Let U and V be independent random variables, both uniformly distributed on [0, 1]. Find the probability that the quadratic equation $x^ 2 + 2Ux + V = 0$ has two real solutions.
My solution: The probability of two real solutions is the probability that $4U^2 - 4V > 0$.
$$ P(4U^2 - 4V > 0) = P(U^2>V)\\ =\int_{0}^{1} P(U^2>V|V=k)f_V(k)dk\\ =\int_{0}^{1} P(U>\sqrt{k})f_V(k)dk\\ =\int_{0}^{1} (1-\sqrt{k})f_V(k)dk\\ =\int_{0}^{1} (1-\sqrt{k}) dk =\frac{1}{3} $$ Does anyone see any problems with this?
A simpler way to compute $\Pr[U^2 > V]$ is to observe that in the unit square $(U,V) \in [0,1]^2$, the region for which $U^2 > V$ is satisfied is given by $$0 \le V < U^2 \le 1,$$ hence $$\Pr[U^2 > V] = \int_{u=0}^1 \int_{v=0}^{u^2} f_{U,V}(u,v) \, dv \, du = \int_{u=0}^1 u^2 \, du = \frac{1}{3}.$$ This is possible because the joint density of $U, V$ is uniform on the unit square.