4.4 Hypergeometric Distributions


Consider trying to form a jury of 12 people from a group of 25 people where 10 of the 25 are women. If we select people are random, what is the probability of selecting a jury that contains exactly 5 women?

Let X be the number of women in our jury of 10 people.

To answer this question we will return to the basic formula for the probability of an even occurring:


Where ­­­­­­­­­­­­__________ represents the number of ways to select the women and ____________ represents the number of ways to select the men.


So



Defn: Consider an experiment involving successive dependent trials with success and failure as the only possible outcomes. Let X be the number of successful trials. Then X follows a hypergeometric distribution.


The jury selection scenario above is example of a hypergeometric distribution.

Note the key features in identifying a hypergeometric distribution are:




Probability in a Hypergeometric Distribution


Let’s now generalize this process. Let the random variable X represent the number of successes in r successive dependent trials. Then the probability of x successes is






Expected Value for a Binomial Distribution


The expected value of a random variable X that follows a geometric distribution is given by:




Example: If 2% of the transactions at a bank are applications for loans, what is the probability that the first loan request on a given day occurs with the 10th customer? How many customers should we expect to see before we see one with a loan request?


Solution: Let X represent the number customers before a customer with a loan request.

p = 0.02, q= 0.98 and the quantity we’re looking for is: P(X=9)

P(X=9) = q9p

= (0.98)9(0.02)

= (0.0166)

So the probability of the first loan request being from the 10th customer is 1.66%

To calculate the expected value:

E[X] = q/p

= 0.98/0.02

= 49

So we would expect 49 customers before getting a loan request.


Homework: pg 394 #1-4, 7-13