 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Gambler's triangulation (Posted on 2010-03-18) I simultaneously toss three standard dice.
I get 3 numbers, 1 to 6, not necessarily distinct. Evaluate the probability that these numbers can represent sides of a triangle.

 Submitted by Ady TZIDON Rating: 3.5000 (2 votes) Solution: (Hide) ANSWER: 37/72 SOLUTION: Denote this probability by p, and let the three numbers that come up be x, y, and z. We will calculate 1-p instead: 1-p is the probability that x >=(y+z); y >=(z+x), or z>=( x+y). Since these events are mutually exclusive, 1-p is just 3 times the probability that x >=(y+z). This happens with probability (0+1+3+6+10+15)/216 = 35/216, so the answer is (1-P)=3*35/216 =35/72. Therefore P= 37/72. Comments: ( You must be logged in to post comments.)
 Subject Author Date Including degenerates Jer 2010-03-19 14:26:56 re(5): Solution/zero area triangle Ady TZIDON 2010-03-19 13:28:03 re(4): Solution Charlie 2010-03-19 10:59:55 re(3): Solution Ady TZIDON 2010-03-19 04:26:27 re(2): Solution Dej Mar 2010-03-19 04:07:19 re: Solution Ady TZIDON 2010-03-18 19:53:34 Solution Dej Mar 2010-03-18 17:57:47 solution Charlie 2010-03-18 15:33:46 General geometric solution Jer 2010-03-18 14:18:10 Please log in:

 Search: Search body:
Forums (1)