Confusion about triangle formation. A stick of unit length is cut in three parts. Whats the probability that it'll form a triangle.  The condition which i used $x_2+x_2\geq x_3$ where they denote sides but i get probability as $\infty$ as $x_3$ can be infinitely small maybe $dx$ and the other two are much larger. Any help would be appreciated.
 A: You need to impose all three constraints:  $x_1+x_2\gt x_3, x_2+x_3\gt x_1,x_3+x_1\gt x_2$. In your example if $x_3$ is very small, the first will be easy to satisfy, but the other two will not.  Alternately, you can sort them so that $x_1 \le x_2 \le x_3$ and then just require $x_1+x_2 \gt x_3$ as the other two are satisfied automatically.
A: Make two cuts (I assume you mean) uniformly along the unit length. Call the smaller $x$ and the larger $y$. Then there are three spacings: $x$, $y-x$, and $1-y$. Then under the conditions of the triangle inequality $(\star)$, the probability of forming a triangle is 
\begin{align*}
P(\triangle) &= P(x+(y-x)>1-y\cup x+(1-y)>y-x \cup (y-x)+(1-y)>x)\tag{$\star$}\\
&= P(y>1-y\cup 2x+1 >2y\cup 1-x>x)\\
&=P\left(y>\frac{1}{2}\cup x+\frac{1}{2}>y \cup \frac{1}{2}>x\right)\\
&=\frac{1}{2}+\left(\frac{1}{2}\right)^3+\frac{1}{2}-\frac{1}{2}-\left(\frac{1}{2}\right)^3-\frac{1}{4}-\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^3\\&=\frac{1}{4}.
\end{align*}
A: Don't get too worried about the infinities.  Let's wave these away by saying that if $x_1 + x_2 = x_3$ we have a triangle.
Let's say that the first cut produces $x_1$ and $x_2 + x_3$.  We then cut the second piece to produce $x_2$ and $x_3$.  Let's say that $x_1 + x_2 + x_3 = 1.$
If $x_1 > 1/2$ then we won't get a triangle.  If it's not, then we may.
Let's say that $x_1 = (1/2) - y$, where $0 < y \leq 1/2.$  Then, we'll get a triangle only when $x_2 > y$ OR $x_3 > y$.
So the probability of getting a triangle, given that $0 \leq y \leq 1/2$ is
$$P(y) = \frac{(1/2)+y-2y}{(1/2) + y} = \frac{1 - 2y}{1+2y}.$$
Then we can integrate the region from $0$ to $1/2$, and take the region from $1/2$ to $1$ to be zero to get the answer:
$$P(y) = \int_0^{1/2}\frac{(1-2y)dy}{1+2y} = \ln 2 - (1/2) \approx 0.193.$$
