Probability #2

There are $\dfrac{n(n - 3)}{2}$ diagonals for a regular polygon with $n$ sides, so for a decagon there are $35$ diagonals, and thus $\dbinom{35}{2} = 595$ "diagonal pairs" to consider for our sample space. Now if a diagonal pair have a shared endpoint then there can be no point of intersection strictly inside the decagon, so for a diagonal pair to have an internal point of intersection it is necessary (but not sufficient) for the diagonals to have endpoints distinct from one another.

Now choose any 4 of the 10 vertices. We can form pairs of diagonals using these vertices in a total of 3 ways, only one of which results in an internal point of intersection. In this way we account for all internally-intersecting diagonal pairs, resulting in a desired probability of $\dfrac{\dbinom{10}{4}}{595} = \dfrac{210}{595} = \dfrac{6}{17}$ , and so $p + q = 6 + 17 = \boxed{23}$ .

First, let's count the total diagonals in an $n$ -gon. A diagonal is determined by any pair of non-adjacent vertices. There are $n \choose 2$ total pairs of vertices, but $n$ of them correspond to adjacent vertices. So, an $n$ -gon has ${n \choose 2} - n$ diagonals. In particular, for the decagon, we have ${10 \choose 2} - 10 = 35$ diagonals.

Second, let's count the number of diagonals that intersect a given diagonal $D$ inside the $n$ -gon. Label one endpoint of $D$ $v_1$ and then consecutively label the remaining vertices of the $n$ -gon $v_2, ... , v_n$ . Suppose the other endpoint of $D$ is $v_k$ , where $3 \leq k \leq n - 1$ . Then $D$ partitions the vertices into three subsets: $\{v_1, v_k\}$ , the endpoints of $D$ , $V = \{v_3, ... , v_{k-1}\}$ , the vertices on one side of $D$ and $W = \{v_{k+1}, ... v_n\}$ , the vertices on the other side of $D$ . A second diagonal intersects $D$ inside the $n$ -gon if and only if it has one endpoint in V and the other endpoint in W. $|V| = k-2$ and $|W| = n - k$ , so there are $(k-2)(n-k)$ such diagonals.

Finally, let's use these two observations to answer the question. At this point we'll use $n = 10$ for the decagon although it's not hard to continue with the general case if desired. Pick a diagonal $D$ of the decagon and label one of its endpoints $v_1$ . There are seven choices $v_3, ... , v_9$ for the other endpoint $v_k$ . According to our second observation, the number of interior-crossing diagonals for each of these choices are:

$k = 3 \rightarrow (3 - 2)(10 - 3) = 7$

$k = 4 \rightarrow (4 - 2)(10 - 4) = 12$

$k = 5 \rightarrow (5 - 2)(10 - 5) = 15$

$k = 6 \rightarrow (6- 2)(10 - 6) = 16$

$k = 7 \rightarrow (7 - 2)(10 - 7) = 15$

$k = 8 \rightarrow (8 - 2)(10 - 8) = 12$

$k = 9 \rightarrow (9 - 2)(10 - 9) = 7$

The decagon has 34 diagonals other than $D$ and each choice of $v_k$ has probability $\frac{1}{7}$ , so the total probability of randomly choosing 2 interior-crossing diagonals is: $\frac{1}{7}\left( \frac{7}{34} + \frac{12}{34} + \frac{15}{34} + \frac{16}{34} + \frac{15}{34} + \frac{12}{34} + \frac{7}{34} \right) = \frac{84}{238} = \ \frac{6}{17}$

The general formula for this probability for an n-sided Polygon is

(n^2 - 3n + 2) / {3*(n^2 - 3n - 2)}