Calculate the exact probability of getting a specific number of successes in a fixed number of independent trials.
The binomial distribution models the probability of getting exactly k successes in n independent trials, where each trial has the same probability p of success (like flipping a coin, passing/failing a test, or a product being defective). It's one of the most commonly used discrete probability distributions in statistics.
P(X = k) = C(n, k) × pk × (1−p)(n−k), where C(n, k) = n! ÷ (k! × (n−k)!) is the binomial coefficient (number of ways to choose k successes from n trials). The cumulative probability P(X ≤ k) sums this formula for every value from 0 up to k. Mean = n × p. Standard Deviation = √(n × p × (1−p)).
Flipping a fair coin (p = 0.5) 10 times, the probability of getting exactly 4 heads is P(X=4) = C(10,4) × 0.54 × 0.56 = 210 × 0.0009766 ≈ 0.205, or about 20.5%. The expected value (mean) is 10 × 0.5 = 5 heads, with a standard deviation of √(10 × 0.5 × 0.5) ≈ 1.58.
Use it whenever you have a fixed number of independent trials, each with only two possible outcomes (success/failure) and the same probability of success each time — common examples include quality control defect rates, A/B test conversions, and simple win/loss probability scenarios.
It calculates the probability of getting exactly a certain number of successes in a fixed number of independent trials, each with the same success probability (like coin flips).
Number of trials (n), probability of success per trial (p), and the number of successes (k) you want the probability for.
Estimating the probability of getting exactly 6 heads in 10 coin flips, or exactly 3 defective items in a batch of 50 with a known defect rate.