How this instrument works
The binomial distribution describes the number of successes in a fixed number of independent trials, n, when each trial has exactly two possible outcomes and the same probability of success, p, every time. Coin flips are the textbook case — heads or tails, same probability each flip — but the same structure shows up whenever you count how many times something happens out of a fixed, known number of attempts: how many of 20 shots go in, how many of 50 emails get opened, how many of 100 components pass inspection.
The probability mass function, P(X = k) = C(n,k) × p^k × (1−p)^(n−k), has three moving parts working together. The binomial coefficient C(n,k) counts how many different orderings of successes and failures produce exactly k successes among n trials — it does the combinatorial bookkeeping so you don't have to. The term p^k is the chance that those k specific trials all succeed, and (1−p)^(n−k) is the chance that the remaining trials all fail. Multiplying the three together gives the probability of exactly k successes, in any order.
For a binomial random variable, the mean is n×p and the variance is n×p×(1−p) — quantities worth knowing even though this instrument focuses on the single-outcome probability P(X = k) rather than the full shape of the distribution. As n grows large, the binomial distribution starts to look increasingly like a smooth bell curve, which is the intuition behind using a normal approximation for large-sample binomial problems.
- Enter the fixed number of trials into Number of trials (n) — how many times the experiment is repeated.
- Enter the chance of success on any single trial into Probability of success on each trial (p), a value between 0 and 1.
- Enter the exact count of successes you want the probability for into Number of successes (k).
- Read the result in P(X = k) — probability of exactly k successes — the exact binomial probability for those inputs.
- k cannot exceed n: you cannot have more successes than trials, and the instrument will flag that combination as invalid.
Worked example — five heads in ten fair coin flips
Enter n = 10 into Number of trials, p = 0.5 into Probability of success, and k = 5 into Number of successes — the classic question of getting exactly 5 heads in 10 fair coin flips. The instrument first finds C(10,5) = 252, the number of distinct orderings of 5 heads and 5 tails among 10 flips. It then multiplies by p^k = 0.5^5 = 0.03125 and (1−p)^(n−k) = 0.5^5 = 0.03125.
P(X = k) reads 0.24609375 — multiplying 252 × 0.03125 × 0.03125, or equivalently 252⁄1024, gives exactly 0.24609375. So even though 5 heads out of 10 is the single most likely outcome for a fair coin, it still only happens about 24.6% of the time; the other roughly 75% of the probability is spread across getting 0 through 4 or 6 through 10 heads instead.
Questions
What conditions does a binomial distribution require?
Four conditions, all needed together: a fixed number of trials n decided in advance, each trial has only two possible outcomes (success or failure), the probability of success p stays the same on every trial, and the trials are independent of one another. Counting how many of 20 independent coin flips land heads fits perfectly; counting cards drawn without replacement from a deck does not, because removing a card changes the odds for the next draw — that situation calls for the hypergeometric distribution instead.
How is the binomial distribution different from the Poisson distribution?
The binomial distribution needs a known, fixed number of trials n and a per-trial success probability p — think 20 coin flips or 50 inspected parts. The Poisson distribution instead counts events over a continuous interval of time or space with no natural 'number of trials' to fix, governed by a single rate λ — think phone calls received per hour. As a technical bridge, if n grows very large while p shrinks so that n×p stays fixed at λ, the binomial distribution converges to the Poisson distribution with that same λ.
What are the mean and variance of a binomial distribution?
The mean is n × p and the variance is n × p × (1 − p). For the worked example (n = 10, p = 0.5), the mean number of heads is 10 × 0.5 = 5 and the variance is 10 × 0.5 × 0.5 = 2.5. The mean tells you the long-run average count of successes across many repeats of the whole n-trial experiment; P(X = k), what this instrument reports, is the probability of landing on one particular count.
Why can't the number of successes k exceed the number of trials n?
Because each success uses up exactly one trial — you cannot succeed more times than you attempted. If k is entered larger than n, the instrument flags the combination as invalid rather than returning a nonsensical probability, since C(n,k) is undefined (equivalently zero) whenever k exceeds n.
Does this instrument give the probability of at least k successes, or exactly k?
Exactly k. P(X = k) is the probability mass function value for one specific count of successes, not a cumulative probability. To get 'at least k' or 'at most k' you would sum P(X = k) across the relevant range of k values — for instance, P(X ≤ 5) with n = 10 and p = 0.5 sums the individual probabilities for k = 0 through k = 5, which comes out well above the 0.246 figure for k = 5 alone.
Can p be exactly 0 or exactly 1?
Yes, though the distribution becomes degenerate at those extremes: with p = 0, every trial fails for certain, so P(X = 0) = 1 and every other k gives 0; with p = 1, every trial succeeds for certain, so P(X = n) = 1 and every other k gives 0. The instrument accepts p anywhere in the closed interval from 0 to 1 inclusive.