91Ó°ÊÓ

We are given,$A{x}^2+Bx+C=0$

where A, B, C are independent random variables and each being distributed over [0,1].

For $(a,b,c)∈{(0,1)}^3$

we have

$F_{A,B,C}(a,b,c)=P\left(A≤a,B≤b,C≤c\right)=P\left(A≤a\right)P(B≤b)P(C≤c)$

where the last equality holds because the variables are independent.

we know that

A, B, C are being distributed over $[0,1]$

then we have $P(A≤a)=a$

and $P(B≤b)=b$

and $P(C≤c)=c$

Therefore, $F_{A,B,C}(a,b,c)=abc$

The discriminant of the equation is greater or equal to zero

such that

$B^2-4ACâ‰\yen 0$

$B^2â‰\yen 4AC$

Now, fid the distribution of the random variable,

$D=AC$

we have

$D∈(0,1)$

and

for $d∈(0,1)$,

$$\begin{gathered} F_D\left(d\right)=P(D≤d)=P\left(AC≤d\right)=∫{∫}_{ac≤d}dadc \\ ={∫}_0^d{∫}_0^{1}dcda+{∫}_d^1{∫}_0^{\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}dcda \\ =d+d{∫}_d^1\frac{1}{a}da \\ =d-d\log \left(d\right) \end{gathered}$$

By differentiation:

We have

$f_D\left(d\right)=\frac{∂}{∂d}{F}_D\left(d\right)=-\log \left(d\right)$

Thus the required probability,

$P(B^2â‰\yen 4D)=\frac{\log \left(2\right)}{6}+\frac{5}{36}$

Where, in order to calculate the second integral,

we have used

$∫s^2\log \left(s\right)ds=\frac{s^3\log \left(s\right)}{3}-\frac{s^3}{9}$