Two diagonals of a regular nonagon (a $9$-sided polygon) are chosen. What is the probability that their intersection lies inside the nonagon?

There are $\frac{9*(9-3)}{2}=27$ diagonals in a nonagon. There are 27C2 ways to choose two diagonals, which is $351$. This is our denominator. The only way the diagonals can intersect inside the nonagon is if they share an endpoint. For each diagonal, there are $5$ other diagonals that share one endpoint, and 5 that share the other for a total of $10$ ways for a certain diagonal to share an endpoint with another. $27$ diagonals means $\frac{10*27}{2}=135$ ways to have adjacent diagonals. The probability for the diagonals to intersect is $\frac{135}{351}=\frac{5}{13}$. Taking the complement of this gives us an answer of $\frac{8}{13}$.

There are $27$ diagonals in a convex nonagon $P$, and you can choose $2$ of them in ${27\choose 2}=351$ ways. In order to intersect in the interior of $P$ the two diagonals have to have different endpoints, a total of four. Conversely any four vertices of $P$ determine exactly one pair of diagonals intersecting in the interior of $P$, hence there are ${9\choose4}=126$ such pairs. The probability in question therefore is ${126\over351}={14\over39}$.