I've a straight stick which has been broken into three random-length pieces.
What is the probability that the pieces can be put together to form a triangle?
If you can answer this at this point, please do.
If not, perhaps this will help: here are several methods to break the stick into the three random length pieces:
- I select, independently, and at random, two points from the points that range uniformly along the stick, then break the stick at these two points.
- I select one point, independently, and at random (again uniformly), and break the stick at this point. I then randomly (with even chances) select one of the two sticks and randomly select a point (again uniformly) along that stick, and break it at that point.
- I select one point, independently, and at random (again uniformly), and break the stick at this point. I then select the larger stick, and randomly select a point (again uniformly) along that stick, and break it at that point.
If this clarifies the problem, please show how this affects your work.
<BLOCKQUOTE>
Let the points be A and B, so 0<=A<=1 and 0<=B<=1. For the three line segments to form a triangle, the sum of the lengths of any two must be greater than or equal to the length of the third.
Case 1: If 0<=B<=A, the three line segments have lengths B, A-B, and 1-A. For these to form a triangle, the following three conditions must be true:
</BLOCKQUOTE>
<BLOCKQUOTE>
B + (A-B) >= 1-A ==> A >= 1/2 (A-B)+(1-A) >= B ==> B <= 1/2 B + (1-A) >= A-B ==> A-B <= 1/2
Case 2: If A<=B<=1, the same conditions must apply as in case 1, except with A substituted for B and B subsituted for A, that is:
A + (B-A) >= 1-B ==> B >= 1/2
(1-B)+(B-A) >= A ==> A <= 1/2
A + (1-B) >= B-A ==> B-A <= 1/2
Plotting these cases on a graph with horizontal axis for A and vertical axis for B, we get:
1 +--+--+
| /| |
|/t| |
1/2 +--+--+
| |t/|
| |/ |
0 +-----+
0 1/2 1
The two triangles labeled t are the only regions of this unit square where A and B create three line segments that agree with the conditions for case 1 (the lower right triangle) or case 2 (the upper left triangle). They total 1/4 of the area of the unit square, so the probability of two randomly chosen points craeting line segments that form a triangle is 1/4.
</BLOCKQUOTE>
solution??
