Suppose you had five sticks of length 1, 2, 3, 4, and 5 inches. If you chose three at random, what is the likelihood tht the three sticks could be put together, tip to tip, so as to form a triangle?

Now suppose you had twenty sticks, of lengths 1 through 20 inches. If you picked three at random, what is the likelihood that the three could be put together, tip to tip, to form a right triangle?

(Assume that a triangle has to have some area)

If you use SilverKnight's (I know it's not his but I first saw the method used by him) 'calculus of differences' method, it becomes clear that one formula cannot generate the possible triangles made from n sticks - however formulas can be found for the odd and even cases seperately.
For the second part of the question the answer is simply found by
(number of pythag' triples under 21)/possibles with 20 stick
= 6/1140
If anyone cares anymore, here's a 2 formulas to find the probability of choosing a legitimate triangle from n sticks
For n even
[2n^3 - 9n^2 + 10n]*[3!(n-3)!] / 24n!
.
For n odd
[2(n-1)^3 - 3(n-1)^2 - 2(n-1)]*[3!(n-3)!] / 24n!
.
pop n=5 in this second formula and you'll get 0.3 (the answer to the first part.
Put that in your excel pipe and smoke it.
I'm off to bed.

Posted by Lee on 2003-11-06 13:18:27