Binomial Probability Calculator
Find the probability of getting exactly k successes, at most k, or more than k in n independent trials. Ideal for quality control, genetics, and game-theory scenarios.
About this calculator
The binomial distribution models repeated independent trials where each trial has exactly two outcomes: success (probability p) or failure (probability 1−p). The probability mass function is P(X = k) = C(n,k) · pᵏ · (1−p)^(n−k), where C(n,k) = n! / (k!(n−k)!) is the binomial coefficient. For cumulative queries, P(X ≤ k) sums the PMF from i = 0 up to k, and P(X > k) = 1 − P(X ≤ k). The mean of the distribution is μ = n·p and the variance is σ² = n·p·(1−p). This model applies whenever trials are independent and the success probability remains constant across every trial.
How to use
Suppose you flip a fair coin (p = 0.5) 10 times and want P(exactly 4 heads). Set trials n = 10, successes k = 4, probability p = 0.5, and calculation type = 'exactly'. The calculator computes C(10,4) · 0.5⁴ · 0.5⁶ = 210 · 0.0625 · 0.015625 = 210 · 0.000977 ≈ 0.2051. So there is roughly a 20.5% chance of getting exactly 4 heads in 10 fair coin flips.
Frequently asked questions
What is the difference between exactly, at most, and at least in binomial probability?
'Exactly k' uses the single-term PMF: P(X = k) = C(n,k)·pᵏ·(1−p)^(n−k). 'At most k' sums P(X = 0) through P(X = k), giving a cumulative probability. 'At least k' (more than k−1) equals 1 minus the cumulative sum up to k−1. Choosing the right option depends on your question: quality inspectors asking 'no more than 2 defects' need at most, while gamblers asking 'at least one win' need the complement form.
When should I use the binomial distribution instead of another probability model?
Use the binomial distribution when you have a fixed number of trials (n), each trial is independent, and the probability of success (p) is the same for every trial. It breaks down when trials influence each other (use the hypergeometric distribution for sampling without replacement) or when n is very large and p is very small (use the Poisson approximation). Classic use cases include coin flips, pass/fail quality tests, and multiple-choice guessing.
How does sample size affect binomial probability calculations?
As n increases, the binomial distribution becomes more bell-shaped and can be approximated by a normal distribution with μ = n·p and σ = √(n·p·(1−p)) when both n·p ≥ 5 and n·(1−p) ≥ 5. Small n values produce skewed, discrete distributions that differ significantly from the normal curve. Larger samples also make extreme outcomes (very high or very low k) exponentially less likely, concentrating probability near the mean.