Binomial Distribution Calculator

Reviewed by CalcMulti Editorial Team·Last updated: ·Statistics Hub

The binomial distribution models the number of successes in n independent trials, each with the same probability of success p. It is the foundation of A/B testing, quality control sampling, clinical trial analysis, and any scenario with exactly two outcomes per trial.

This calculator computes the exact probability P(X=k), cumulative probability P(X≤k), and the complementary P(X>k) for any combination of n, k, and p. It also shows the full distribution table and key statistics: mean, variance, and standard deviation.

Formula

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

n
number of independent trials
k
number of successes (0 ≤ k ≤ n)
p
probability of success on a single trial (0 < p < 1)
C(n,k)
binomial coefficient — n! / (k! × (n−k)!)

Integer 1–1000

Integer 0 to n

Enter as percentage

Binomial vs Poisson vs Normal Approximation — Which to Use?

ScenarioUseConditionsWhy
Fixed n trials, known p, counting successesBinomial (exact)Any n, any p — always validExact model for the problem
Very large n (>1000), small p (<0.05), np < 10Poisson (λ=np)n→∞, p→0, np finiteComputationally simpler, essentially exact
Large n, np ≥ 5 and n(1−p) ≥ 5Normal approxn×p ≥ 5 and n×(1−p) ≥ 5Central limit theorem applies
Small n (< 20) or p near 0 or 1Binomial (exact)Skewed distributionNormal approx fails at extremes
No upper bound on count, counting in intervalPoissonContinuous interval, rare eventsBinomial requires fixed n
Proportion CI or z-test for conversion rateNormal (CLT)n large, p not extremeAllows use of z-table / confidence intervals

Case Study: Semiconductor Quality Control

A chip fabrication engineer samples 50 chips from each production batch. The known defect rate is p=4% per chip. She needs to know: what is the probability of finding more than 5 defective chips in a sample of 50, so she can set a reject threshold without over-triggering false alarms?

Using n=50, p=0.04: mean = 2.0, SD ≈ 1.39. P(X=5) ≈ 0.0295 (2.95%). P(X≤5) ≈ 0.9806. P(X≥6) ≈ 0.0194 (1.94%). This means a batch with normal 4% defect rate will trip the "≥6 defects" alarm only 1.94% of the time — acceptable false alarm rate.

The engineer sets the reject threshold at ≥6 defects (P≈2% false alarm) rather than ≥5 (P≈10%), reducing unnecessary batch holds by 80% while still catching batches with genuinely elevated defect rates. Exact binomial calculation was critical — the normal approximation (np=2, n(1−p)=48 → np < 5) would have been unreliable at this low defect rate.

Disclaimer

This calculator is for educational purposes only and does not constitute professional advice. Results are based on standard mathematical formulas. Always verify critical calculations with a qualified professional before making important decisions.

Frequently Asked Questions