Probability, Diagonals of a convex polygon

[Saitama]

Homework Statement

In a convex polygon of 6 sides, 2 diagonals are selected at random. The probability that they intersect in the interior of the polygon is

Homework Equations

The Attempt at a Solution

There are 9 diagonals in a polygon of 6 sides. Therefore the total cases are 9C2. But how would i go about finding the favourable cases? I don't have a clue on how to proceed further. I basically can't understand how can i show that the diagonals intersect in the interior of the polygon.

Any help is appreciated!

 

[ehild]

What about a drawing? ehild

 

[haruspex]

There are 9 diagonals in a polygon of 6 sides.

Depends how you define a diagonal. There are 6 'degenerate' diagonals. But ok, let's not count those.
Try breaking it into three cases, according to how many endpoints the two diagonals share.
For the case where they share none, think of it as choosing the four points and then deciding how to pair them into diagonals.

 

[Saitama]

For the case where they share none, think of it as choosing the four points and then deciding how to pair them into diagonals.

Its a little difficult for me to understand your post completely because of the language barrier.

To select 4 points, i will use 6C4, but i am not sure how would i pair the points into diagonals.

 

[haruspex]

Its a little difficult for me to understand your post completely because of the language barrier.

To select 4 points, i will use 6C4, but i am not sure how would i pair the points into diagonals.

Select 2 (how many ways?) for one pair of endpoints, leaving 2 for the other. In what fraction of cases would they cross?

 

[Sourabh N]

For the first diagonal: Start from a vertex. What are the possible diagonals?
Case 1. Directly opposite (one of that kind)
Case 2. Not directly opposite (two of those kind).

For the second diagonal:
Case 1. 2 points on either side of the diagonal. How do you choose 2 points out of these to make second diagonal cross the first one?
Case 2. 1 point on one side, 3 on other side. How do you choose 2 points out of these to make second diagonal cross the first one?

 

[haruspex]

For the first diagonal: Start from a vertex. What are the possible diagonals?
Case 1. Directly opposite (one of that kind)
Case 2. Not directly opposite (two of those kind).

Yes.

For the second diagonal:
Case 1. 2 points on either side of the diagonal. How do you choose 2 points out of these to make second diagonal cross the first one?

You have only 2 points left, and it seems to me that in this case the diagonals must cross.

Case 2. 1 point on one side, 3 on other side. How do you choose 2 points out of these to make second diagonal cross the first one?

Again, you have only 2 points left, but in this case they cannot cross, right?
So, given the four points, what is the probability that the diagonals cross?