A department store is holding a drawing to give away free shopping sprees. There are 9 customers who have entered the drawing: 2 live in the town of Gaston, 2 live in Pike, and 5 live in Wells. Two winners will be selected at random. What is the probability that the first winner lives in Gaston and the second lives in Pike? Write your answer as a fraction in simplest form.
There are 9 ways to choose the first winner and 8 ways to choose the second winner after the first winner has been chosen, for a total of $9 \times 8 = 72$ pairs of winners (ignoring the order in which they are chosen). Since only a particular order is wanted for this problem, we multiply by 2 to consider the order in which the winners occur; there are a total of $72 \times 2 = 144$ possible (ordered) outcomes. If the Gaston winners are labeled G1 and G2, the Pike winners as P1 and P2, and the Wells winners as W1, W2, W3, W4, and W5, then there are 4 successful outcomes: $(G1,P1)$, $(G1,P2)$, $(G2,P1)$, and $(G2,P2)$. Thus the probability that the first winner lives in Gaston and the second lives in Pike is $\dfrac{4}{144} = \boxed{\dfrac{1}{36}}$.
