# Homework Help: Probability, Diagonals of a convex polygon

1. Oct 30, 2012

### Saitama

1. The problem statement, all variables and given/known data
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

2. Relevant equations

3. 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!

Last edited: Oct 30, 2012
2. Oct 30, 2012

### ehild

ehild

3. Oct 30, 2012

### haruspex

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.

4. Oct 30, 2012

### Saitama

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.

5. Oct 30, 2012

### haruspex

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

6. Oct 30, 2012

### 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?

7. Oct 30, 2012

### haruspex

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