Binomial Distribution Calculator: Probability and Formula

A binomial distribution calculator finds the probability of getting a specified number of successes in a fixed number of independent trials, when each trial has the same success probability. Enter the number of trials, the success probability, and the success count; the calculator returns the probability of exactly that many successes, along with useful cumulative probabilities.

Formula

The probability of exactly k successes in n trials is:

P(X = k) = C(n, k) p^k (1 − p)^{n − k}

• n is the total number of trials, a nonnegative whole number.
• k is the number of successes being counted, from 0 through n.
• p is the probability of success on each trial, from 0 to 1.
• C(n, k), also written “n choose k,” counts how many different arrangements have k successes among n trials.

Probability has no physical unit. The calculator displays it as a decimal between 0 and 1; multiply by 100 to express it as a percentage. “At most k” means X ≤ k and includes k. “At least k” means X ≥ k and also includes k.

Worked Example

Using the calculator’s defaults, n = 10, p = 0.5, and k = 5:

1. There are C(10, 5) = 252 ways to arrange five successes in ten trials.
2. The probability of each particular arrangement is 0.5⁵ × 0.5⁵ = 0.0009765625.
3. Multiply: P(X = 5) = 252 × 0.0009765625 = 0.24609375, which rounds to 0.246094.

For these same inputs, the probability of at most five successes is 0.623047, and the probability of at least five is 0.623047. The equality of those cumulative results follows from the distribution’s symmetry when p = 0.5 and n = 10.

What the Calculator Tells You

The first result is the chance of exactly the selected number of successes. For the default example, the probability of exactly five successes is about 24.61%. This does not mean that five successes are guaranteed or that the same outcome will occur in every group of ten trials. It describes the long-run probability under the stated model assumptions.

The cumulative results answer common threshold questions without requiring you to add individual probabilities by hand. For example, if a process has ten attempts and a success chance of 0.5 on each, “at most five” includes outcomes of zero, one, two, three, four, or five successes. “At least five” includes five through ten.

For a range such as “between three and six successes, inclusive,” subtract cumulative probabilities: P(3 ≤ X ≤ 6) = P(X ≤ 6) − P(X ≤ 2). The calculator provides the components, but you must enter the relevant k values and do the subtraction.

When a Binomial Model Applies

A binomial model is appropriate only when its conditions fit the situation:

• The number of trials is fixed in advance.
• Each trial has two outcomes classified as success or failure. “Success” is just the outcome of interest, not necessarily a desirable result.
• Trials are independent: the result of one does not change the probability of another.
• The probability of success, p, is the same for every trial.

For example, if a fair coin is tossed ten times, “heads” can be defined as success. Then n = 10 and p = 0.5. By contrast, drawing several items from a small collection without replacing them can change the chance of success from draw to draw. That situation is generally not binomial unless the draws are independent or the changing probabilities are negligible for the question at hand.

Common Input and Interpretation Mistakes

• Entering a percentage as a whole number: use 0.25 for 25%, not 25. The success probability must be between 0 and 1.
• Using a noninteger count: n and k must be whole numbers, and k cannot exceed n. A count such as 3.5 successes has no meaning in this model.
• Confusing exactly with at least: “exactly four” is P(X = 4); “four or more” is P(X ≥ 4). These can be very different.
• Assuming independence or a constant probability without checking: repeated measurements, sampling without replacement, and changing conditions may violate the binomial assumptions.
• Rounding intermediate results: rounding each term before adding can distort cumulative probabilities. Keep full precision during the calculation and round only the displayed result.

For extremely large n, direct calculations may exceed ordinary calculator precision. Statistical software often uses numerically stable algorithms for such cases. Also, the calculator cannot decide whether a real-world definition of “success” is appropriate or whether the trials truly meet the assumptions; those are modeling decisions.

Related Measures and Alternatives

The expected number of successes is np, and the variance is np(1 − p); the standard deviation is √[np(1 − p)]. These summarize the distribution’s center and spread, but neither is the probability of one particular outcome. A mean can be fractional even though the observed success count must be an integer.

Other probability models may be more suitable when the binomial assumptions do not hold. A hypergeometric model describes a fixed number of draws without replacement from a finite population. A Poisson model can approximate counts of relatively rare events under suitable conditions, while a geometric model concerns the number of trials until a first success. Choosing among them depends on how the data are generated, not merely on which formula is easiest to use.

Frequently Asked Questions

How do I calculate binomial probability?
Use P(X = k) = C(n, k)p^k(1 − p)^{n − k}, where n is the number of trials, k is the success count, and p is the per-trial success probability.

How do I find the probability of at least k successes?
Add the probabilities from k through n, inclusive. Equivalently, calculate 1 − P(X ≤ k − 1). The calculator’s “at least” output includes k.

Can the probability of exactly k successes be larger than 50%?
Yes. A particular outcome can be more likely than all other individual counts, especially when the distribution is concentrated. The probabilities for all possible counts together still sum to 1.

What does n choose k mean?
It is the number of distinct ways to select which k of the n trials are successes. Its formula is C(n, k) = n! / [k!(n − k)!].

What if my trials are not independent?
A standard binomial calculation may not represent the situation accurately. Consider a model that accounts for dependence or changing probabilities, such as the hypergeometric distribution for sampling without replacement.

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in n | Number of trials | - | 10 | 0
in p | Probability of success | - | 0.5 | 0
in k | Number of successes | - | 5 | 0
out exact | Probability of exactly k successes | - | ncr(n,k) * p^k * (1-p)^(n-k) | 6
out atmost | Probability of at most k successes | - | sum(i,0,k,ncr(n,i) * p^i * (1-p)^(n-i)) | 6
out atleast | Probability of at least k successes | - | sum(i,k,n,ncr(n,i) * p^i * (1-p)^(n-i)) | 6