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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.

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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 $(1/8)/(1/2)=1/4$.

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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 $\frac{1}{4}$.

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  • $\begingroup$ yes it was a nice explaination .. so answer will be 1/4 because in probability is askes of not forming the triangle ... thank you sir very helpful $\endgroup$ – kumar Sep 5 '13 at 9:50
  • $\begingroup$ True, I put the probability of forming the triangle. I added the reverse to the answer for the sake of completeness. $\endgroup$ – Arash Sep 5 '13 at 9:52

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