Question

What is the probability that all the roots of x2 + bx + c = 0 are real? [0,1]?

Answer 1

$x^2+bx+c$ will have real roots when the discriminant of the quadratic, $\Delta=b^2-4c$, is non-negative, i.e.

$b^2-4c\ge0\implies b^2\ge4c$

Your question about probability is currently impossible to answer without knowing exactly what the experiment is. Are you picking $b,c$ at random from some interval? Is the choice of either distributed a certain way?

I'll assume the inclusion of "[0,1]" in your question is a suggestion that both $b,c$ are chosen indepently of one another from [0, 1]. Let $B,C$ denote the random variables that take on the values of $b,c$, respectively. I'll assume $B,C$ are identical and follow the standard uniform distribution, i.e. they each have the same PDF and CDF as below:

$f_X(x)=\begin{cases}1&\text{for }0<x<1\\0&\text{otherwise}\end{cases}$

$F_X(x)=\begin{cases}0&\text{for }x<0\\x&\text{for }0\le x<1\\1&\text{for }x\ge1\end{cases}$

where $X$ is either of $B,C$.

Then the question is to find $P(B^2\ge4C)$. We have

$P(B^2\ge4C)=P\left(C\le\dfrac{B^2}4\right)$

and we can condition the random variable $C$ on the event of $B=b$ by supposing

$P\left(C\le\dfrac{B^2}4\right)=P\left(\left(C\le\dfrac{B^2}4\right)\land(B=b)\right)=P\left(C\le\dfrac{B^2}4\mid B=b\right)\cdot P(B=b)$

then integrate over all possible values of $b$.

$=\displaystyle\int_{-\infty}^\infty P\left(C\le\dfrac{b^2}4\right)f_B(b)\,\mathrm db$

$=\displaystyle\int_{-\infty}^\infty F_C\left(\dfrac{b^2}4\right)f_B(b)\,\mathrm db$

$=\displaystyle\int_0^1\dfrac{b^2}4\,\mathrm db=\frac1{12}$