Probability of lines in a pentagon intersecting internally or at vertices.

We need to organize the situation properly. There are three types of intersections:- side-side ($SS$), side-diagonal ($SD$), diagonal-diagonal ($DD$).

Now a very important observation : every point of intersection counts a pair of lines.

The inside five points result from $DD$. These are $5$ pairs. Hence probability $5/45=1/9$.

The vertices result from one $SS$ pair, one $DD$ pair and four $SD$ pairs. For example $A$ is intersection point of $SS$ - $(AE,AB)$; $DD$ - $(AD,AC)$; and $SD$ - $(AE,AD)$, $(AE,AC)$, $(AB,AD)$, $(AB,AC)$. Hence probability of

$$\frac{5+5+4\cdot 5}{45}=\frac{30}{45}=\frac{2}{3}$$

We either have side-side (S-S) intersection, side-diagonal (S-D) intersection or diagonal-diagonal (D-D) intersection.

For S-S intersection, if you take any side, out of $4$ possible intersections,
i) it intersects with $2$ adjoining sides on pentagon $(\frac{1}{2})$
ii) it intersects with $2$ of them externally $(\frac{1}{2})$

For S-D intersection, if you take any side, out of $5$ possible intersections,
i) it intersects with $4$ diagonals on pentagon $(\frac{4}{5})$
ii) it never intersects with $1$ which is parallel to it $(\frac{1}{5})$

For D-D intersection, if you take any diagonal, out of $4$ possible intersections,
i) it intersects with $2$ diagonals at an interior point $(\frac{1}{2})$
ii) it intersects with other $2$ diagonals on the pentagon. $(\frac{1}{2})$

Now, probability of choosing two sides $ = \displaystyle \frac{5 \choose 2}{10 \choose 2} = \frac{2}{9}$

Probability of choosing two diagonals is the same $\frac{2}{9}$.

Probability of choosing a side and a diagonal $ \displaystyle = \frac{{5 \choose 1}^2}{10 \choose 2} = \frac{5}{9}$

So as you can see that only two diagonals intersect internally and so the probability for the first is $ = \frac{1}{2} \times \frac{2}{9} = \frac{1}{9}$.

Probability for the second is $ \displaystyle = \frac{2}{9} \times \frac{1}{2} + \frac{2}{9} \times \frac{1}{2} + \frac{5}{9} \times \frac{4}{5} = \frac{2}{3}$.