Chapter 5.2, Problem 43BB

Hypergeometric Distribution If we sample from a small finite population without replacement, the binomial distribution should not be used because the events are not independent. If sampling is done without replacement and the outcomes belong to one of two types, we can use the hypergeometric distribution. If a population has A objects of one type (such as lottery numbers you selected), while the remaining B objects are of the other type (such as lottery numbers you didn’t select), and if n objects are sampled without replacement (such as six drawn lottery numbers), then the probability of getting x objects of type A and n − x objects of type B is

$P(x)=\frac{A!}{(A-x)!x!}•\frac{B!}{(B-n+x)!(n-x)!}÷\frac{(A+B)!}{(A+B-n)!n!}$

In New Jersey’s Pick 6 lottery game, a bettor selects six numbers from 1 to 49 (without repetition), and a winning six-number combination is later randomly selected. Find the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use A = 6. B = 43, n = 6, and x = 2.)