Binomial Calculator
Calculate binomial coefficients (combinations), binomial probabilities, and related statistics.
Calculation Mode
Pascal's Triangle Row 10
Formulas
Binomial Coefficient
C(n,k) = n! / (k!(n-k)!)
Binomial Probability
P(X=k) = C(n,k) x p^k x (1-p)^(n-k)
Binomial Coefficient C(n,k)
120
Ways to choose 3 from 10
Distribution Statistics
Expected Value E[X]
5.0000
n x p = 10 x 0.5
Variance Var(X)
2.5000
n x p x (1-p)
About Binomial Distribution
Uses
- Coin flip experiments
- Quality control (defect rates)
- Election polling
- Medical trials
Requirements
- Fixed number of trials (n)
- Two possible outcomes (success/failure)
- Constant probability (p)
- Independent trials
What Is a Binomial Calculator?
A binomial calculator performs two related computations: binomial coefficients C(n,k) — the number of ways to choose k items from n without regard to order — and binomial probabilities P(X=k) = C(n,k) × p^k × (1−p)^{n−k} — the probability of exactly k successes in n independent Bernoulli trials each with success probability p. The calculator supports both modes with a simple toggle.
Binomial coefficients appear throughout combinatorics, algebra, and probability: they are the entries in Pascal's triangle, the coefficients in the binomial expansion (a+b)^n, and the building blocks of the binomial distribution. The calculator also computes the distribution statistics — expected value (np) and variance (np(1−p)) — and displays Pascal's triangle row for the chosen n with the selected k highlighted.
Binomial Formulas
The binomial coefficient is a ratio of factorials. The binomial probability mass function multiplies this by success and failure probabilities.
Binomial Coefficient and Probability
Where:
- n= Total number of trials or items — a non-negative integer up to ~170 for factorial computation
- k= Number of successes or selections — an integer between 0 and n
- p= Probability of success on a single trial — must be between 0 and 1
Understanding the Results
| Output | Description |
|---|---|
| C(n,k) | Number of combinations — the binomial coefficient. For n=10, k=3: 120 ways. |
| P(X=k) | Probability of exactly k successes — shown as a percentage |
| Pascal's Triangle Row | All C(n,0) through C(n,n) — the selected k is highlighted |
| E[X], Var(X), σ | Distribution parameters; expected value = n×p, variance = n×p×(1−p) |
How to Use This Calculator
- Select mode: Choose Coefficient for combinations only, or Probability for the full binomial distribution computation.
- Enter n: Total number of trials or items. Maximum ~170 for exact factorial computation.
- Enter k: Number of successes or selections. Must be between 0 and n.
- If probability mode, enter p: Success probability — a decimal between 0 and 1 (e.g., 0.5 for a fair coin).
- Read results: The highlighted result shows C(n,k) or P(X=k)%. Pascal's triangle row and distribution statistics complete the picture.
Real-World Applications
Binomial coefficients are ubiquitous in combinatorics and probability. In quality control, the binomial distribution models the number of defective items in a sample of size n — critical for acceptance sampling plans. In clinical trials, the number of patients responding to a treatment follows a binomial distribution with p equal to the drug's true efficacy rate.
In game design and sports analytics, the probability of winning exactly k out of n matches against an opponent with a known win probability follows the binomial distribution. In genetics, the probability of inheriting k copies of a specific allele follows a binomial pattern. Pascal's triangle — whose entries are binomial coefficients — also appears in the expansion of (a+b)^n, foundational to algebra.
Worked Examples
Combinations (Lottery)
Problem:
How many ways to choose 6 numbers from 49?
Solution Steps:
- 1Set n=49, k=6 in coefficient mode.
- 2C(49,6) = 49!/(6!×43!) = (49×48×47×46×45×44)/(720).
- 3Compute: 10,068,347,520 / 720 = 13,983,816.
Result:
C(49,6) = 13,983,816. Each lottery ticket has a 1 in ~14 million chance of matching all 6 numbers.
Binomial Probability (Coin Flips)
Problem:
A fair coin is flipped 10 times. What is the probability of exactly 3 heads?
Solution Steps:
- 1Set n=10, k=3, p=0.5 in probability mode.
- 2C(10,3) = 120. P = 120 × 0.5³ × 0.5⁷ = 120 × 0.125 × 0.0078125.
- 3P = 120 × 0.0009765625 = 0.1171875 = 11.72%.
Result:
P(X=3) ≈ 11.72%. Expected heads: E[X] = 10×0.5 = 5. Standard deviation: σ ≈ 1.58.
Tips & Best Practices
- ✓Binomial coefficients are symmetric: C(n,k) = C(n,n−k). Choosing 3 from 10 is the same as choosing 7 from 10.
- ✓The expected value is the peak of the binomial distribution — the most likely range of outcomes clusters around n×p.
- ✓Pascal's triangle row sums to 2^n — the total number of subsets of an n-element set.
- ✓For very large n, the binomial distribution approximates a normal distribution with mean np and variance np(1−p).
- ✓In probability mode, p=0.5 gives a symmetric distribution; p ≠ 0.5 produces a skewed distribution.
- ✓The factorial function uses a simple loop — for n > 170, intermediate products may overflow, so results use exponential notation.
Frequently Asked Questions
Sources & References
Last updated: 2026-06-06
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Editorial Note
MyCalcBuddy Editorial Team
This page is maintained as an educational calculator reference.
Formula Source: Handbook of Mathematical Functions
by Abramowitz & Stegun