Binomial Distribution Calculator

Exact probability P(X = k): The chance of observing precisely k successes.

Four cumulative probabilities: P(X ≤ k), P(X < k), P(X ≥ k), and P(X > k), so you can answer at most, fewer than, at least, and more than questions without re-entering data.

Distribution summary: The mean np, the variance np(1 − p), and the standard deviation.

The binomial model applies whenever a random process satisfies four conditions, sometimes remembered by the acronym BINS:

• Binary outcomes: Each trial ends in exactly one of two states — defective or acceptable, click or no click, cured or not cured.
• Independence: The outcome of one trial does not change the probability of any other trial.
• Number fixed in advance: You decide on n before collecting data, rather than sampling until a success appears.
• Same probability: The success probability p stays constant from the first trial to the last.

If the number of trials is not fixed but the observation window is — for example, counting arrivals per hour — the Poisson distribution is usually the better model. If you sample without replacement from a small population, successive trials are no longer independent and the hypergeometric distribution applies.

The calculator evaluates the probability mass function and its cumulative sums:

The cumulative probability P(X ≤ k) adds the mass function from 0 up to k. The remaining outputs follow from complements: P(X > k) = 1 − P(X ≤ k) and P(X ≥ k) = 1 − P(X ≤ k − 1).

Before software made exact computation trivial, statisticians replaced the binomial with easier distributions, and the approximations still matter for intuition and for large-n shortcuts.

Normal approximation: When both np ≥ 10 and n(1 − p) ≥ 10, the binomial is closely matched by a normal distribution with mean np and standard deviation √(np(1 − p)). A continuity correction — treating P(X ≤ k) as the normal area below k + 0.5 — improves accuracy near the center.

Poisson approximation: When n is large and p is small (a common rule of thumb is n ≥ 100 with np ≤ 10), the binomial is well approximated by a Poisson distribution with rate λ = np. This is why rare-event counts such as manufacturing defects are often modeled directly with the Poisson.

The binomial distribution shows up wherever repeated yes/no trials are audited or planned.

• Quality control: Estimate how likely an inspection lot is to pass when each unit has a known defect rate, and set acceptance sampling plans accordingly.
• Marketing and A/B testing: Model the number of conversions among n visitors when each converts with probability p.
• Medicine: Predict how many patients out of a trial cohort will respond to a treatment with a known response rate.
• Education: Compute the chance of passing a multiple-choice exam by random guessing, which sets a floor for interpreting test scores.

A quality inspector samples 10 items from a production line where each item independently has a 5% chance of being defective (p = 0.05). What is the probability of finding exactly 1 defect?

1. Count the arrangements: C(10, 1) = 10, because the single defect could be any of the ten items.
2. Probability of that pattern: one defect contributes 0.05¹ = 0.05, and the nine good items contribute 0.95⁹ ≈ 0.630249.
3. Multiply: P(X = 1) = 10 × 0.05 × 0.630249 ≈ 0.315125, or about a 31.5% chance.

The complementary results are just as useful. The chance of a clean sample is P(X = 0) = 0.95¹⁰ ≈ 0.598737, so the probability of finding at least one defect is P(X ≥ 1) = 1 − 0.598737 ≈ 0.401263. The cumulative probability of at most one defect is P(X ≤ 1) ≈ 0.913862, leaving P(X > 1) ≈ 0.086138 for two or more defects.

The distribution summary reads: mean np = 10 × 0.05 = 0.5 defects per sample, variance np(1 − p) = 0.475, and standard deviation √0.475 ≈ 0.6892. Entering n = 10, p = 0.05, k = 1 into the calculator reproduces every one of these numbers.