 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Random chord problem 2 (Posted on 2016-06-09) Consider the ellipse x2/20 + y2/16 = 1

Six points A, B, C, D, E and F are selected randomly on the ellipse forming the chords AB, CD and EF.

Evaluate the probability that:

(i) The chords AB, CD and EF intersect at a single point.

(ii) Any two of the three chords intersect at a point.

 No Solution Yet Submitted by K Sengupta No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution Comment 1 of 1
I almost made this problem too difficult.  How are the points chosen?  Random x-coordinate from -sqrt(20) to sqrt(20) and then random +/-y?  Random angle from the origin?  It doesn't matter.

The relative positions are all that matters.

(i) is clearly zero.

(ii) requires the order of the points not be pairwise (AB)(CD)(EF)
Lets consider picking the points and then applying labels in such a way that no pairs meet.
There are 6!/2/2/2=90 ways to do this.
If we call one A then there are 2 choices for B.
We have 4 choices for C but this fixes D.
Finally 2 choices for E fixes F.
2*4*2=16 so there's a 16/90=8/45 chance that two two meet and a 37/45 chance that at least two meet.

 Posted by Jer on 2016-06-09 14:19:55 Please log in:

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