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:
- 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.
- 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.
- 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.
(In reply to re(3): help me! correction
"...and (1/9999)*[(3/9999) + (5/9998) + (7/9997) + (9/9996) + ... + (5001/5001)] does appear to be approximately 1/4. "
As Lee has pointed out, the 2500th term would be 5001/7500. At that point, the total divided by 9999 would be only about 0.075367. If you continued on to the 4999th term, however, that would be 9999/5001, and the total divided by 9999 would indeed be close to the answer for case 3, not case 1. It would come out to 0.3863022960678033686.
I don't know which case you were addressing in your original post, but your series does not approximate 1/4. It is about 0.3863, as mentioned, which is close to the actual answer found by calculus (not computer, unless you count taking the natural log of 2 via calculator program as finding the answer by computer) for case 3: = 2ln2 - 1 = .38629436111989....
I also don't know how you came up with the approximate total of 1/4 for this series, but it makes a lot of sense to use a computer:
10 for X=1 to 4999
50 print T/9999
Note that the last term within parentheses within the square brackets is in fact 9999/5001.
I don't know how it is intended to evaluate the series without a computer.
And, while approximation was good enough for Archimedes, he did not have the advantage of standing on Newton's and Leibnitz's shoulders.
Of course we also could have left the answer to case 3 as 2 ln(2) - 1, and be in the same quandary as to its numerical value, but it helps to see the numbers.
Posted by Charlie
on 2003-12-08 10:01:12