The binomial probability mass function has two distinct components. The combination C(n,k), also written as "n choose k", counts how many different ways k successes can be arranged across n trials. For example, if you flip a coin 4 times and want exactly 2 heads, the heads can appear in positions (1,2), (1,3), (1,4), (2,3), (2,4), or (3,4), that is C(4,2) = 6 arrangements. Each of those arrangements has the same probability, p^k × (1-p)^(n-k). Multiply the count by the probability of any single arrangement and you get the total probability.
The formula works because the trials are independent, the outcome of any one flip does not affect any other. This independence assumption is what makes the probabilities multiplicative, and it is the condition most commonly violated in real-world applications.
Worked example, probability of exactly 3 heads in 10 coin flips (p = 0.5):
P(X = k) is the exact probability, the chance of getting precisely k successes, no more and no fewer. P(X ≤ k) is the cumulative probability, the chance of getting k or fewer successes. Most practical decisions involve the cumulative form. A quality control manager asking "will we accept this batch?" is asking whether defects are at most some threshold, not whether defects are exactly that threshold. A student asking "what are the chances I pass?" is asking P(X ≥ passing score), which is 1 − P(X ≤ passing − 1).
| Question Type | Formula | Example |
|---|---|---|
| Exactly k successes | P(X = k) | Exactly 3 defects in a batch of 20 |
| At most k successes | P(X ≤ k) | 3 or fewer defects, acceptance sampling |
| At least k successes | 1 − P(X ≤ k−1) | At least 7 correct on a 10-question test |
| Between j and k successes | P(X ≤ k) − P(X ≤ j−1) | Between 3 and 6 heads in 10 flips |
| More than k successes | 1 − P(X ≤ k) | More than 2 free throws made in 5 attempts |
For quality control problems with acceptance sampling, the cumulative form P(X ≤ c) is the standard, where c is the acceptance number. For a batch of 20 items with a 5% defect rate, P(X ≤ 2) = 92.5%, meaning there is a 92.5% chance of accepting the lot if your threshold is 2 defects. This calculator computes both exact and cumulative values simultaneously so you can read both without a second calculation.
Researches and verifies the formulas, methodology, and source data behind each calculator on CalculatorFlux. All tools are built and checked against the cited references before publication.