If the coefficients of the quadratic equation $ax^2+bx +c$ are u.i.i.d ran variates in $(0,1)$ what is the probability of roots being real? If the coefficients a,b,c(taken in order ,c being the constant term) of a quadratic equation are randomly and independenly chosen in the open interval(0,1) what is the probability that both the roots are real?

$a,b,c$ are limited to the interval $(0,1)$, so that
$$
P(b^2 < 4ac) = \int_{a=0}^1\int_{c=0}^1\int_{b=0}^{\min(2\sqrt{ac},1)}da\,db\,dc =
\int_{a=0}^1\int_{c=0}^1 \min(2\sqrt{ac},1)\,da\,dc,
$$
so that
$$
P(b^2 < 4ac) = \int_{a=0}^1\int_{c=0}^{\min(1/(4a),1)}2\sqrt{ac}\,da\,dc
+ \int_{a=0}^1\int_{c=\min(1/(4a),1)}^1\,da\,dc.
$$
The first integral is
$$
\begin{multline}
\int_{a=0}^1 2\sqrt{a}\,da\int_{c=0}^{\min(1/(4a),1)}\sqrt{c}\,dc = \frac{4}{3}\int_{a=0}^1 \sqrt{a}\min\left(\frac{1}{8a^{3/2}},1\right)\,da\\
= \frac{4}{3}\int_{a=0}^{1/4}\sqrt{a}\,da + \frac{1}{6}\int_{a=1/4}^1\frac{1}{a}\,da = \frac{1}{9} - \frac{1}{6}\ln(1/4)\tag{1}.
\end{multline}
$$
The second integral is
$$
\int_{a=0}^1 \left[1- \min\left(\frac{1}{4a},1\right)\right]\,da = \int_{a=1/4}^1\left(1 - \frac{1}{4a}\right)\,da = \frac{3}{4} + \frac{1}{4}\ln(1/4).\tag{2}
$$
Combining $(1)$ and $(2)$, we get
$$
P(b^2 < 4ac) = \frac{31}{36} + \frac{1}{12}\ln(1/4),
$$
and
$$
P(b^2 > 4ac) = \frac{5}{36} - \frac{1}{12}\ln(1/4) = \frac{5}{36}+\frac{1}{6}\ln 2.
$$
