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Binomial Probability Calculator

Calculate exact and cumulative binomial probabilities given the number of trials, probability of success, and target successes. Used in A/B testing, quality control, and statistical inference.

Experiment Details

Number of Trials (n)

Prob of Success (p)

Target Number of Successes (k)

Calculated Probabilities

Exact Probability P(X = 5)24.609%Decimal: 0.24609

At Most 5P(X ≤ 5)

62.305%0.62305

At Least 5P(X ≥ 5)

62.305%0.62305

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Binomial Probability

The Binomial Distribution describes the probability of having exactly k successes in n independent yes/no experiments, where the probability of success p is the same for every trial.

The Formula

The exact probability of exactly k successes is given by the binomial probability mass function (PMF):

P(X = k) = ⁿCₖ · pᵏ · (1 - p)ⁿ⁻ᵏ

n: Number of Trials (e.g., flipping a coin 10 times).
p: Probability of Success on a single trial (e.g., 0.5 for heads).
k: Target Number of Successes (e.g., exactly 4 heads).
ⁿCₖ: "n choose k", the binomial coefficient representing the number of ways to arrange the successes.

Cumulative Probabilities

Often, you don't just want the probability of exactly k successes, but rather the probability of answering questions like "What are the odds of getting at least 7 heads?" or "at most 2 defects?".

P(X ≤ k): Cumulative probability of getting k or fewer successes.
P(X ≥ k): Cumulative probability of getting k or more successes.

What is Binomial Probability Distribution?

The binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success. It applies whenever you have: (1) a fixed number of trials n, (2) each trial is independent, (3) each trial has exactly two outcomes (success/failure), and (4) the probability p of success is constant across all trials. Examples: coin flips, A/B test conversions, defect counts in a production batch, drug trial responses.

Mathematical Foundation

Binomial PMF (Exact Probability)

P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}

n

= Number of trials (fixed, positive integer). Must be determined before observing data.

k

= Number of successes to calculate probability for (0 ≤ k ≤ n).

p

= Probability of success on each single trial (0 ≤ p ≤ 1).

\binom{n}{k}

= Binomial coefficient = n! / (k!(n−k)!). The number of distinct ways to choose k successes from n trials (combinations). Also written C(n,k) or nCk.

p^k (1-p)^{n-k}

= Probability of any specific sequence of k successes and (n−k) failures. Multiplied by C(n,k) to count all such sequences.

Cumulative Distribution Function (CDF)

P(X \leq k) = \sum_{i=0}^{k} \binom{n}{i} p^i (1-p)^{n-i}

P(X \leq k)

= Probability of k or fewer successes. Used for 'at most k' questions and left-tail hypothesis tests.

P(X \geq k)

= = 1 − P(X ≤ k−1). Probability of k or more successes. Used for 'at least k' questions and right-tail tests (e.g., 'at least 10 defects').

Laws & Principles

Mean and variance: E[X] = np; Var(X) = np(1−p); σ = √(np(1−p)). For n=20 coin flips (p=0.5): mean = 10, σ = √5 = 2.24. Expect the number of heads to fall within 10 ± 2.24 about 68% of the time.
Normal approximation: When np ≥ 5 AND n(1−p) ≥ 5, the binomial is well-approximated by Normal(μ=np, σ²=np(1−p)). Use the continuity correction P(X = k) ≈ P(k−0.5 < Z < k+0.5) for improved accuracy. For n=100, p=0.5: np=50, n(1−p)=50 — Normal approximation is excellent.
Poisson approximation: When n is large AND p is very small (np ≤ 10), the binomial is approximated by Poisson(λ=np). Example: defect rate p=0.001 in n=1,000 units → Poisson(λ=1) is a good model.
Independence requirement: The binomial distribution strictly requires independent trials. Sampling without replacement from a finite population violates this; use the hypergeometric distribution instead. For large populations (N > 20n), the approximation is negligible and binomial is acceptable.

Step-by-Step Example Walkthrough

" A/B test: 200 visitors see a new checkout design. Historical conversion rate is 8% (p=0.08). What is the probability of getting exactly 20 conversions? What is the probability of getting 20 or more? "

n = 200, p = 0.08, k = 20
C(200, 20) = 200! / (20! × 180!) ≈ 1.614 × 10²³
P(X=20) = C(200,20) × (0.08)²⁰ × (0.92)¹⁸⁰
P(X=20) ≈ 0.0623 (6.23% chance of exactly 20 conversions)
Expected value: E[X] = 200 × 0.08 = 16 conversions
Standard deviation: σ = √(200 × 0.08 × 0.92) = √14.72 = 3.84

Final Result: P(X = 20) ≈ 6.23%. P(X ≥ 20) = 1 − P(X ≤ 19) ≈ 16.8% — there is a 16.8% chance of observing 20 or more conversions by chance alone if the true rate is 8%. To claim statistical significance at α=0.05, you would need to observe a result where P(X ≥ k) < 0.05, which happens at k = 24 or more conversions (P(X ≥ 24) ≈ 2.9%).

Quick Answer: How do you calculate binomial probability?

P(X = k) = C(n,k) × p^k × (1−p)^n−k, where C(n,k) = n! / (k!(n−k)!) is the number of combinations. Example: flipping a fair coin 10 times (n=10, p=0.5), what is the probability of getting exactly 7 heads (k=7)? C(10,7) = 120. P(X=7) = 120 × (0.5)⁷ × (0.5)³ = 120 × (0.5)¹⁰ = 120/1024 = 11.72%. For cumulative probability (“7 or more heads”): P(X≥7) = P(7) + P(8) + P(9) + P(10) = 11.72% + 4.39% + 0.98% + 0.098% = 17.19%.

Binomial Distribution Reference: n=10, Various p

P(X = k) for 10 trials at different success probabilities. Rows are number of successes k; columns are probability of success per trial p. Values rounded to 4 decimal places.

k (successes)	p = 0.10	p = 0.25	p = 0.50	p = 0.75	p = 0.90
k = 0	0.3487	0.0563	0.0010	0.0000	0.0000
k = 1	0.3874	0.1877	0.0098	0.0001	0.0000
k = 2	0.1937	0.2816	0.0439	0.0004	0.0000
k = 3	0.0574	0.2503	0.1172	0.0031	0.0001
k = 4	0.0112	0.1460	0.2051	0.0162	0.0001
k = 5	0.0015	0.0584	0.2461	0.0584	0.0015
k = 6	0.0001	0.0162	0.2051	0.1460	0.0112
k = 7	0.0000	0.0031	0.1172	0.2503	0.0574
k = 8	0.0000	0.0004	0.0439	0.2816	0.1937
k = 9	0.0000	0.0001	0.0098	0.1877	0.3874
k = 10	0.0000	0.0000	0.0010	0.0563	0.3487
Note the symmetry: P(X=k \| p) = P(X=n\u2212k \| 1\u2212p). The distribution at p=0.25 is the mirror of p=0.75. The distribution at p=0.50 is symmetric around k=5 (the peak, 0.2461). All columns sum to 1.0000.

Pro Tips & Common Binomial Probability Mistakes

Do This

✓Use the cumulative CDF for hypothesis testing, not the exact PMF. In A/B testing and quality control, the question is rarely “what is the probability of exactly k events?” — it is usually “what is the probability of k or more events if the null hypothesis is true?” This is the p-value: P(X ≥ k_observed | H₀). If this right-tail probability is below your significance level α (typically 0.05), reject H₀. Using P(X = k) alone will give you too small a value and lead to premature rejection of the null hypothesis.
✓Always verify the four binomial conditions before applying the formula. The four required conditions: (1) fixed n, (2) independence, (3) binary outcomes, (4) constant p. If p changes per trial (e.g., sampling without replacement, or a drug whose effect changes over time), the binomial model is incorrect. Sampling without replacement from a population where n is more than 5% of the total population requires the hypergeometric distribution. Non-constant p may require a beta-binomial or logistic regression model instead.

Avoid This

✗Don't stop an A/B test early when you see significance — this is “peeking” and inflates Type I error dramatically. If you check significance after every new conversion and stop when p < 0.05, your actual Type I error rate can exceed 30% even with α = 0.05. The binomial p-value is only valid if you predetermined the sample size n before starting the test. If you need early stopping, use Sequential Probability Ratio Test (SPRT) or a Bayesian approach with pre-registered stopping rules. “Two-week test, check once at the end” is the simplest correct approach for practitioners.
✗Don't confuse P(X = k) with P(X ≤ k) or P(X ≥ k) — they answer different questions. P(X = 7) for coin flips = 11.72% (exactly 7 heads). P(X ≤ 7) = 94.5% (7 or fewer heads). P(X ≥ 7) = 17.2% (7 or more heads). These are fundamentally different probabilities. In quality control: “probability that exactly 2 units are defective” (PMF) vs “probability that 2 or more units are defective” (1 − CDF of 1) vs “probability that at most 2 are defective” (CDF of 2). Specify precisely which quantity you need before calculating.

Frequently Asked Questions

How do I use binomial probability for A/B testing?

In a one-proportion binomial test: H₀: p = p₀ (control rate) vs H_a: p ≠ p₀ (two-tailed) or p > p₀ (one-tailed for detecting improvement). The p-value = P(X ≥ k_observed | n, p₀) for a right-tail test. Example: historical conversion rate 8% (p₀=0.08), n=200 visitors, k=22 conversions observed. P(X ≥ 22 | n=200, p=0.08) ≈ 6.4%. Since 6.4% > 5%, do not reject H₀ at α=0.05 — the improvement is not statistically significant. If k=24 is observed, P(X ≥ 24) ≈ 2.9% < 5% → statistically significant. For two-group A/B tests comparing two proportions, use the two-proportion z-test or chi-squared test instead — those account for variance in both groups simultaneously.

When should I use the Normal approximation instead of the exact binomial?

Use the Normal approximation when np ≥ 5 AND n(1−p) ≥ 5. With modern calculators, there is rarely a practical reason to use the approximation — the exact binomial is computable for any feasible n. However, the approximation remains important for understanding: z-tests for proportions use X ~ N(μ=np, σ²=np(1−p)) exactly. Apply the continuity correction: P(X = k) ≈ P(k−0.5 < Z < k+0.5); P(X ≤ k) ≈ P(Z < k+0.5). Without the continuity correction, the approximation error can be 10–20% for moderate n. Use the Poisson approximation instead when n > 100 and p < 0.01 (rare events) — Poisson(λ=np) often gives a better approximation than Normal for skewed rare-event distributions.

What is the binomial coefficient C(n,k) and how do I calculate it for large n?

C(n,k) = n! / (k!(n−k)!) counts the number of distinct ways to choose k items from n without regard to order. For large n, factorial directly overflows double-precision floating point (n=171! exceeds 10³⁰⁸). Compute in log space instead: ln(C(n,k)) = ln(n!) − ln(k!) − ln((n−k)!) using the log-gamma function (lgamma in most math libraries): log_C = lgamma(n+1) − lgamma(k+1) − lgamma(n−k+1). Then C(n,k) = exp(log_C) and P(X=k) = exp(log_C + k×ln(p) + (n−k)×ln(1−p)). This approach computes exact binomial probabilities for any n without overflow. Alternatively, use Pascal's triangle identity C(n,k) = C(n−1,k−1) + C(n−1,k) for recursive computation.

How is the binomial distribution used in quality control (AQL sampling)?

Acceptance Quality Limit (AQL) sampling (per ISO 2859/ANSI Z1.4) uses the binomial distribution to determine rejection thresholds. A sampling plan specifies: sample n units from a lot; accept the lot if fewer than c defects are found (reject if ≥ c). The Producer's Risk (α) is the probability of rejecting a good lot: α = P(X ≥ c | p = AQL). The Consumer's Risk (β) is the probability of accepting a bad lot: β = P(X < c | p = LTPD), where LTPD is an unacceptable defect rate. Example: n=50, c=2, AQL=2%, LTPD=10%. α = P(X ≥ 2 | n=50, p=0.02) = 1 − P(X=0) − P(X=1) = 1 − 0.3642 − 0.3716 = 0.264 (26.4% producer's risk). β = P(X < 2 | n=50, p=0.10) = P(X=0) + P(X=1) = 0.005 + 0.029 = 3.4% consumer's risk. AQL tables in ISO 2859 are pre-calculated binomial CDFs organized by batch size and quality level.