Chapter 19: Hypergeometric r.v.’s

Scenario: There are a fixed number of successes and failures (which are known in advance), from which we make \(n\) draws without replacement. We are counting the number of successes from the \(n\) trials.

Example 1.   A wildlife biologist is using mark-recapture to research a wolf population. Suppose a specific study region is known to have 24 wolves, of which 11 have already been tagged. If 5 wolves are randomly captured, what is the probability that 3 of them have already been tagged?

Solution:

Properties of Hypergeometric r.v.’s

  • There is a finite population of \(N\) items.

  • Each item in the population is either a success or a failure, and there are \(M\) successes total.

  • We randomly select (sample) \(n\) items from the population.

Hypergeometric vs. Binomial r.v.’s

Suppose a hypergeometric r.v. \(X\) has the following properties:

  • the population size \(N\) is really big,

  • the number of successes \(M\) in the population is relatively large,

    • \(\frac{M}{N}\) shouldn’t be close to 0 or 1

  • and the number of items \(n\) selected is small.

Then, in this case, making \(n\) draws from the population doesn’t change the probability of success much, and the hypergeometric r.v. can be approximated by a binomial r.v.

Example 2.   Suppose a specific study region is known to have 2400 wolves, of which 1100 have already been tagged.

  1. If 50 wolves are randomly captured, what is the probability that 20 of them have already been tagged?

  2. Approximate the probability in part (1) using the binomial distribution.

Solution: