The Binomial Approximation to the Hypergeometric

  Suppose we still have the population of size N with M units labelled as ``success'' and N-M labelled as ``failure,'' but now we take a sample of size n is drawn with replacement. Then, with each draw, the units remaining to be drawn look the same: still M ``successes'' and N-M ``failures.'' Thus, the probability of drawing a ``success'' on each single draw is

$$p \; = \; M/N ,$$

and this doesn't change. When we were drawing without replacement, the proportions of successes would change, depending on the result of previous draws. For example, if we were to obtain a ``success'' on the first draw, then the proportion of ``successes'' for the second draw would be (M-1)/(N-1), whereas if we were to obtain a ``failure'' on the first draw the proportion of successes for the second draw would be M/(N-1).

The random variable Y is defined as the number of ``successes'' in the sample, when we are drawing with replacement. Then Y is a binomial random variable:

$$Y \; \sim \; Bin(n,p) .$$

The probability mass function for Y is

$$p_Y (y) \; = \; \left( \begin{array} {c} n \\ y \end{array} \right) p^y (1-p)^{(n-y)} , \quad y = 0 , 1 , \ldots , n ,$$

and otherwise.

Proposition: If the population size $N \rightarrow \infty$ in such a way that the proportion of successes $M/N \rightarrow p$,and n is held constant, then the hypergeometric probability mass function approaches the binomial probability mass function:

$$h(x;N,M,n) \; \rightarrow \; b(x;n,p) .$$

In practice, this means that we can approximate the hypergeometric probabilities with binomial probabilities, provided $N \gg n$.As a rule of thumb, if the population size is more than 20 times the sample size (N > 20 n), then we may use binomial probabilities in place of hypergeometric probabilities.

We next illustrate this approximation in some examples.