5.5 Hypergeometric Distribution

The hypergeometric distribution is a discrete probability distribution used to find the probability of success when there are two outcomes to each trial, and there are a set number of dependent trials. A hypergeometric probability distribution results from a random experiment that meets all of the following requirements. ₁2

The procedure has a fixed sample size n taken from a population N.
The trials are taken without replacement (dependent).
Each trial has exactly two outcomes, the number of successes = a and the number of failures = b. Note a + b = N.

If a random experiment satisfies all of the above, the distribution of the random variable X, where X counts the number of successes, is called a hypergeometric distribution, we write X ~ H(n, a, N).

Example 1

A committee of 5 people is to be formed from 15 volunteers and 6 appointed positions. Compute the probability that the committee will consist of 4 randomly selected volunteers.

# The calculator has no hypergeometric key, so the section sends you to an
# outside applet. Here it is built in - and, more usefully, you can DRAW the
# committees and watch sampling without replacement produce the same number.
# Committee: N = 21 (m = 15 volunteers, n = 6 appointed), sample k = 5.
dhyper(4, m = 15, n = 6, k = 5)      # P(X = 4) -> book's 0.4025

set.seed(5)
people <- rep(c('volunteer', 'appointed'), c(15, 6))
draws  <- replicate(20000, sum(sample(people, 5) == 'volunteer'))
mean(draws == 4)                     # simulated, converging on 0.4025

# The section's other three answers, exactly:
dhyper(5, m = 40, n = 30, k = 12)    # jury, 5 with a degree -> 0.1259
dhyper(2, m = 3,  n = 5,  k = 4)     # wallet, two $100 bills -> 0.4286
1 - phyper(1, m = 4, n = 21, k = 5)  # lot returned, X >= 2  -> 0.1664

Show solution

There are a total of N = 15 + 6 = 21 people. A sample of n = 5 people are selected without replacement. The success is choosing a volunteer, so a = 15. Of the 15 volunteers, we are asked to find the probability of selecting x = 4. Substitute in each value: P(X = 4) = $\frac{_{15} C_{4} \cdot_{6} C_{1}}{_{21} C_{5}} = 0.4025$ .

Note: the calculator does not have the hypergeometric distribution shortcut. The following online calculator will calculate the probability: https://homepage.divms.uiowa.edu/~mbognar/applets/hg.html. The M = a is the number of successes a from our example, or use Excel. Figure 5-7 may help you decide when to use the “True” for cumulative in Excel. Figure 5-7 A bookstore owner examines 5 books from each lot of 25 to check for missing pages. If they find at least 2 books with missing pages, the entire lot is returned. If, indeed, there are 4 books with missing pages, find the probability that the lot will be returned. Solution Sometimes the hardest part is determining which of the numbers goes in the right place in the formula. The population size is always the largest number, N = 25. A success is what you are trying to find the probability for, which is getting a book with a missing page. The number of successes out of the population is a = 4 (the 4 books with the missing pages). This means b = 25 – 4 = 21 (the number of books without missing pages). Note a + b = N. The sample size is the n = 5 is the number of books the owner examines. The owner will send the lot back if X is at least 2 missing pages. We can stop at X = 4 since there are only 4 books with missing pages. Your stopping point will be the smaller of n or a. P(X ≥ 2) = P(X = 2) + P(X = 3) + P(X = 4) = $\frac{_{4} C_{2} \cdot_{21} C_{3}}{_{25} C_{5}} + \frac{_{4} C_{3} \cdot_{21} C_{2}}{_{25} C_{5}} + \frac{_{4} C_{4} \cdot_{21} C_{1}}{_{25} C_{5}} = 0.1502 + 0.0158 + 0.0004 = 0.1664$ . In Excel use the formula: =1- HYPGEOM.DIST(1,5,4,25,TRUE). Figure 5-8 shows the online calculator at https://homepage.divms.uiowa.edu/~mbognar/applets/hg.html. Online calculator for the hypergeometric distribution X ~ HG(n, N, M) with inputs n = 5, N = 25, M = 4, and x = 2, showing P(X ≥ x) = 0.1664; below, a bar chart of the distribution has tall blue bars at x = 0 and 1 and small red-shaded bars from x = 2 upward marking the probability region. Figure 5-8

Adapted from Mostly Harmless Statistics by Rachel Webb (Portland State University), hosted on LibreTexts (stats.libretexts.org) and licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0.