what's the probability $\text{'}\mathbb{P*}\text{'}$ that $x^2-2Ax+B$ has $2$ distinct real roots? let $A, B $ be two independent random variables uniformly distributed on $[0,1]$
so $\mathbb{P*} = \mathbb{P}(A^2-B > 0) = \int_{\mathbb{R_{+}}} f_{A^2-B = U}(u)du$
I tried proceeding but got stuck en route :
let $g(U,V) = (A^2-B, A)$ then $(A,B) = (V, V^2-U)$
so $$f_{U,V}(u,v)= |J(a,b)|^{-1}f_{A,B}(v,v^2-u) = |2v|^{-1}\chi_{(v^2-1,v^2)}(u)\cdot \chi_{(0,1)}(v)$$
Hence 
$$f_U(u) = \int_{\mathbb{R}}|2v|^{-1}\chi_{(v^2-1,v^2)}(u)\cdot \chi_{(0,1)}(v)dv$$
how do you compute this integral ? also, is there asimpler approach ?
 A: Here is how I would find $P*$.  It can also be thought of as the $P(B<A^2)$.  The joint pdf for A and B is 1 for 0≤a≤1 and 0≤b≤1, 0 otherwise since they are iid uniform (0,1).  So to find the probability you want, integrate the joint probability over the region where 0≤b≤a^2≤1.  Easiest to integrate first wrt b from 0 to a^2, then with respect to a from 0 to 1.  
A: $$\mathbb{P}(A^2 > B) = \mathbb{E}\left[\mathbb{E}\left[\mathbf{1}_{B < A^2}|A\right]\right] = \mathbb{E}[A^2] = \frac{1}{3}$$
A: By the law of total probability, conditioning on $A^2 = x^2$ we have
\begin{align*}
\mathbf P\left[B < A^2\right] & = \int_{0}^1 \mathbf P\left[B < x^2 \right] \mathbf P\left[A^2 \in d x^2\right] dx \\
& = \int_{0}^1 \mathbf P\left[B < x^2 \right] \mathbf P\left[A \in dx \right] dx \\
& = \int_{0}^1 x^2 dx \\
& = 1/3,
\end{align*}
where we used the fact that since $A,B$ are uniform, the probability density function satisfies
 $$\mathbf P[A \in dx] = f(x) \equiv 1,$$
whilst the cumulative distribution function satisfies
$$\mathbf P[B < x^2] = F(x^2) = x^2.$$
