The distribution of the number of successes in n independent trials that each succeed with the same probability p: P(k) = C(n, k) pᵏ (1 − p)ⁿ⁻ᵏ. Its mean is np and its variance np(1 − p).
Card 1 of 244. Answer: The distribution of the number of successes in n independent trials that each succeed with the same probability p: P(k) = C(n, k) pᵏ (1 − p)ⁿ⁻ᵏ. Its mean is np and its variance np(1 − p).
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A binomial count needs a fixed number of trials, two outcomes per trial, independence between trials and the same success probability every time. One trial is a Bernoulli variable; the binomial is the sum of n of them. The variance np(1 − p) is largest at p = 0.5.
Example: an email converts 10% of recipients. Out of 20 recipients, the chance that exactly 2 convert is C(20, 2) × 0.1² × 0.9¹⁸ ≈ 28.5%, and the chance that none do is 0.9²⁰ ≈ 12.2%. Expect 2 conversions, with a standard deviation of √(20 × 0.1 × 0.9) ≈ 1.34.
Approximations come up in interviews: when np and n(1 − p) are both at least about 10, a normal distribution with the same mean and variance works, which is the basis of the usual A/B test on conversion rates; when n is large and p small, the Poisson with λ = np does. If you sample without replacement from a small population, the trials aren't independent and the hypergeometric distribution is the right one.
