Picking a discrete set in a continuous probability distribution maybe this a stupid question, however I could not solve it properly. What´s the general method to solve problems relating the probability of a given event in a set of discrete events picked from a non-discrete set of events? 
For instance, given $n$ points belonging to $[0, 1]$, what´s the probability of there exists 3 points $a < b < c$ (among the $n$ points choosed) such that the equation $ax^2 +bx + c = 0$ has real solutions?
Or, given 4 points in a square, what´s the probability of they form a convex polygon with area ranging from $A_1$ to $A_2$ and perimeter ranging from $p_1$ to $p_2$?
Thanks in advance.
 A: For the first question, note that $ax^2+bx+c=0$ has real solutions if $b^2\ge 4ac$. 
The following will investigate the case $n=3$ exactly and give lower estimates for bigger $n$:
If $0<x_1<x_2<\ldots <x_n<1$ are our random numbers,
we investigate the case $(a,b,c)=(x_1,x_{n-1},x_n)$ further as that is the constellation where we expect $b$ to be as relatively close to $c$ as possible and at the same time $a$ as small as possible. In other words, if any triple with $c=x_n$ exists, then replacing $b$ with $x_{n-1}$ gives us another such triple and so does replacing $a$ with $x_1$.
For $0<r<1$, the probability that $x_n<r$ equals $r^n$ (the probability that $n$ numbers are $<r$). If we know $x_n$, then $x_1, \ldots, x_{n-1}$ are uniformly distributed in $[0,x_n]$. Therefore, 
$$\tag1P(x_{n-1}<rx_n)=r^{n-1}.$$
Similarly, $$\tag2P(x_1<sx_{n-1})=1-(1-s)^{n-2}.$$
Now with independent(!) random variables $R=\frac{x_{n-1}}{x_n}=\frac bc$ and $S=\frac{x_1}{x_{n-1}}=\frac ab$, we want to calculate $P(R\ge 4S)$. 
With the densitiy function for $R$ obtained from $(1)$, this turns out to be
$$\tag3\begin{align} &\int_0^1\left(1-\left(1-\frac r4\right)^{n-2}\right)\cdot (n-1)r^{n-2}\,\mathrm dr.\end{align}$$
If $n=3$, then $(3)$ gives the exact answer $\frac16$.
If $n=4$, then $(3)$ evaluates to $\frac{27}{80}$, but gives only a lower estimate for the probability asked for (we are missing the cases where $x_3^2<4x_1x_4$, but $x_2^2>4x_1x_3$). By numerical experimentation, the correct answer seems to be $\approx 0.352$ (compared to $\frac{27}{80}=0.3375 $).
If $n=5$, then $(3)$ evaluates to $\frac{271}{560}\approx0.48$ as lower bound (numerical estimation for correct value: $0.505$) and for $n>19$ the result of $(3)$ is $>0.99$.
A better approximation in case $n>3$ may be obtained by investigating three independent random variables $R_1=\frac{x_{n-1}}{x_n}$, $R_2=\frac{x_{n-2}}{x_{n-1}}$, $S=\frac{x_1}{x_{n-2}}$ and find $P(R_2>4S\lor R_1>4R_2S)$ by integration.
