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, while the remaining B objects are of the other type, and if n objects are sampled without replacement, then the probability of getting x objects of type A and n-x objects of type B under the hypergeometric distribution is given by the following formula. In a lottery game, a bettor selects four numbers from 1 to 47 (without repetition), and a winning four-number combination is later randomly selected. Find the probabilities of getting exactly two winning numbers with one ticket. (Hint: Use A = 4, B = 43, n = 4, and x = 2.)
$P(x) = \frac{A!}{(A-x)!x!} \times \frac{B!}{(B-n+x)!(n-x)!} \times \frac{(A+B)!}{(A+B-n)!n!}$
P(2) =
(Round to four decimal places as needed.)
