There are 8 people fishing at Lake Connor: 5 have fishing licenses, and 3 do not. An inspector chooses two of the people at random. What is the probability that the first person chosen has a license and the second one does not? Write your answer as a fraction in simplest form.
There are a total of $\binom{8}{2} = 28$ pairs of people. We count the number of ways there can be one person with a license and one person without a license. There are 5 people with a license and 3 people without a license so there are $5\cdot3=15$ ways such a pair can be formed. Further, for both pairs, the question uniquely specifies which person has and does not have a license (for the first pair, the person with a license is chosen first; for the second pair, the person without a license is chosen first). So, there are $2\cdot15=30$ ways in which a successful pair can be chosen. However, this double-counts pairs where the first person does not have a license and the second person does have a license. This occurs when both people have a license or both people do not, which happens in $3\cdot2=6$ ways. Thus there are $30-6=24$ possible successful pairs. The desired probability is $\dfrac{24}{\binom{8}{2}}=\boxed{\dfrac{6}{7}}$.
