Probability of a triangle given 3 lengths from a PDF I have a probability density function $f_{X}(x)=5(1-x)^4$ and want to calculate the probability of several conditions.  For example,

Find the probability that three randomly chosen lengths form a triangle.  That is, having selected lengths $A$, $B$, and $C$, what is the probability that
$$(A+B<C) \lor (A+C<B) \lor (B+C<A)$$

I am not sure how to construct an expression for this set of constraints using the PDF.
I know that I can find the probability of a particular interval, i.e., $\int_{a}^{b}f_{X}(x) \ dx$.  However, I would love to be able to construct any logical expression on multiple samples from the PDF!  (resources appreciated)
 A: Let $X=A+B-C,Y=A+C-B,Z=B+C-A$. Then $A=\frac{X+Y}2\in(0,1)\iff X+Y\in(0,2)$. Similarly for $Y+Z,X+Z$. The joint PDF of $X,Y,Z$ is given by$$\begin{align*}f_{X,Y,Z}(x,y,z)&=f_{A,B,C}(a(x,y,z),b(x,y,z),c(x,y,z))\left|\frac{\partial(a,b,c)}{\partial(x,y,z)}\right|\\&
=5^3[(1-a)(1-b)(1-c)]^4\left|\begin{vmatrix}1/2&1/2&0\\1/2&0&1/2\\0&1/2&1/2\end{vmatrix}\right|\\&
=\frac{5^3}{2^{14}}[(2-x-y)(2-y-z)(2-x-z)]^4
\end{align*}$$ with the support $0<X+Y<2,0<Y+Z<2,0<X+Z<2$.
We need to find the probability $P(X>0\wedge Y>0\wedge Z>0)$ which will be obtained by integrating the joint PDF over the intersection of the support and first octant:$$\begin{align*}P&=\int_{x=0}^{x=2}\int_{y=0}^{y=x}\int_{z=0}^{z=2-x} f_{X,Y,Z}(x,y,z)~dz~dy~dx+\int_{y=0}^{y=2}\int_{x=0}^{x=y}\int_{z=0}^{z=2-y} f_{X,Y,Z}(x,y,z)~dz~dy~dx\\&=2\int_{x=0}^{x=2}\int_{y=0}^{y=x}\int_{z=0}^{z=2-x} f_{X,Y,Z}(x,y,z)~dz~dy~dx\end{align*}$$
I have attached a visual of the region of integration. The horizontal plane is the $xy$ plane and the vertical axis is $z$ axis.

Mathematica gives me the answer $\frac{73501}{252252}\approx0.291$, so happy integrating!
A: The system of inequalities
$$
\left\lbrace
\begin{align*}
&0 < a < b < c < 1\\[4pt]
&a+b > c\\[4pt]
\end{align*}
\right.
$$
is equivalent to the system of inequalities
$$
\left\lbrace
\begin{align*}
&c-b < a < b
\;\;\;\;\;\;
\\[4pt]
&\frac{c}{2} < b < c\\[4pt]
&0 < c < 1\\[4pt]
\end{align*}
\right.
$$
hence if $a,b,c\in (0,1)$ are chosen randomly and independently from the given pdf,
then since the $6$ possible strict orderings of $a,b,c$ are equally likely and mutually exclusive, the probability that a triangle can be formed by edges of lengths $a,b,c$ is equal to
$$
6\int_0^1\int_{\frac{c}{2}}^c\int_{c-b}^b
\left(5(1-a)^4\right)
\left(5(1-b)^4\right)
\left(5(1-c)^4\right)
\,da\,db\,dc
$$
which evaluates to
$
{\large{\frac{73501}{252252}}}
$, or approximately $.2913792557$.
A: It looks like for the triangle probability you can work with the order statistics for a sample of size $3$ from your distribution. If $a \le b\le c$ are these order statistics, you want $a+b \ge c$ to have a triangle formed.
"Order statistics" could be looked up on wiki or otherwise googled.
A: Checking by simulation in R:  The given density function is
that of the distribution $\mathsf{Beta}(1,5).$ A triangle
is formed if the sum of the smaller two numbers exceeds the maximum
of the numbers. [@ coffeemath's (+1) order statistics.] With
a million iterations we can expect three place accuracy:
$0.2915 \pm 0.0009.$ [Agrees with answers of @quasi (+1) & @ShubhamJohn (+1).]
set.seed(2021)
m = 10^6;  tri=logical(m)
for(i in 1:m) {
 s = sort(rbeta(3, 1, 5))      # random sides of triangle
 tri[i] = s[1] + s[2] > s[3]
 }
mean(tri)
[1] 0.291507      ## aprx 73501/292232

73501/292232      ## from @quasi & @ShubhamJohn
[1] 0.2515159    
2*sd(tri)/1000
[1] 0.0009089134  ## aprx 95% margin of simulation error

Notes: At the end of the program, the logical vector
tri contains a million TRUEs and FALSEs, according as
the ith iteration produces a triangle or not. The mean
of a logical vector is the proportion of its TRUEs.
