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Binomial Distribution Calculator

Calculate probabilities for the binomial distribution

Results:

P(X = 5): 24.6094%

P(X ≤ 5): 62.3047%

P(X > 5): 37.6953%

Binomial Coefficient C(10,5): 252

Mean (Expected Value): 5.0000

Variance: 2.5000

Standard Deviation: 1.5811

About Binomial Distribution

The binomial distribution models the number of successes in n independent trials, each with probability p of success.

PMF: P(X = k) = C(n,k) × p^k × (1-p)^(n-k)

Mean: μ = np

Variance: σ² = np(1-p)

What Is the Binomial Distribution?

The binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success. It is the extension of the Bernoulli distribution from a single trial to n trials. If you flip a fair coin 10 times, the number of heads follows a Binomial(n=10, p=0.5) distribution.

The distribution is defined by two parameters: n (the number of trials) and p (the probability of success on each trial). It answers questions like "What is the probability of getting exactly 5 heads in 10 flips?" or "What is the probability of at most 3 defects in a batch of 100 items?"

This calculator computes the exact probability P(X=k), the cumulative probability P(X≤k), the complementary probability P(X>k), the binomial coefficient, mean, variance, and standard deviation — everything you need for binomial probability analysis.

The Binomial Probability Formula

The probability mass function (PMF) of the binomial distribution combines the binomial coefficient — which counts the number of ways to choose k successes from n trials — with the probability of each specific sequence of k successes and (n−k) failures.

Binomial PMF

P(X = k) = C(n,k) × p^k × (1−p)^(n−k)

Where:

  • n= Number of independent trials
  • p= Probability of success on each trial (0 to 1)
  • k= Number of successes observed (0 to n)
  • C(n,k)= Binomial coefficient = n! / (k! × (n−k)!) — number of ways to choose k from n

Interpreting Binomial Probabilities

The binomial probability P(X=k) gives the exact likelihood of observing k successes in n trials. The cumulative probability P(X≤k) sums probabilities for 0 through k successes — useful for "at most" questions. P(X>k) = 1−P(X≤k) answers "more than" questions.

Scenario Formula Use Case
Exactly kP(X=k)"What is the chance of exactly 5 heads?"
At most kP(X≤k)"Probability of 3 or fewer defects?"
More than k1−P(X≤k)"Chance of more than 8 correct answers?"

The mean np tells you the expected number of successes — on average, how many successes you'd see if you repeated the n-trial experiment many times. The variance np(1−p) reaches its maximum at p = 0.5, reflecting maximum unpredictability when success and failure are equally likely.

How to Use This Calculator

Enter three values to get complete binomial distribution results:

  1. Number of Trials (n): How many independent experiments or observations are you making? Enter a non-negative integer — for example, 10 coin flips, 100 survey respondents, or 50 manufactured items.
  2. Probability of Success (p): The probability that any single trial results in success. Enter as a decimal between 0 and 1. For a fair coin, p = 0.5; for a process with 2% defect rate, p = 0.02.
  3. Number of Successes (k): The specific outcome you want to evaluate. Must be between 0 and n. The calculator shows P(X=k), P(X≤k), and P(X>k) for this value.
  4. Read the results: The calculator outputs exact probability, cumulative probability, complementary probability, binomial coefficient, mean, variance, and standard deviation.

Real-World Applications

The binomial distribution powers decision-making across many fields. In quality control, manufacturers sample n items from a production run and use the binomial distribution to determine whether the number of defects k is consistent with an acceptable defect rate p. Acceptable Quality Level (AQL) sampling plans are built on binomial probability calculations.

In medical research, the binomial distribution models the number of patients who respond to a treatment out of n trial participants. Sample size calculations for clinical trials rely on binomial power analysis to ensure the study can detect a clinically meaningful response rate.

In finance and credit scoring, banks model the number of loan defaults in a portfolio of n loans using the binomial distribution. This feeds into loan loss reserves and risk-based pricing models. In genetics, the binomial distribution calculates the probability of inheriting a specific number of dominant alleles in offspring, forming the basis of Mendelian inheritance predictions.

Worked Examples

Coin Flips — Exactly 5 Heads

Problem:

You flip a fair coin (p = 0.5) 10 times. What is the probability of getting exactly 5 heads?

Solution Steps:

  1. 1Step 1: Enter n = 10, p = 0.5, k = 5 into the calculator.
  2. 2Step 2: The binomial coefficient C(10,5) = 10!/(5!×5!) = 252. This is the number of distinct sequences with exactly 5 heads.
  3. 3Step 3: P(X=5) = 252 × 0.5^5 × 0.5^5 = 252 × 0.5^10 = 252/1024 ≈ 0.2461.
  4. 4Step 4: CDF P(X≤5) sums probabilities for k = 0 through 5, giving approximately 0.6230 — a 62.3% chance of 5 or fewer heads.

Result:

The probability of exactly 5 heads in 10 flips is 24.61%. The probability of 5 or fewer heads is 62.30%. The most likely outcome (5 heads) still occurs less than 25% of the time — binomial outcomes are spread across a range of possible k values.

Quality Control Sampling

Problem:

A factory produces items with a 2% defect rate. In a sample of 100 items, what is the probability of finding 0 defects? Of finding at most 3 defects?

Solution Steps:

  1. 1Step 1: Enter n = 100, p = 0.02, k = 0 to find P(X=0).
  2. 2Step 2: P(X=0) = C(100,0) × 0.02^0 × 0.98^100 = 1 × 1 × 0.1326 ≈ 0.1326, or 13.26%.
  3. 3Step 3: For P(X≤3), sum P(X=0)+P(X=1)+P(X=2)+P(X=3). Set k = 3 and use the CDF output.
  4. 4Step 4: C(100,1)=100, C(100,2)=4950, C(100,3)=161700. Compute each term: P(X=1)=100×0.02×0.98^99≈0.2707; P(X=2)=4950×0.0004×0.98^98≈0.2734; P(X=3)=161700×0.000008×0.98^97≈0.1823.

Result:

P(X=0) ≈ 13.26%. P(X≤3) ≈ 0.1326+0.2707+0.2734+0.1823 ≈ 0.859, or 85.9%. A sample of 100 items has an 85.9% chance of containing 3 or fewer defects, but also a 14.1% chance of 4 or more — important for setting inspection thresholds.

Multiple Choice Guessing

Problem:

A multiple-choice test has 20 questions, each with 4 options. A student guesses randomly on every question (p = 0.25). What is the probability of getting at least 10 correct?

Solution Steps:

  1. 1Step 1: Enter n = 20, p = 0.25, k = 9 into the calculator to find P(X≤9).
  2. 2Step 2: The calculator computes the cumulative probability for 0 through 9 successes. With n=20 and p=0.25, the mean is np = 5 — the expected score from random guessing.
  3. 3Step 3: P(X≥10) = 1 − P(X≤9). Use the CDF at k=9 to find P(X≤9), then subtract from 1.
  4. 4Step 4: The probability of scoring 10 or above by pure guessing is very small — only about 0.6%.

Result:

P(X≥10) ≈ 0.006, or 0.6%. A student would need to correctly guess 10+ out of 20 questions, and the chance of this occurring randomly is less than 1%. This illustrates why random guessing is a poor strategy — the expected score is only 5 out of 20.

Tips & Best Practices

  • ✓The binomial distribution is symmetric only when p = 0.5 — for p ≠ 0.5, the distribution is skewed.
  • ✓Always check that your trial conditions are genuinely independent — dependent outcomes require different models.
  • ✓Use P(X≤k) for 'at most' questions and 1−P(X≤k) for 'at least' questions — the calculator gives both.
  • ✓The variance np(1−p) peaks at p = 0.5 — extreme p values (near 0 or 1) produce less variable outcomes.
  • ✓For very large n, the binomial coefficient grows combinatorially — the calculator handles the computation efficiently.

Related Calculators

Bernoulli Distribution Calculator

Model a single binary trial (the building block of binomial)

Binomial Probability Calculator

Alternative binomial calculator with additional visualization options

Geometric Distribution Calculator

Count trials until the first success

Poisson Distribution Calculator

Approximate binomial for rare events with large n

Frequently Asked Questions

Four conditions must hold: each trial must have exactly two outcomes (success/failure), the number of trials n must be fixed in advance, each trial must be independent of the others, and the probability of success p must be the same for every trial. If these conditions are violated, the beta-binomial distribution or other models may be more appropriate.
The Bernoulli distribution models a single trial (n=1), while the binomial distribution models the sum of n independent Bernoulli trials. The binomial is the generalization — Binomial(1, p) is identical to Bernoulli(p). The mean scales linearly with n: E[Binomial] = np versus E[Bernoulli] = p.
For large n, the normal distribution with mean np and variance np(1−p) approximates the binomial well. The rule of thumb is that both np ≥ 10 and n(1−p) ≥ 10 should hold. The approximation simplifies calculations for very large n where exact binomial coefficients become computationally unwieldy. This calculator uses exact computation, so the approximation is rarely needed.
This is a consequence of the Central Limit Theorem: the sum of many independent, identically distributed Bernoulli random variables converges to a normal distribution as n increases. The deformation from discrete to continuous becomes negligible when n is large, and the binomial probability histogram takes on the familiar bell curve shape.
The most probable number of successes (the mode) is either floor((n+1)p) or ceil((n+1)p)−1, which is approximately np for large n. When (n+1)p is an integer, both that value and (n+1)p−1 are joint modes. For example, with n=10 and p=0.5, the mode is 5 — exactly half the trials as expected for a symmetric distribution.

Sources & References

Last updated: 2026-06-06

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Editorial Note

MyCalcBuddy Editorial Team

This page is maintained as an educational calculator reference.

Source

Formula Source: Standard Mathematical References

by Various

UpdatedLast reviewed: May 2026
CheckedFormula checks are based on standard references and internal QA review.