What is the probability for a wood stick of any real number length breaking in three piece that can forming precisely a triangle without any extra segment extend from the side of the triangle?
My friend told me it can be solve in a calculus way and a algebra-precalculus way, what is the calculus way?
I found a relevant of this problem, but it is not the same to me : Broken stick probability problem
Last but not least: happy puzzle solving!
Picking two numbers independently that are uniformly distributed in $[0,1]$ is the same as picking a uniformly distributed point in the unit square. Call the smaller one $x$ and the bigger one $y$. The graph of $0<x<y<1$ is a triangular region and we are choosing a point within it. The lengths of the three sides will then be $x$, $y-x$, and $1-y$. We need three instances of the triangle inequality: \begin{align} x + (y-x) & > 1-y, \\ x + (1-y) & > y-x, \\ (y-x)+(1-y) & > x. \end{align} Draw the graph of the set of inequalities above within the aforementioned triangular region and you'll see the answer.
In the comments you asked me by name for a calculus way. Personally I would not do it this way.
Let us assume that the stick is broken at two random points $X, Y$ where the two points are independently uniformly distributed on $[0,l]$, so pieces with lengths $\min(X,Y), \max(X,Y)-\min(X,Y), l-\max(X,Y)$ and a triangle cannot be constructed if any of the lengths is $l/2$ or more (almost certainly mutually exclusive events).
Then $\displaystyle \Pr(\min(X,Y) \ge l/2)=\int_{x=l/2}^{l} \int_{y=l/2}^{l} \frac1{l^2} \,dy \,dx = \frac14$
$\displaystyle \Pr\left(\max(X,Y)-\min(X,Y) \ge l/2\right)=\int_{x=l/2}^{l} \int_{y=0}^{x-l/2} \frac1{l^2} \,dy \,dx + \int_{x=0}^{l/2} \int_{y=x+l/2}^{l} \frac1{l^2} \,dy \,dx = \frac14$
$\displaystyle \Pr(l-\max(X,Y) \ge l/2)=\int_{x=0}^{l/2} \int_{y=0}^{l/2} \frac1{l^2} \,dy \,dx = \frac14$.
The probability you want is $1-\dfrac14-\dfrac14-\dfrac14=\dfrac14$.
