1 ruler has length=L , devide this ruler: 3 sections. probability for 3 sections can be 3 edges of 1 triangle?

Question

a.1/2 b1/3 c 1/4 d 1/5

Answer 1

I believe it's 1/2, because the only way the segments could not form a triangle is if the sum of the lengths of any two of them was not greater than the other one. The only way that can happen is if both of the breaks are on the same side of the middle of the ruler or if one break is precisely in the middle of the ruler (which we will consider to be impossible). The probability of that (for one side) is (1/2)^2, or 1/4. Because both breaks may happen together on either side of the middle of the ruler, the probability of a triangle not being formed is doubled, thus yielding a probability of 1/2 for the possibility of the segments not forming a triangle. The other 1/2 of the time a triangle will be formed, so 1/2 is the answer. Good luck!

Edit: shyam: You have a good approach, but your result is wrong. In the graph:
x+y>L/2
0 0 it is obvious that 1/2 of the total region specified is shaded and not 1/4 of the region. So the answer is really 1/2. You can see this because x+y > L/2 has an x-intercept of L/2 and a y-intercept of L/2, which in this context makes it the diagonal of a square in which everything above the diagonal is shaded. I hadn't thought of it that way. Thanks for enlightening us!

Answer 2

Let one side be 'x'
the 2nd side be 'y'
third side is L-x-y

Now x,y>0 and x,y
Conditions for x,y,L-x-y to be the sides of a triangle:
x+y>L-x-y => x+y>L/2
x+L-x-y>y => x y+L-x-y>x => y
Now in a graph, shade the region common to
x,y>0
x,y x+y>L/2
x y
Now the reqd ans = area of shaded region / area of total region
=1/4