If We Sample from a Small Finite Population Without Replacement $$P ( x ) = \frac { A ! } { ( A - x ) ! x ! } \cdot \frac { B ! } { ( B - n + x ) ! ( n - x ) ! } \div \frac { ( A + B ) ! } { ( A + B - n ) ! n ! }$$

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 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 is $$P ( x ) = \frac { A ! } { ( A - x ) ! x ! } \cdot \frac { B ! } { ( B - n + x ) ! ( n - x ) ! } \div \frac { ( A + B ) ! } { ( A + B - n ) ! n ! }$$ In a relatively easy lottery, a bettor selects 4 numbers from 1 to 14 (without repetition) , and a winning 4-number combination is later randomly selected. What is the probability of getting all 4 winning numbers?

A) 0
B) 0.002
C) 0.001
D) 0.003