The binomial distribution counts successes in a fixed number of independent yes/no trials, such as coin tosses or defective items in a batch. Enter n, k and p to get the exact and cumulative probabilities.
Enter the number of trials and successes.
Enter the probability of success per trial.
Read the probabilities.
Exactly k: P(X = k) = C(n, k) pᵏ (1 − p)ⁿ⁻ᵏ
Mean: μ = np
The binomial distribution answers a common question: if you repeat the same yes-or-no trial n times, each with success probability p, what is the chance of getting exactly k successes? This calculator gives that probability, plus the chance of at most k and at least k, and the mean and variance of the count.
It fits many everyday situations. How likely is a student who guesses on 10 multiple-choice questions to get 4 right? If 10% of bulbs from a line are faulty, what is the chance that a box of 15 has two or fewer duds? How many seeds out of 20 will germinate? Class 12 probability, college statistics and quality-control sampling plans all rely on it.
1. Check the conditions: a fixed number of trials n, two outcomes per trial, the same success probability p each time, and independent trials.
2. Compute the number of arrangements C(n, k) = n! ÷ (k! (n − k)!).
3. Multiply by pᵏ for the k successes and by (1 − p)ⁿ⁻ᵏ for the n − k failures: P(X = k) = C(n, k) pᵏ (1 − p)ⁿ⁻ᵏ.
4. For at most k, add P(X = i) for i = 0, 1, …, k.
5. For at least k, use 1 − P(X ≤ k − 1), which equals 1 − P(X ≤ k) + P(X = k).
6. Mean = np and variance = np(1 − p).
Take one specific sequence with k successes and n − k failures, such as success, success, failure, and so on. Because the trials are independent, its probability is the product of the individual probabilities: p multiplied k times and (1 − p) multiplied n − k times. Every other sequence with the same number of successes has exactly the same probability, just in a different order. The number of such sequences is the number of ways to choose which k of the n positions are successes, C(n, k). Adding all of them gives the formula.
Think of the count X as a sum of n separate trials, each contributing 1 for success and 0 for failure. One trial has mean p and variance p(1 − p). Because the trials are independent, means add and variances add, giving np and np(1 − p). The variance is largest when p = 0.5 and shrinks towards zero as p approaches 0 or 1, when the outcome becomes almost certain. The cumulative 'at least' figure includes P(X = k) itself, which is why the calculator adds it back.
The binomial model breaks when trials are not independent or p changes. Drawing 5 cards from a deck without replacement is the classic case; the correct model is the hypergeometric distribution, although the binomial is close when the sample is a small fraction of a large lot. For large n with small p, the Poisson distribution with λ = np is a good approximation. For large n with p not near 0 or 1, roughly when np and n(1 − p) both exceed 5, the normal curve with mean np and SD √(np(1 − p)) works well.
A quality inspector at an LED bulb factory in Pune picks 15 bulbs from a line where about 10% are known to be defective. She wants the chance of finding exactly 2 defectives, and of finding 2 or fewer.
Binomial formula: P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ = C(15, 2) × 0.1^2 × 0.9^13 = 105 × 0.01 × 0.254187 = 0.266896
At most k: P(X ≤ 2) = Σ P(X = i) for i = 0…2 = 0.815939
At least k: P(X ≥ 2) = 1 − P(X ≤ 2) + P(X = 2) = 0.450957
Mean and variance: μ = np = 1.5, σ² = np(1 − p) = 1.35
Answer: P(X = 2) 0.266896; P(X ≤ 2) 0.815939; P(X ≥ 2) 0.450957
Forgetting the C(n, k) factor, which gives the probability of one particular order only.
Entering p as a percentage, such as 10 instead of 0.1.
Treating 'at least 2' as 1 − P(X ≤ 2), which wrongly excludes exactly 2.
Using the binomial model when trials are dependent, such as drawing without replacement from a small batch.
Swapping the meanings of success and failure halfway through a problem.
Designing acceptance sampling plans in manufacturing quality control.
Estimating the chance of passing a multiple-choice test by guessing.
Predicting the number of seeds germinating, patients responding or customers converting.
Class 12 probability questions on Bernoulli trials.
Checking whether an observed count of successes is unusual under an assumed rate.
What is the probability of exactly 5 heads in 10 tosses?
C(10, 5) × 0.5¹⁰ ≈ 0.2461.
Can p be a percentage?
Enter it as a decimal, for example 0.3 for 30%.