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This question already has an answer here:

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?

My approach is let be the larger part be $x$ and so the smaller part be $(1-x)$, now the two other sides be $\frac{4x}{7}$ and $\frac{3x}{7}$, now I know a triangle has a property its two sides sum is larger than the third side, but how to apply that in this question, I am confused. The answer given is $1/4$.

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marked as duplicate by Parcly Taxel, user99914, Theo Bendit, Taroccoesbrocco, stressed out Jul 13 '18 at 10:22

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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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