Broken Stick

When the stick is broken into 2 parts, let the length of the shorter part be $x$ , where $0 < x \le \dfrac{1}{2}$ . For the longer part, its length is $1-x$ . If the longer part is further divided to 2 parts into ratio 4:3, so the lengths are $\dfrac{3(1-x)}{7}$ and $\dfrac{4(1-x)}{7}$ respectively. If this 3 sticks CANNOT form a triangle, the sum of the two shorter parts is less than the longest part. This means that $x+\dfrac{3(1-x)}{7}<\dfrac{4(1-x)}{7}$

Solve this inequality, we obtain $x<\dfrac{1}{8}$ . Then the probability that the three sticks that are left CANNOT form a triangle is $\dfrac{\,\,\,\dfrac{1}{8}\,\,\,}{\dfrac{1}{2}}=\boxed{\dfrac{1}{4}}$