Binomial Distribution Calculator

Last updated: April 5, 2026

The Binomial Distribution Calculator computes P(X=k), cumulative distribution, mean, variance, and standard deviation from n trials, probability p, and success count k. The foundational discrete probability distribution for sequences of independent yes/no trials with a constant success probability.

Calculator

Number of Trials (n)

Target Successes (k)

Success Probability (p)

Results

Exact Probability P(X = k)

Expected Successes

Variance

2.5

Standard Deviation

1.5811

Z-Score of k

-1.2649

Results

Exact Probability P(X = k)

Expected Successes

Variance

2.5

Standard Deviation

1.5811

Z-Score of k

-1.2649

Quality control engineers testing whether a manufacturing process produces an acceptable defect rate, A/B testing analysts measuring whether a new landing page has a higher conversion rate, epidemiologists tracking disease transmission rates — all are working with the binomial distribution, whether they name it or not. The binomial distribution calculator computes all key statistics for any binomial scenario from three parameters: trials, probability, and success count.

The Binomial Probability Formula

The probability of exactly k successes in n independent trials, each with success probability p:

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

where C(n, k) = n! / (k!(n−k)!) is the binomial coefficient (ways to choose k positions from n for the successes). Three key statistics:

Mean: μ = n × p
Variance: σ² = n × p × (1−p)
Standard deviation: σ = √(n × p × (1−p))

For a quality control scenario: testing 20 items, each with 5% defect rate (p=0.05). P(exactly 2 defects) = C(20,2) × 0.05² × 0.95¹⁸ = 190 × 0.0025 × 0.3972 = 0.1887 ≈ 18.9%. Mean defects = 20 × 0.05 = 1.0. Use this online calculator for any n, k, and p combination. The binomial probability calculator provides an alternative interface focusing on probability comparisons.

Cumulative Probability: P(X ≤ k) and P(X ≥ k)

Most practical binomial questions involve cumulative probabilities rather than exact counts:

P(X ≤ k): probability of k or fewer successes — the CDF; computed as the sum of P(X=0) + P(X=1) + ... + P(X=k)
P(X ≥ k): probability of k or more successes = 1 − P(X ≤ k−1)
P(X = k) "exactly k": single point probability; often smaller than intuition suggests for large n

Practical example: a pharmaceutical trial with 100 patients, each having 70% probability of responding to a drug. What is the probability that at least 65 respond? P(X ≥ 65) = 1 − P(X ≤ 64). With n=100, p=0.70: mean = 70, σ = 4.58; using the normal approximation, P(X ≥ 65) ≈ P(Z ≥ (64.5−70)/4.58) ≈ P(Z ≥ −1.20) ≈ 0.885 = 88.5%.

Binomial Distribution in A/B Testing

Web conversion rate testing is a classic binomial application: each visitor either converts (success) or does not (failure), with some underlying conversion rate p. Testing whether a new variant's conversion rate p_B differs from control rate p_A:

Null hypothesis: p_A = p_B (no difference)
For n_A = 1,000 control visitors with 50 conversions (p̂_A = 5%) and n_B = 1,000 variant visitors with 65 conversions (p̂_B = 6.5%): is this difference statistically significant?
The two-sample proportion z-test uses the pooled proportion: p_pool = (50+65)/2,000 = 5.75%; SE = √(p_pool × (1−p_pool) × (1/n_A + 1/n_B)) = 0.00330; z = (0.065−0.05)/0.00330 = 4.55; p-value ≈ 0.000005 — highly significant

The Bayesian updating calculator offers an alternative Bayesian approach to A/B testing. The probability distribution calculators cover the complete discrete distribution toolkit.

Normal Approximation to the Binomial

For large n with p not too close to 0 or 1, the binomial distribution is well approximated by a normal distribution with the same mean and variance. The rule of thumb for when the approximation is valid: n×p ≥ 5 AND n×(1−p) ≥ 5. With continuity correction: P(X ≤ k) ≈ P(Z ≤ (k + 0.5 − np) / √(np(1−p))). The continuity correction (+0.5) accounts for the approximation of a discrete distribution by a continuous one and significantly improves accuracy for moderate n. The Poisson approximation (λ = np) works well when n is large and p is small (rare events): n ≥ 20 and p ≤ 0.05.

Visual Analysis

How It Works

Enter number of trials (n), success count of interest (k), and probability of success per trial (p, between 0 and 1). The calculator computes: P(X=k) = C(n,k) × pᵏ × (1-p)^(n-k); cumulative P(X≤k) by summing all terms from 0 to k; P(X≥k) = 1 − P(X≤k-1); mean = n×p; variance = n×p×(1-p); standard deviation = √variance.

Understanding Your Results

The PMF P(X = k) is the exact probability of getting exactly k successes in n trials. For instance, if PMF = 0.1172 for k=3, n=10, p=0.5, there is an 11.72% chance of getting exactly 3 heads in 10 coin flips. The mean tells you the expected number of successes on average. The variance and standard deviation measure how much the actual number of successes typically deviates from the mean. A higher variance means more variability in outcomes. Note: the PMF returns 0 if k > n, since you cannot have more successes than trials.

Worked Examples

Fair Coin: 3 Heads in 10 Flips

Inputs

n10

p0.5

Results

pmf0.1171875

mean val5

variance2.5

std dev1.5811

The probability of getting exactly 3 heads in 10 fair coin flips is about 11.72%. The expected number of heads is 5 (half of 10), with a standard deviation of about 1.58.

Quality Control: 2 Defects in 20 Items

Inputs

n20

p0.05

Results

pmf0.1886768

mean val1

variance0.95

std dev0.9747

With a 5% defect rate, the probability of finding exactly 2 defective items in a batch of 20 is about 18.87%. On average, you expect 1 defect per batch.

Frequently Asked Questions

Use the binomial distribution when: you have a fixed number of independent trials n; each trial has exactly two outcomes (success/failure); the probability p is constant across trials; and you want the probability of exactly k successes. Use the Poisson distribution when: events occur randomly in time or space with a constant average rate; the number of trials is very large and p is very small (rare events); and you know the expected count λ = np rather than n and p separately. As n→∞ and p→0 with np = λ constant, the binomial converges to Poisson — practically, if n ≥ 20 and p ≤ 0.05, the Poisson approximation is accurate.

The binomial coefficient C(n,k), also written as 'n choose k' or nCk, counts the number of ways to select k items from n without regard to order. Formula: C(n,k) = n! / (k! × (n−k)!). Examples: C(5,2) = 5!/(2!×3!) = 120/12 = 10 ways to choose 2 items from 5. C(10,0) = 1 (only one way to choose nothing). C(n,1) = n (n ways to choose 1 item). C(n,n) = 1 (only one way to choose all items). Pascal's triangle provides a visual recursive construction: C(n,k) = C(n-1,k-1) + C(n-1,k). For large n and k, use the logarithmic form: ln C(n,k) = ln n! − ln k! − ln(n-k)!, computed with Stirling's approximation for numerical stability.

The binomial distribution's shape depends on both p and n: when p = 0.5, the distribution is perfectly symmetric around np regardless of n. When p < 0.5, the distribution is right-skewed (tail toward higher values); when p > 0.5, it is left-skewed. Skewness = (1−2p)/√(np(1−p)). As n increases for any fixed p, the distribution becomes more symmetric and bell-shaped (central limit theorem). For p very close to 0 or 1, the distribution becomes highly skewed and concentrated near 0 or n respectively. A binomial with p = 0.01 and n = 100 has most probability mass at k = 0 and k = 1, looking similar to a geometric or Poisson distribution.

A fair coin flip is the canonical binomial example: n flips, p = 0.5 probability of heads per flip, X = number of heads. Key probabilities for 10 fair flips: P(exactly 5 heads) = C(10,5) × 0.5¹⁰ = 252/1024 ≈ 24.6% (the most likely single outcome but still only 1 in 4 chance). P(at least 7 heads) = P(7) + P(8) + P(9) + P(10) = (120 + 45 + 10 + 1)/1024 = 176/1024 ≈ 17.2%. P(exactly 10 heads in a row) = 0.5¹⁰ = 1/1024 ≈ 0.098%. The gambling fallacy ('I've had 8 heads in a row, tails is due') contradicts the independence assumption — each flip has exactly 50% probability regardless of history.

Finding n given a desired probability P(X ≥ 1) ≥ target requires solving: P(X = 0) ≤ 1 − target, which gives (1−p)ⁿ ≤ 1 − target, so n ≥ ln(1−target) / ln(1−p). For example, to have 95% probability of detecting at least one defective item when the defect rate is 2%: n ≥ ln(0.05)/ln(0.98) = −2.996/(−0.0202) = 148.4, so test at least 149 items. This is the foundation for acceptance sampling in quality control — standards like MIL-STD-1916 and ISO 2859 use this calculation to define sample sizes that provide specified assurance levels for given acceptable quality levels.

The binomial distribution is the exact basis for one-proportion z-tests and sign tests. For testing whether an observed conversion rate p̂ = k/n differs significantly from a hypothesized rate p₀: the p-value is P(X ≥ k) under the null hypothesis Binomial(n, p₀). For n = 100 visitors and k = 12 conversions, testing H₀: p = 0.08 (8% baseline): P(X ≥ 12 | n=100, p=0.08) = sum of binomial PMF from 12 to 100. If this probability is below 0.05, reject H₀ at 5% significance. For large n, the z-test approximation z = (p̂ − p₀) / √(p₀(1−p₀)/n) provides the same conclusion faster. Fisher's exact test uses the hypergeometric distribution for 2×2 contingency tables and is the non-approximated alternative for small samples.

Sources & Methodology

Wackerly, D., Mendenhall, W., Scheaffer, R. (2008). Mathematical Statistics with Applications, 7th ed. Thomson Brooks/Cole. Ross, S.M. (2019). Introduction to Probability Models, 12th ed. Academic Press.

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How It Works

Understanding Your Results

Worked Examples

Fair Coin: 3 Heads in 10 Flips

Inputs

n10

p0.5

Results

pmf0.1171875

mean val5

variance2.5

std dev1.5811

The probability of getting exactly 3 heads in 10 fair coin flips is about 11.72%. The expected number of heads is 5 (half of 10), with a standard deviation of about 1.58.

Quality Control: 2 Defects in 20 Items

Inputs

n20

p0.05

Results

pmf0.1886768

mean val1

variance0.95

std dev0.9747

With a 5% defect rate, the probability of finding exactly 2 defective items in a batch of 20 is about 18.87%. On average, you expect 1 defect per batch.

Frequently Asked Questions