HI Friends

 

The question is, A stick  of length 1 m is cut into two parts. The larger part is again cut into two .

 

Find the probability of the three parts making a triangle.

 

Try this with caution regarding the order of breaking sticks.

 

Cheers!

The Best Solution I got was this !

 

To make a triangle out of three pieces of a stick of unit length,
the longest piece must have a length which is less than half a unit.

Let's make our first break and let x be the length of the shorter
piece. x can vary between 0 and 0.5. Let's pick values in this
range and consider the probability that we can make a triangle for
that value of x.

Suppose x = 0.2; then the long stick has length 0.8. The break on
this stick must be made so that the longest piece has length less
than 0.5. This means the break must be made between 0.3 and 0.5.
So the probability of being able to make a triangle with the three
pieces if the first break is at 0.2 is 0.2/0.8.

Now let's pick one more value for where we make the first break.
Suppose x = 0.4; then the long stick has length 0.6. Again, the
break on this stick must be made so that the longest piece has
length less than 0.5. This means the break must be made between 0.1
and 0.5. So the probability of being able to make a triangle with
the three pieces if the first break is at 0.4 is 0.4/0.6.

Using our two specific examples, we can see that, in the general
case, the probability of being able to make a triangle out of the
three pieces, as a function of the length x of the short piece we
get on our first break, is

x/(1-x)

Our variable x can range from 0 to 0.5, so the overall probability
that we can make a triangle with the three pieces of the stick,
given your statement of the problem, is

2 *integral from 0 to 0.5 of (x/(1-x))  ( as the points can be selected in two ways)

This turns out to be

-.5 - ln(.5) = 2 ln 2 -1 = -.5 + .69314718.... = .19314718...

 

Cheers!