04 // Interactive Formula Compendium // STATISTICSP(X = k) = C(n, k) · p^k · (1 − p)^(n − k)
Binomial Probability Mass Function Formula
The binomial PMF P(X=k) = [n!/(k!(n−k)!)]·p^k·(1−p)^(n−k) gives the exact probability of k successes out of n independent trials with constant trial probability p.
Formula Reference
Main Formula
P(X = k) = C(n, k) · p^k · (1 − p)^(n − k)
Rearranged Forms
E[X] = n · p
Var(X) = n · p · (1 − p)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| P | Exact Probability P(X = k) | probability (0–1) |
| n | Number of Trials n | trials |
| k | Observed Successes k | successes |
| p | Success Probability p | 0–1 |
Inputs & Parameters
[trials]
[successes]
[0–1]
Calculated Result (Exact Probability P(X = k))
P =0.250282probability (0–1)
Formula Notes & Background
- Any specific sequence of k successes and n−k failures has probability p^k(1−p)^(n−k), and there are C(n,k) mutually exclusive orderings.
- Sums to 1 by the Binomial Theorem: ∑_{k=0}^n C(n,k) p^k (1−p)^(n−k) = (p + 1 − p)^n = 1.
Step-by-Step Examples
#1: Probability of getting exactly k = 3 heads in n = 10 tosses of a biased coin (p = 0.25).
n = 10k = 3p = 0.25
C(10,3) × (0.25)³ × (0.75)⁷ = 120 × 0.015625 × 0.133484
→ P(X = 3) = 0.2503 (25.03%)
#2: Mean and standard deviation of defective items in a batch of n = 400 with p = 0.02.
n = 400p = 0.02
μ = 400 × 0.02 = 8; σ = √(400 × 0.02 × 0.98) = √7.84
→ μ = 8.0 defects, σ = 2.80 defects
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Frequently Asked Questions
What are the four requirements for a binomial model (BINS)?+
Binary outcomes, Independent trials, Fixed Number of trials n, and Same probability p on every trial.
When should I use Hypergeometric instead of Binomial?+
When sampling without replacement from a finite population where sample size n exceeds 5% of total population N.
How does Poisson relate to Binomial?+
As n → ∞ and p → 0 with λ = np held constant, Binomial(n,p) converges to Poisson(λ).