Binomial distribution

Bernoulli trials

A Bernoulli trial has exactly two outcomes, called success and failure. A sequence of trials is a set of Bernoulli trials if:

• the number of trials n is fixed (finite),
• the trials are independent,
• each trial has only two outcomes, and
• the probability of success p is the same in every trial.

Let q = 1 − p be the probability of failure.

The binomial distribution

If X is the number of successes in n Bernoulli trials, then P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, for r = 0, 1, 2, …, n. We write X ~ B(n, p). These probabilities are the terms of the expansion of (q + p)ⁿ, which is why they add to 1.

Mean and variance

• Mean = np
• Variance = npq, standard deviation = √(npq).

Since q < 1, the variance is always less than the mean. So a "binomial distribution with mean 6 and variance 8" is impossible.

Examples

• A fair coin is tossed 5 times. P(exactly 3 heads) = ⁵C₃(1/2)⁵ = 10/32 = 5/16.
• n = 10, p = 0.4: mean = 4, variance = 10 × 0.4 × 0.6 = 2.4.
• Mean 4 and variance 3: q = 3/4, p = 1/4, n = 4/(1/4) = 16.
• A die is thrown 4 times. P(at least one six) = 1 − (5/6)⁴ = 671/1296.

Key points

• Four conditions: fixed n, independent trials, two outcomes, constant p.
• P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ.
• Mean np > variance npq.
• "At least one" = 1 − P(X = 0) = 1 − qⁿ.

Exam tips

• Divide variance by mean to get q quickly.
• Drawing without replacement from a small lot is not binomial, since p changes.