Binomial distribution calculator

Count successes in a fixed number of yes/no trials. The bars show every possible count; the ones you asked about light up.

Example: 10 multiple-choice questions with 4 options each, guessing: n = 10, p = 0.25.

Probability of
P(X = 3)0.1172
Mean5
Std deviation1.5811
0: 0.001001: 0.009812: 0.043923: 0.117234: 0.205145: 0.246156: 0.205167: 0.117278: 0.043989: 0.0098910: 0.001010

P(X = 3) = C(10, 3) × 0.5^3 × 0.5^7 = 120 × 0.125 × 0.007813 = 0.1172

Full probability table
kP(X = k)P(X ≤ k)
00.00100.0010
10.00980.0107
20.04390.0547
30.11720.1719
40.20510.3770
50.24610.6230
60.20510.8281
70.11720.9453
80.04390.9893
90.00980.9990
100.00101.0000

When to use the binomial distribution

Use it when you count successes in a fixed number of independent yes/no trials that all have the same chance of success. A handy checklist is BINS: Binary outcomes, Independent trials, fixed Number of trials, Same probability each time.

P(X = k) = C(n, k) · pᵏ · (1 − p)ⁿ⁻ᵏ

What each piece of the formula does

  • pᵏ(1 − p)ⁿ⁻ᵏ is the probability of one particular sequence with k successes, such as SSFFF…
  • C(n, k) counts how many different sequences have exactly k successes. This is the same path-counting that makes the Galton board produce a bell.

Worked example

A student guesses all 10 questions of a 4-option multiple-choice quiz. The chance of exactly 5 right is C(10, 5) × 0.25⁵ × 0.75⁵ = 252 × 0.000977 × 0.2373 ≈ 0.0584. The chance of passing with at least 6 is P(X ≥ 6) ≈ 0.0197, about 1 in 50.

Normal approximation

When np and n(1 − p) are both at least 10, the binomial is close to a normal distribution with mean np and SD √(np(1 − p)). Set n = 100 above and the bars trace out a bell. For exact answers, this calculator never needs the approximation.

Still not clicking? Ask the tutor

An AI tutor that sees the numbers you entered above and explains them. It can make mistakes, so check anything important against the worked steps.

Questions students ask

What is the binomial probability formula?

P(X = k) = C(n, k) × p^k × (1 − p)^(n − k), where C(n, k) = n! / (k!(n − k)!) counts the ways to arrange k successes among n trials.

When is a situation binomial?

When there is a fixed number of trials n, each trial has two outcomes, the probability of success p is the same every time, and the trials are independent (the BINS checklist).

What are the mean and standard deviation of a binomial distribution?

Mean = np and standard deviation = √(np(1 − p)). For 20 coin flips, the mean is 10 heads and the SD is about 2.24.

How do I calculate "at least" probabilities?

P(X ≥ k) = 1 − P(X ≤ k − 1). For "at least one", that is 1 − (1 − p)^n.

Keep exploring