Let $X$ and $Y$ bet independent random variables so $X\sim Uni(2,5)$ and $Y\sim Uni(1,3)$. Lets define $U=\frac{X^2}{4Y}$. Calculate $P(U>1)$.
In the solution they did: $$ P\left(U>1\right)=P\left(0.25X^{2}>Y\right)=\int_{2}^{5}\int_{1}^{\min\{3,0.25x^{2}\}}\frac{1}{6}\,dx\,dy $$ I don't understand why it's true. I understand that $P\left(U>1\right)=P\left(0.25X^{2}>Y\right)$. I also understand that $X$ and $Y$ are independent so $f_{X,Y}(x,y)=\frac{1}{3}\cdot\frac{1}{2}=\frac{1}{6}$. But what formula did they use to get to $\int_{2}^{5}\int_{1}^{\min\{3,0.25x^{2}\}}\frac{1}{6}\,dx\,dy$? How do you calculate the probability where you have two random variables sepreated with $\leq, =,\geq $? Is it true to say that they missed the step $P(0.25X^2>y, Y=y)$? if so, how from that they got to the integral? I only know the following formula: $$ P(X\leq x, Y\leq y)=\int_{-\infty}^{x}\int_{-\infty}^y f_{X,Y}(u,v)\,du\,dv $$
Edit: The full solution they did:
\begin{align} \mathbb P\{ 0.25X^2>Y\}&\underset{total\ probability}{=}\frac{1}{2}\int_1^3\mathbb P\{x^2>Y\mid X=x\}\,\mathrm d x\\ &\underset{independant}{=}\frac{1}{2}\int_2^5\mathbb P\{Y<0.25x^2\}\,\mathrm d x\\ &=\frac{1}{6}\int_2^5\int_1^{0.25x^2}\boldsymbol 1_{[1,3]}(y)\,\mathrm d y\,\mathrm d x\\ &=\frac{1}{6}\int_2^5\int_1^{\min\{3,0.25x^2\}}\,\mathrm d y\,\mathrm d x\\ &=\frac{1}{6}\int_2^{\sqrt{12}}\int_0^{0.25x^2}\,\mathrm d y\,\mathrm d x+\frac{1}{6}\int_{\sqrt{12}}^5\int_1^3\,\mathrm d y\,\mathrm d x\\ &=... \end{align}