Assume the endpoints of the stick are at the points 0 and 1.

Let $x$ and $y$ be the two breaking points, chosen at random from the values in the interval $\left[0\, 1\right]$.

Let us assume at first that $y\> x$. The three sides of the triangle have lengths $x$, $y-x$, and $1-y$. The three conditions that must be satisfied for a triangle to be formed are thus:

$x\+\left(y-x\right) \> 1-y$,

$\left(y-x\right)\+\left(1-y\right) \> x$,

$\left(1-y\right) \+ x\> y-x$.

Simplifying these inequalities gives us:

$x < \frac{1}{2}$, $y \&gt; \frac{1}{2}$, and $y-x < \frac{1}{2}$.

We can rearrange this to give:

$x < \frac{1}{2}$, $\frac{1}{2} < y < \frac{1}{2}\&plus;x$.

The total probability for this case can thus be given by the integral:

$P\&equals;{\int }_0^{\frac{1}{2}}{\int }_{\frac{1}{2}}^{\frac{1}{2}\&plus;x} \&DifferentialD;y\&DifferentialD;x\&equals;{\int }_0^{\frac{1}{2}}x \&DifferentialD;x \&equals; \frac{1}{8}$.

Considering that the other case (that $y < x$) is equally likely, the total probability that a triangle can be formed is $\frac{1}{8}\&plus;\frac{1}{8}\&equals; \frac{1}{4}$.