What Is the Binomial Distribution?

The binomial distribution models the number of successes in a fixed number of independent yes/no trials. It applies whenever four conditions hold: a fixed number of trials n, exactly two outcomes per trial (success/failure), a constant success probability p, and independence between trials. Coin flips, defective items on a production line, and free-throw attempts all fit this pattern.

The Probability Mass Function

The probability of exactly k successes in n trials is:

P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ

where C(n, k) = n! / (k!(n − k)!) is the binomial coefficient — the number of ways to choose which k of the n trials are successes.

Worked Example: Coin Flips

What is the probability of exactly 6 heads in 10 fair coin flips? Here n = 10, k = 6, p = 0.5. The coefficient C(10, 6) = 210, so P(X = 6) = 210 × 0.5⁶ × 0.5⁴ = 210 / 1024 ≈ 0.2051, about 20.5%. For "at least 6 heads," use the Cumulative tab with the at-least option: P(X ≥ 6) ≈ 0.377.

Mean, Variance, and Shape

The Statistics tab reports the distribution's key summary numbers: mean μ = np, variance σ² = np(1 − p), and standard deviation σ = √(np(1 − p)). For n = 100 trials with p = 0.3, expect 30 successes on average with a standard deviation of √21 ≈ 4.58. When both np and n(1 − p) are at least about 10, the distribution is nearly symmetric and well approximated by a normal curve — a useful bridge to Z-score methods.

Real-World Uses

• Quality control: probability of finding more than 2 defects in a sample of 50 when the defect rate is 1%.
• Medicine and A/B testing: how likely is a result this extreme if the true response rate is p?
• Sports analytics: chance an 80% free-throw shooter makes at least 8 of 10 attempts (about 67.8%).
• Exam strategy: probability of passing by guessing — e.g. getting at least 12 of 20 four-choice questions right by luck is under 0.1%.