Binomial Distribution Calculator

Binomial Distribution Guide

What should I know about Binomial Distribution Conditions?

The binomial distribution B(n, p) applies when: fixed number of trials n; each trial is independent; only two outcomes (success/failure); constant probability of success p. P(X = r) = C(n,r) × p^r × (1-p)^(n-r). Where C(n,r) = n!/(r!(n-r)!) = binomial coefficient. Mean μ = np. Variance σ² = np(1-p). Standard deviation σ = √(np(1-p)). Examples: number of heads in 20 coin flips, number of defective items in a batch, number of patients responding to treatment.

What's the key thing to understand about Cumulative Binomial Probabilities?

P(X ≤ r) = Σ P(X = k) for k = 0 to r. This is the cumulative distribution function (CDF). P(X ≥ r) = 1 − P(X ≤ r-1). P(X > r) = 1 − P(X ≤ r). In practice, statistical tables give cumulative binomial probabilities — the tables typically show P(X ≤ r) for various n, p, and r combinations. Calculator approach: sum individual probabilities P(X=0) + P(X=1) + ... + P(X=r). Most scientific calculators have a built-in binomial CDF function.

What do I need to know about Normal Approximation to Binomial?

When n is large and p is not close to 0 or 1, the binomial distribution approximates a normal distribution N(np, np(1-p)). Rule of thumb: use the normal approximation when np ≥ 5 AND n(1-p) ≥ 5. Continuity correction: P(X ≤ r) in binomial ≈ P(X ≤ r + 0.5) in normal. The correction accounts for the fact that normal is continuous and binomial is discrete. For p close to 0.5 and large n, the approximation is very accurate. For p close to 0 or 1, a Poisson distribution may be a better approximation.

What's the key thing to understand about Real-World Applications?

Quality control: if a manufacturing process produces 2% defective items (p=0.02), what is the probability that a sample of 100 items contains more than 5 defectives? Genetics: in a dominant/recessive cross (Aa × Aa), each offspring has a 3/4 probability of showing the dominant phenotype. In a family of 6 children, P(exactly 4 dominant) = C(6,4) × (3/4)⁴ × (1/4)² = 0.297. Medical trials: a drug has 70% success rate. In 15 patients, P(at least 12 succeed) = P(X≥12) = 1 − P(X≤11). Sports: a basketball player with a known free-throw percentage can use the binomial distribution to estimate the probability of making a specific number of shots out of a set attempted.

What should I know about Mean and Variance of a Binomial Distribution?

For a binomial distribution with n trials and success probability p, the mean (expected number of successes) is simply μ = np, and the variance is σ² = np(1−p), giving standard deviation σ = √(np(1−p)). For flipping a fair coin 100 times (n=100, p=0.5): expected heads = 50, variance = 100×0.5×0.5=25, standard deviation = 5. This means outcomes between about 40 and 60 heads (within 2 standard deviations) are common, while 70 heads or more would be a notably unusual result worth questioning the fairness of the coin. These formulas let you estimate spread without computing every individual probability, and they're the bridge to the normal approximation: when np and n(1−p) are both reasonably large (commonly ≥5), the binomial distribution's shape is close enough to normal that z-score methods can approximate binomial probabilities quickly.

What do I need to know about Expected Value in Repeated Trials?

Expected value (np) answers 'on average, what result should I expect if I repeated this experiment many times?' — it is not a prediction of any single outcome, since the actual result in any one trial will typically differ. For a basketball player with a 70% free-throw success rate (p=0.7) taking 20 shots (n=20), the expected makes is 14, but any specific game might see 11, 15, or 17 successes purely from normal variation. This distinction matters in quality control (expected defect rate in a batch), genetics (expected ratio of offspring phenotypes), and A/B testing (expected conversions from a known conversion rate) — the expected value sets the benchmark, while the binomial probability calculation tells you how likely any specific deviation from that benchmark actually is.

What's the difference between Binomial and Geometric Distribution?

Both the binomial and geometric distributions model repeated independent trials with two outcomes (success/failure) and a fixed probability p of success, but they answer different questions. The binomial distribution asks "how many successes occur in a fixed number of trials, n?" — for example, the number of heads in 10 coin flips. The geometric distribution asks "how many trials does it take to get the first success?" — for example, how many times you flip a coin before you first get heads. The geometric distribution's probability formula, P(X = k) = (1-p)^(k-1) × p, reflects k-1 failures followed by one success, and unlike the binomial distribution it has no upper bound on k, since in theory the first success could take arbitrarily long to arrive. Recognizing which question is being asked — "how many successes in n tries" versus "how long until the first success" — determines which distribution applies.

Binomial Probability Table for n = 10, p = 0.3

Using P(X = k) = C(n,k) pᵏ(1-p)ⁿ⁻ᵏ with n = 10 trials and success probability p = 0.3 (for example, 10 independent sales calls each with a 30% conversion chance):

The expected value is np = 10 × 0.3 = 3, which matches the table's peak probability landing at k = 3. By k = 6 the cumulative probability already exceeds 98.9%, so getting 7 or more successes out of 10 calls happens barely more than 1% of the time.

Worked Example: Binomial Probability in a Quality Control Inspection

A factory's defect rate is 5% (p = 0.05). An inspector randomly samples 20 units from a batch (n = 20). What's the probability that exactly 2 units are defective? P(X = 2) = C(20,2) × (0.05)² × (0.95)¹⁸ C(20,2) = 190. (0.05)² = 0.0025. (0.95)¹⁸ = 0.39718 (since 0.95¹⁸ = e^(18 ln 0.95) = e^(-0.9233) ≈ 0.3972). P(X = 2) = 190 × 0.0025 × 0.3972 = 0.475 × 0.3972 = 0.1887, or about 18.9%. For comparison, the probability of zero defects in the sample is P(X = 0) = (0.95)²⁰ = 0.3585 (about 35.9%), and the probability of at least one defect is 1 − 0.3585 = 0.6415 (64.2%). This is the statistical backbone of acceptance sampling plans used in manufacturing: a factory can decide, before shipping a batch, exactly how large a sample to pull and how many defects to tolerate in that sample to keep the chance of accepting a genuinely bad batch below an agreed threshold.

Using the Poisson Distribution to Approximate a Binomial with Large n, Small p

When n is large and p is small (commonly n ≥ 100 and p ≤ 0.05, or np ≤ 10), the binomial distribution can be approximated by a Poisson distribution with λ = np, avoiding the need to compute large binomial coefficients. For example, a factory produces 200 items with a 2% defect rate (n = 200, p = 0.02). The exact binomial mean is np = 4, matching the Poisson parameter λ = 4. Using the Poisson approximation, P(X = 3 defects) ≈ e^(-4) × 4³/3! = 0.018316 × 64/6 = 0.018316 × 10.667 = 0.1954 (19.5%). The true binomial probability, C(200,3)(0.02)³(0.98)¹⁹⁷, computes to approximately 0.1951 — within 0.3 percentage points of the Poisson estimate, confirming the approximation is excellent here. This Poisson shortcut is widely used in quality control and insurance risk modeling specifically because it avoids computing enormous binomial coefficients like C(200,3), while still giving accurate results whenever the "many trials, rare event" condition holds. The approximation breaks down once p grows past roughly 0.05-0.1 even with large n, because the binomial's variance np(1-p) starts diverging noticeably from the Poisson's variance (which is simply λ = np, with no (1-p) correction factor), so the two distributions' spreads no longer match closely enough for the shortcut to stay reliable. As a practical guideline, many statistics texts recommend using the Poisson approximation only when n ≥ 100 and np ≤ 10, and falling back on the exact binomial formula or the normal approximation (valid when np ≥ 5 and n(1-p) ≥ 5) outside that range. Choosing the right approximation matters in practice because insurers, manufacturers, and epidemiologists routinely need fast estimates for rare-event counts across thousands of trials, where the exact binomial sum would be impractical to compute by hand.