+1245 A line segment is broken at two random points along its length. What is the probability that the three new segments can be arranged to form a triangle?
+753 Suppose the points where we break the line are x and y, where 0 < x < y < 1. So our "possible" region is the following triangle on the plane:
The three sides have lengths x, y-x, and 1-y. We must satisfy the Triangle Inequality on all three sides, meaning that the sum of any two of these lengths must be greater than the third. This gives us
1. 
: This gives y>1-y, so y> 1/2.
2. 
: This gives 1-x> x, so x<1/2.
3. 
: This gives 
, so 
.
Adding these lines to the diagram gives us the darker "success" region.
The darker region makes up 1/4 of the total region, so the probability is 1/4.
+118673
A line segment is broken at two random points along its length. What is the probability that the three new segments can be arranged to form a triangle?
I would do this with a boolean contour map.
The line segment can be any length so let it be 1
Let the segment by cut into segments that are x, 1-y and y-x units long.
(If you add these you can see that the total length is 1 unit.)
No each of these lengths must be between 0 and 1 unit long.
so
\(0 x \quad (3) \qquad and \qquad y
I need to plot these to get the full sample space. The dark triangle is the sample space.
Now in order for the 3 peices to from a triangle, the sum of the two little ones must be greater than the length of the long one.
Remembering that the segments are x, 1-y and y-x units long.
\(x+(1-y)>y-x\\ 2x+1>2y\\ y
AND
\(x+(y-x)>1-y\\ y>1-y\\ 2y>1\\ y>0.5 \qquad(6)\)
AND
\((1-y)+(y-x)>x\\ 1-x>x\\ 1>2x\\ x<0.5 \qquad (7)\)
So a quarter of the sample space is the area of the desired region.
P(the three pieces form a triangle) = 0.25
