Probability of a Triangle

Date: 12/10/97 at 06:29:33
From: Chez Kwiatkowski
Subject: Probability

If I cut a 12-inch ruler with two arbitrary cuts, what is the 
probability that from the three pieces created I will be able to 
create a triangle?

I came up with 25 percent probability.  Is that correct?

Chez

Date: 12/10/97 at 08:32:13
From: Doctor Anthony
Subject: Re: Probability

These problems are best done by considering two random variables 
x and y on the usual rectangular axes, with both x and y uniformly 
distributed between 0 and 12. The probability space is a square of 
side 12, and we need to consider regions of the square that lead to 
three lengths which can form a triangle.

Suppose that the breaks are made at points distance x and y from the 
one end. Then if y > x (we can consider the alternative later) the 
three lengths will be x, y-x and 12-y. For a triangle any two of these 
must be greater than the third, so

  x + y-x > 12-y         y-x + 12-y > x           x + 12-y > y-x

        y > 6                  x < 6                    y < x+6


If you plot these three boundaries you get a triangle with vertices at

  (6,6) (0,6) (6,12)  and area equal 1/8 of the square.

This assumed y > x.  If instead x > y then we get a second region 
symmetrically placed below the diagonal y=x of the square and again 
having an area 1/8 the area of the square. So the total area of the 
square giving values of x and y which allow for a triangle to be 
formed is 1/8 + 1/8 = 1/4 .

So the probability that a triangle can be formed is 1/4.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/