binomial distribution and the law of large numbers

What this tool helps you do

Estimate a success count in a fixed number of independent trials with two outcomes and the same chance each time. This is a probability model, not a count of arrangements or an automatic estimate of an unknown success chance.

What to enter

Use whole numbers for trials and successes, with successes no greater than trials. Enter percentages: 50 for 50%, or 0.5 for 0.5%. Define the event consistently; if you count failures as successes, use the probability of that event.

How it is calculated

For k successes in n trials, P(X=k) = C(n,k) × p^k × (1−p)^(n−k), with p equal to the entered percentage divided by 100. At most adds counts 0 through k; at least adds k through n. Each tail is summed directly using logarithms for numerical stability.

Worked example

Synthetic example: ten fair coin tosses with a 50% chance of heads. Exactly five heads has probability 252/1024 = 24.609375%. At least five and at most five each have probability 638/1024, displayed as approximately 62.304688%. Both include exactly five.

Understanding the answer

Exactly refers only to the chosen count; at least and at most include it and overlap. Do not add the three answers. The theoretical mean n×p is a long-run average, not the count guaranteed in each batch. Zero trials can only have zero successes.

Scope and limitations

The model does not fit dependent trials, draws without replacement, guaranteed outcomes after retries or changing chances. Values are floating-point approximations: a positive chance below 0.000001% is shown as <0.000001, and a non-certain chance close to 100% as >99.999999. Save the event definition and probability source with your plan. Arrangement counts and observed-data statistics belong to the related tools. No actual outcome, investment gain or medical conclusion is guaranteed.

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