Probability of lines in a pentagon intersecting internally or at vertices. I'm studying maths as a hobby.
The points A,B,C,D & E are the vertices of a regular pentagon. All possible lines joining the pairs of these points are drawn. If two of these lines are chosen at random, what is the probability that their point of intersection is (a) inside the pentagon, (b) one of the points A,B,C,D,E?
I start out by saying the number of ways of choosing 2 lines from 10 is $\binom{10}{2} = 45$
Then I started getting confused. I can see there are 5 internal points of intersection, which I mark in red, so that would give me $\frac{5}{45} = \frac{1}{9}$ probability of choosing lines which meet internally. And this is indeed the answer my text book gives. But I'm not sure if this is just coincidence.
As for (b), I'm not sure how to proceed. The book gives the answer as $\frac{2}{3}$

 A: 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
$$\frac{5+5+4\cdot 5}{45}=\frac{30}{45}=\frac{2}{3}$$
A: 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}$.
A: Your approach for (a) is correct. I would clarify that the intersection points in the inside of the pentagon are formed by unique pairs of diagonals, which means that there would be $5$ pairs of lines that would intersect inside the pentagon. If, for example, two distinct pairs of diagonals intersected at a single intersection point inside the pentagon(which is impossible), then it would not be $5$ possible pairs.
For (b), we would first calculate the number of possible pairs, which is $45$ as you stated. Now assume that the intersection point is $A$ for now. There are $4$ lines extending out from $A$, and choosing any two of those $4$ lines will result in a pair of lines with an intersection point at $A$. Therefore there is $\binom42=6$ pairs that would intersect at $A$. By symmetry, the other vertices will have the same amount of pairs that would intersect at that vertex, so we can multiply by $5$, the number of vertices in a pentagon, to obtain $5\cdot6=30$ total pairs.
Thus, the final probability is $\frac{30}{45}=\frac23$ as stated in your question.
A: a) There are 5 intersections and each of the intersection is an intersection of two line segments. So you need to choose both of them to make them itersect at that point. So you have one possibility for each intersection. That is, 5 possibilities out of 45. The probability is $\frac19$.
b) There are 5 intersections and each of them is an intersection of four line segments. To make the choosen line segments intersect in a given intersection, you have to choose 2 out of 4 line segments, i.e. there are $\binom42=6$ possibilities. And there are 5 intersections, so you have 30 possibilities out of 45 in total. That means, the probability is $\frac23$.
A: Notice that there is exactly one pair of lines for each of the red dots, so there are in fact only five pairs that intersect inside the pentagon.
As for the vertices, there are four lines for each one, so for each vertex there are six ways to choose a pair of lines that meet at that vertex. Now, if two lines meet at one vertex they will not meet at any other vertex, so the total number of lines that meet at a vertex is 6*5 = 30 and the probability is then 30/45 = 2/3.
