# What is the probability of forming a triangle? [duplicate]

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 $1/4$.
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 $\frac{1}{4}$.