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.
firstly if 3 lengths are to form a triangle then they must satisfy a property stated as " the sum of any two of the three must be greater than the third "
Now let us see the problem as a step procedure...
to break the stick into 3 pieces we'll have to make 2 random cuts.
The probability of getting the first random cut exactly at the centre of the stick is very very low, say d (d<<1).
therefore the probability of getting the cut anywhere else on the stick is 1-d (which will be almost 1).
so this cut will break the stick into two parts, which are:
a)EQUAL, if the cut is at centre (which is with
probability d). And in this case making the
second cut anywhere on the two pieces wont make
a triangle possible.
b)UNEAQUAL, if the cut is off the centre(with probability almost 1). Now its easy to see that triangle is only possible if the second cut is made on the piece which is the larger of the 2 pieces obtained after the first random cut.
And here the longer piece can be selected with probability 1/2 from the two equally likely pieces (the position of the cut on this piece doesn't matter).
So, the triangle is only possible with probability (1-d)(1/2) which is almost equal to
Posted by vikas
on 2004-01-07 13:53:09