probability that broken sticks will not form a triangle. A stick of unit length is broken into two at a point chosen at random. Then, the larger part of the
stick is further divided into two parts in the ratio 4:3. What is the probability that the three sticks that are
left CANNOT form a triangle?
in this problem i am not getting how to proceed like  what is  sample space?it is quiet confusing to me. Please give any simple view for this problem.
 A: Choose a point $x$ between $(0,0.5)$. Then this point will be the breaking point on the stick. $x$ will be the length of smaller segment. Then you break the bigger part $1-x$ which is the one you have to break by 4:3. This means that you get two other sticks with lengths $(1-x)\frac{4}{7}$ and $(1-x)\frac{3}{7}$. The conditions for having triangle will be:
$$
(1-x)\frac{4}{7} < (1-x)\frac{3}{7}+x
$$
$$
(1-x)\frac{3}{7} < (1-x)\frac{4}{7}+x
$$
$$
x < (1-x)\frac{3}{7}+(1-x)\frac{4}{7}
$$
All of them except the first one is automatically satisfied. For the first one you should have:
$$
x>\frac{1}{8}
$$
Now the probability will be equal to picking a point in $(0,0.5)$ bigger that $\frac{1}{8}$. Assuming uniform distribution you get:
$$
\mathbb{P}(x>\frac{1}{8})=\frac{3}{4}
$$
which gives you the probability of making a triangle so the inverse will be $\frac{1}{4}$. 
A: The only way three pieces of a broken stick of unit length can fail to make a triangle is if the length of the longest piece is greater than $1/2$.  By the set-up of the problem, the length of the shorter piece from the first break is a random number $a$ between $0$ and $1/2$.  The stipulation of a $4:3$ ratio means that the longer piece is broken (deterministically) into pieces of length $b$ and $c$, with $b={3\over4}c$.  So the only way to not get a triangle is to have $c\gt1/2$, which makes $b\gt3/8$, which means $a=1-b-c\lt1/8$.  The probability of this happening is $(1/8)/(1/2)=1/4$.
