Binomial Probability Calculator — P(X=k)
The binomial distribution calculates the probability of getting exactly k successes in n independent trials, each with a constant probability p of success. Enter your values below and get P(X=k), the expected value (mean), and the variance instantly. The formula — P(X=k) = C(n,k) · pᵏ · (1−p)^(n−k) — applies in statistics, quality control, genetics, epidemiology, finance, and everyday probability questions.
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 ways to choose k from n. Example: flipping a fair coin 10 times and getting exactly 3 heads → P(X=3) = C(10,3) × 0.5³ × 0.5⁷ = 120 × 0.125 × 0.0078125 ≈ 0.1172 (11.72%). Mean = n×p, Variance = n×p×(1−p).
When to use this calculator
- Quality control: probability of exactly 2 defective items in a batch of 50 (defect rate 4%)
- Genetics: Mendelian inheritance — probability of exactly 3 out of 5 children inheriting a recessive gene (p=0.25)
- Epidemiology: probability of exactly 5 infections in a group of 20 (infection rate 15%)
- Education: probability of guessing exactly 6 correct answers on a 10-question multiple-choice test (4 options, p=0.25)
- Finance and risk: probability of exactly k defaults in a portfolio of n independent loans
Worked example: 10 coin flips, exactly 3 heads
- n = 10 (trials), k = 3 (successes), p = 0.5 (fair coin)
- C(10,3) = 10! / (3! × 7!) = 120
- P = 120 × 0.5³ × 0.5⁷ = 120 × 0.125 × 0.0078125
- P(X=3) ≈ 0.11719 (11.72%)
- Mean μ = 10 × 0.5 = 5 heads expected
- Variance σ² = 10 × 0.5 × 0.5 = 2.5
How it works
2 min readHow Binomial Probability Works
Binomial probability answers: "In n independent trials, each with probability p of success, what is the exact probability of k successes?"
The formula:
P(X = k) = C(n, k) × p^k × (1 − p)^(n − k)
Where:
C(n, k) = n! / (k! × (n − k)!) ← binomial coefficient
n = total number of trials (fixed)
k = exact number of successes wanted
p = probability of success on each trial
(1 − p) = q = probability of failure
Mean (expected value): μ = n × p
Variance: σ² = n × p × (1 − p)
Standard deviation: σ = √(n × p × (1 − p))Step-by-step: n=10, k=3, p=0.5
1. C(10,3) = (10×9×8) / (3×2×1) = 120
2. p³ = 0.5³ = 0.125
3. (1−0.5)⁷ = 0.5⁷ = 0.0078125
4. P = 120 × 0.125 × 0.0078125 = 0.1172 (11.72%)
5. Mean = 10 × 0.5 = 5; Variance = 10 × 0.5 × 0.5 = 2.5
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Reference Table: P(X=k) for n=10
Probabilities for n = 10 trials at different success probabilities p:
| k (successes) | p = 0.10 | p = 0.25 | p = 0.50 | p = 0.75 | p = 0.90 |
|---|---|---|---|---|---|
| 0 | 0.3487 | 0.0563 | 0.0010 | 0.0000 | 0.0000 |
| 1 | 0.3874 | 0.1877 | 0.0098 | 0.0003 | 0.0000 |
| 2 | 0.1937 | 0.2816 | 0.0439 | 0.0016 | 0.0000 |
| 3 | 0.0574 | 0.2503 | 0.1172 | 0.0031 | 0.0001 |
| 4 | 0.0112 | 0.1460 | 0.2051 | 0.0162 | 0.0015 |
| 5 | 0.0015 | 0.0584 | 0.2461 | 0.0584 | 0.0015 |
| 6 | 0.0001 | 0.0162 | 0.2051 | 0.1460 | 0.0112 |
| 7 | 0.0000 | 0.0031 | 0.1172 | 0.2503 | 0.0574 |
| 8 | 0.0000 | 0.0004 | 0.0439 | 0.2816 | 0.1937 |
| 9 | 0.0000 | 0.0000 | 0.0098 | 0.1877 | 0.3874 |
| 10 | 0.0000 | 0.0000 | 0.0010 | 0.0563 | 0.3487 |
> Note: the distribution is perfectly symmetric when p=0.50. It shifts right when p>0.50.
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Key Conditions for Using Binomial
The binomial model is valid when ALL four conditions hold:
1. Fixed n: number of trials is decided before the experiment
2. Binary outcomes: each trial results in success or failure only
3. Independence: the result of one trial does not affect others
4. Constant p: same probability of success on every trial
If sampling without replacement from a small population, use the hypergeometric distribution instead.
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Common Worked Examples
Quality control (n=20, k=0, p=0.05)
A factory produces parts with a 5% defect rate. What is the probability of zero defects in a batch of 20?
P = C(20,0) × (0.05)⁰ × (0.95)²⁰ = 1 × 1 × 0.3585 = 0.3585 (35.85%)
Mean = 20 × 0.05 = 1 defective part expectedMultiple-choice guessing (n=10, k=6, p=0.25)
A student guesses all 10 questions (4 options each, p=0.25). Probability of exactly 6 correct?
C(10,6) = 210
P = 210 × (0.25)⁶ × (0.75)⁴ = 210 × 0.000244 × 0.3164 ≈ 0.0162 (1.62%)---
Common Mistakes
1. Confusing P(X=k) with P(X≤k): The formula gives exact probability. Cumulative probability requires summing P(X=0)+P(X=1)+...+P(X=k).
2. Rounding p too early: Use the exact probability value throughout the calculation to avoid compounding error for large n.
3. Ignoring independence: Binomial requires independent trials. Sampling without replacement from small populations violates this.
4. Wrong definition of "success": Clearly define which outcome is "success" before calculating — the result flips if you swap success/failure.
Frequently asked questions
What is binomial probability and when do I use it?
Binomial probability calculates P(X=k): the probability of exactly k successes in n independent trials, each with constant probability p. Use it when: (1) n is fixed beforehand, (2) each trial has only two outcomes (success/failure), (3) trials are independent, (4) p is the same for every trial. Classic examples: coin flips, quality inspection, genetics (Mendelian), and survey yes/no responses.
How do I calculate C(n,k) — the binomial coefficient — by hand?
C(n,k) = n! / (k! × (n−k)!). Shortcut for computation: C(10,3) = (10×9×8)/(3×2×1) = 720/6 = 120. Pascal's Triangle also gives binomial coefficients directly — row n=10, position k=3 (zero-indexed) equals 120. For large n (above 20), use logarithms or a calculator to avoid overflow.
What is the difference between P(X=k) and P(X≤k)?
P(X=k) is the point probability: exactly k successes. P(X≤k) is the cumulative distribution function (CDF): the probability of k or fewer successes. P(X≤k) = P(X=0)+P(X=1)+...+P(X=k). For n=10, p=0.5: P(X=3) ≈ 0.1172, but P(X≤3) = 0.0010+0.0098+0.0439+0.1172 ≈ 0.1719 (17.19%).
When can I approximate binomial with the normal distribution?
Apply the normal approximation when both np ≥ 5 AND n(1−p) ≥ 5. Then use μ=np and σ=√(np(1−p)) with a continuity correction (+0.5 or −0.5 depending on direction). Example: n=100, p=0.3 → μ=30, σ≈4.58. For rare events (p<0.05, n>50), the Poisson approximation (λ=np) is more precise.
What is the difference between binomial and Poisson distribution?
Binomial uses parameters n (fixed trials) and p, suitable when n is known and finite. Poisson uses λ (average rate) and models rare events over continuous intervals (time, area) where n→∞ and p→0 but np=λ stays constant. Rule of thumb: use Poisson when n>50 and p<0.05 — for example, defects per meter of fabric, or calls per minute at a call center.
What is a Bernoulli distribution and how does it relate to binomial?
Bernoulli is the special case of binomial with n=1: a single trial with probability p of success. Its outcome is 0 or 1. The binomial distribution is the sum of n independent Bernoulli trials. So binomial(n=1, p) = Bernoulli(p).
Why do all binomial probabilities sum to exactly 1?
Because Σ C(n,k) pᵏ (1−p)^(n−k) for k=0 to n = (p+(1−p))ⁿ = 1ⁿ = 1, by the Binomial Theorem. This guarantees the distribution is complete: some number of successes between 0 and n will always occur. This property is a useful check — if your probabilities don't sum to 1 (even approximately), there is a calculation error.
How is binomial probability used in quality control?
In acceptance sampling (ISO 2859 / ANSI/ASQ Z1.4), the binomial distribution determines the probability of accepting a lot given the Acceptable Quality Level (AQL), sample size n, and acceptance number c. P(X≤c | n, p) gives the probability of acceptance. For example, n=50, c=2, p=0.04: P(X≤2) ≈ 0.677 — a 67.7% chance of accepting a lot with 4% defectives.
What happens to the distribution shape as p changes?
When p=0.5, binomial is symmetric around n/2. When p<0.5, the distribution is right-skewed (long tail toward high k values). When p>0.5, it is left-skewed. Skewness = (1−2p)/√(np(1−p)). As n increases, the distribution approaches normality for any fixed p, provided np≥5 and n(1−p)≥5.