The binomial distribution

Many questions count successes in n independent trials, each with the same chance p: heads in 10 flips, defective parts in a box of 20, correct guesses on a quiz.

The number of successes X follows the binomial distribution:

P(X = k) = C(n, k) × pᵏ × (1 − p)ⁿ⁻ᵏ

• pᵏ(1 − p)ⁿ⁻ᵏ is the chance of one particular order of k successes and n − k failures.
• C(n, k), "n choose k", counts how many such orders there are.

Its mean is n × p and its variance n × p × (1 − p).

from math import comb

n, p = 10, 0.5
for k in range(n + 1):
    prob = comb(n, k) * p ** k * (1 - p) ** (n - k)
    print(k, "heads:", round(prob, 4))
print("mean n*p =", n * p)

For 10 fair coins, 5 heads is most likely, but only about 25% of the time. Exactly 10 heads happens once in 1,024 tries.

Common mistakes

• Using it when trials are not independent, or p changes (drawing without replacement from a small group).
• Forgetting C(n, k): the chance of "HHT" is not the chance of "two heads in three flips".

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