All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 I've a broken stick (Posted on 2003-12-07)
I've a straight stick which has been broken into three random-length pieces.
What is the probability that the pieces can be put together to form a triangle?
If you can answer this at this point, please do.
If not, perhaps this will help: here are several methods to break the stick into the three random length pieces:
1. I select, independently, and at random, two points from the points that range uniformly along the stick, then break the stick at these two points.
2. I select one point, independently, and at random (again uniformly), and break the stick at this point. I then randomly (with even chances) select one of the two sticks and randomly select a point (again uniformly) along that stick, and break it at that point.
3. I select one point, independently, and at random (again uniformly), and break the stick at this point. I then select the larger stick, and randomly select a point (again uniformly) along that stick, and break it at that point.
If this clarifies the problem, please show how this affects your work.

 No Solution Yet Submitted by SilverKnight Rating: 3.6000 (5 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 No Subject | Comment 9 of 26 |
Penny,
"...and (1/9999)*[(3/9999) + (5/9998) + (7/9997) + (9/9996) + ... + (5001/5001)] does appear to be approximately 1/4. "
Your bracketed term has a numerator that increases by 2 per term and a denominator that decreases by one - you would never have the term
5001/5001
the 2500th term would be
5001/7500 wouldn't it?
The final term would be
20001/1

 Posted by Lee on 2003-12-08 03:12:51

 Search: Search body:
Forums (0)