# Some probability

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• Apr 19th 2010, 09:51 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, 09:53 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 $>$ and $\geq$ doesn't matter here.

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

Triangle inequality: $a+b > c$

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

(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 $p = \frac{2}{8} = \frac{1}{4}$.

• Apr 20th 2010, 10:55 AM
dwsmith
Quote:

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

Spoiler:

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

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

Triangle inequality: $a+b > c$

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

(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 $p = \frac{2}{8} = \frac{1}{4}$.

Good job.
• Apr 21st 2010, 12:11 AM
simplependulum
I let $x+y<1$
, $x+y > \frac{1}{2}$
$x,y < \frac{1}{2}$

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