$A,B\sim\mathscr{U}(0,1)$ and independent. We consider:
$$x^2+2Ax+B=0$$
Given that both of the roots of this equation are real, what is the probability that they lie in the unit disc?
Thoughts: if $\lambda, \mu$ are the roots then I have found that ${\mathbb{P}(\lambda,\mu \in\mathbb{R})}=1/3$ so by Bayes' law:
$$\mathbb{P}(|\lambda|\leq 1,\; |\mu|\leq 1)=3\cdot\mathbb{P}(\lambda,\mu\in [-1,1])$$ Update: following rogerl's suggestion: $\mathbb{P}(\lambda,\mu \in [-1,1])=\mathbb{P}(2A-1-B\leq 0)$.
Set $W=2A-1-B$ and $Z=B$ then the Jacobian is $\frac{1}{2}$ so:
$$f_{W,Z}(w,z)=\frac{1}{2}f_{A,B}(a(w,z),b(w,z))=\frac{1}{2}$$ Then:
$$\mathbb{P}(W\leq 0)=\frac{1}{2}\int_{?}^?\int_?^? dzdw$$ Is this alright? I am stuck on the limits of integration; my thoughts on these were that the unit square in the $A-B$ plane get mapped to the following figure in the $W-Z$ plane:
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$
The region of integration is then the region below the $Z$ axis, which gives $\mathbb{P}(W\leq 0)=3/4.$ This can't be right though, because then the desired probability exceeds $1$. Any help?
For your equation $x^2 + 2Ax + B = 0$, the event of the two roots being real is the discriminant should be positive or $\boxed{A^2 > B}$
In addition, the two roots should lie inside the unit circle so we get two inequalities:
$$ -1 < -A - \sqrt{A^2 - B} < -A + \sqrt{A^2 - B} < 1$$
In addition, consider $(A,B)$ as a uniformly chosen point of the unit square $[0,1]^2$. You are trying to find the area outlined by these inequalities.
Numerically, I plotted the $A,B$ satisfying your condition. Your probability is the area of the shape below, roughly 8.4%
The area under the parabola is $\mathbb{P}[\text{real roots}]=\int_0^1 x^2 \, dx = \frac{1}{3}$ the area of the triangle is $$\mathbb{P}\big[\text{roots }\notin [-1,1] \bigcap \text{real roots}\big]=\frac{1}{2} \times \frac{1}{2} \times 1 = \frac{1}{4}$$ Subtract one from the other $\boxed{\mathbb{P} = \tfrac{1}{3} - \tfrac{1}{4} = \tfrac{1}{12} = 0.083\dots}$