Home > Probability
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.)
|
|
Please log in:
Forums (0)
Newest Problems
Random Problem
FAQ |
About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On
Chatterbox:
|