Probability in geometry: finding the probability of selecting a $5$-tuple such that a pentagon can be formed from its sides Consider all tuples of  $5$ real numbers each less than $5$.
Find the probability of selecting a tuple such that a pentagon could be formed from its sides.
My attempt : 
I tried to visualize the problem for a triangle and could come up with a solution with graphs but I don't know how to approach this for the above one .
Part 2 : 
Solve the above problem if the sum of the elements of the tuple is 5.
 A: Let $[5] = \{1,2,3,4,5\}$ and $X_i \sim \mathcal{U}([0,5)), i \in [5]$ be five iid random variables uniformly sampled over $[0,5)$. Let $Y_1 \ge Y_2, \cdots, Y_5$ be any reordering of them such that $Y_1$ is the largest number.
In order for $X_1,\ldots, X_5$ to form a pentagon, the necessary and sufficient condition is $$X_i < \sum\limits_{j=1,\ne i}^5 X_i\quad\text{ for }i \in [5]\quad\iff\quad
Y_1 < Y_2 + Y_3 + Y_4 + Y_5\tag{*1}$$ 
Aside from events of probability zero, $Y_1 \in (0,5)$. When $Y_1 > 0$, define
$$Z_2 = \frac{Y_2}{Y_1},\quad Z_3 = \frac{Y_3}{Y_1},\quad Z_4 = \frac{Y_4}{Y_1}, \quad Z_5 = \frac{Y_5}{Y_1}$$
Under the assumption $Y_1 > 0$, it is easy to see $Z_2,\ldots,Z_5 \sim \mathcal{U}([0,1])$. As a result, the probability we seek equals to
$$
\verb/Pr/[ 1 < Z_2 + Z_3 + Z_4 + Z_5]
= 1 - \underbrace{\verb/Pr/[Z_2 + Z_3 + Z_4 + Z_5 \le 1 ]}_{\text{ volume of a right-angled 4-simplex of sides 1}}
= 1 - \frac{1}{4!} = \frac{23}{24}$$
Update
For part 2, let $X = \sum_{i=1}^5 X_i$. Under the constraint $X = 5$, the condition on LHS of $(*1)$ becomes
$$X_i < X - X_i,\;\forall i \in [5]\quad\iff\quad X_i < \frac{5}{2},\; \forall i \in [5]$$
Notice up to events of probability zero, the events $X_i \ge \frac{5}{2}$ for different $i$ are disjoint.
This means the conditional probability for forming a pentagon can be evaluated as
$$\begin{align}\verb/Pr/\left[
\bigcap_{i=1}^5 \left( X_i < \frac{5}{2} \right) : X = 5 
\right]
&= 1 - \verb/Pr/\left[
\bigcup_{i=1}^5 \left(X_i \ge \frac{5}{2}\right) : X = 5 \right]\\
&= 1 - \sum_{i=1}^5 \verb/Pr/\left[ X_i \ge \frac{5}{2} : X = 5\right]\\
&= 1 - 5 \verb/Pr/\left[ X_1 \ge \frac{5}{2} : X = 5 \right]
\end{align}
$$
Geometrically, the region determined by the relations
$\frac{5}{2} \le X_1, 0 \le X_2, X_3, X_4, X_5$ and $X = 5$
is a 4-simplex whose sides is one half of the 4-simplex determined
by the relations $0 \le X_1, X_2, X_3, X_4, X_5$ and $X = 5$.
This implies
$$\verb/Pr/\left[ X_1 \ge \frac{5}{2} : X = 5 \right] = \frac{1}{2^4}$$
and as a result,
$$\verb/Pr/\left[
\bigcap_{i=1}^5 \left( X_i < \frac{5}{2} \right) : X = 5 
\right] = 1 - \frac{5}{16} = \frac{11}{16}$$
