Some probability

• Apr 19th 2010, 08:51 PM
dwsmith
Some probability
A unit length is broken off into 3 pieces. What is the probability of them forming a triangle?

Moderator Edit: Approved Challenge question.
• Apr 20th 2010, 08:53 AM
undefined
Quote:

Originally Posted by dwsmith
A unit length is broken off into 3 pieces. What is the probability of them forming a triangle?

Moderator Edit: Approved Challenge question.

The length of the segment is irrelevant (provided it is nonzero).

Spoiler:

The distinction between $\displaystyle >$ and $\displaystyle \geq$ doesn't matter here.

WLOG let the segments be $\displaystyle a, b, c$ with $\displaystyle a \leq b \leq c$.

Triangle inequality: $\displaystyle a+b > c$

In fact this is equivalent to $\displaystyle c < \frac{1}{2}$, for the given constraints. ($\displaystyle a+b+c=1$ so we can substitute using $\displaystyle a+b=1-c$.)

(Edited out a false start.)

So we have two variables x, y with uniform probability distribution in [0,1]. I made a sketch:

http://i39.tinypic.com/2is9wf4.jpg

The shaded area is the probability we want, which is $\displaystyle p = \frac{2}{8} = \frac{1}{4}$.

• Apr 20th 2010, 09:55 AM
dwsmith
Quote:

Originally Posted by undefined
The length of the segment is irrelevant (provided it is nonzero).

Spoiler:

The distinction between $\displaystyle >$ and $\displaystyle \geq$ doesn't matter here.

WLOG let the segments be $\displaystyle a, b, c$ with $\displaystyle a \leq b \leq c$.

Triangle inequality: $\displaystyle a+b > c$

In fact this is equivalent to $\displaystyle c < \frac{1}{2}$, for the given constraints. ($\displaystyle a+b+c=1$ so we can substitute using $\displaystyle a+b=1-c$.)

(Edited out a false start.)

So we have two variables x, y with uniform probability distribution in [0,1]. I made a sketch:

http://i39.tinypic.com/2is9wf4.jpg

The shaded area is the probability we want, which is $\displaystyle p = \frac{2}{8} = \frac{1}{4}$.

Good job.
• Apr 20th 2010, 11:11 PM
simplependulum
I let $\displaystyle x+y<1$
, $\displaystyle x+y > \frac{1}{2}$
$\displaystyle x,y < \frac{1}{2}$

The area of the region bounded is also one fourth of the triangle .