1. ## Breaking a stick.

What is the probability that when a stick is broken into 3 pieces that a triangle can be formed. Assume the 2 breaks to occur at the same time and are independent of each other.

2. Let A and B the randomly selected points on [0,1]. Assume B>A for beginning.
Triangular inequalities will reveal:
A<1/2 and
B>1/2 and
B-A<1/2

I am skipping several steps here but due to conditional probability that should be equivalent to:
$
\int_{1/2}^{1} (x-1/2) dx \;=\frac{1}{8}
$

We also have the symmetric case where A>B. So,
the answer = 2 * 1/8 = 1/4

-O believes.

3. Very good oswaldo.
Just to make picture even more clear:
Triangle will only be formed when sum of length of two sticks is more that the length of third stick. So each part should have length less than half the stick

4. Originally Posted by VENI
What is the probability that when a stick is broken into 3 pieces that a triangle can be formed. Assume the 2 breaks to occur at the same time and are independent of each other.
Stick Broken Into Three Pieces from Interactive Mathematics Miscellany and Puzzles

The Broken Stick Problem

Random Triangles

Of related interest: Problem 8: Random Triangles and an Introduction to Density