Suppose I would like to compute the probability $\hat{p} :=P(\max\{XY-1,0\} > 0, Y \in [3,5])$ where $X \sim U(-1,1)$ and $Y \sim U(-5,5)$ are independent. To do this first, I considered $$ \hat{p} = P(\max\{XY-1,0\} > 0 | Y \in [3, 5])P(Y\in [3,5]) $$ Note that $P(Y\in [3,5]) = 2/10 = 1/5$. However, the resulting probability is a function of $Y$, not a number; $$ P(\max\{XY-1,0\} > 0 | Y \in [3, 5]) = P(X > 1/Y | Y \in [3, 5]) = (1-1/Y)\cdot \mathcal{I}_{[3,5]}(Y) $$ where $\mathcal{I}_A(x)$ is an indicator function. It seems that I can obtain the answer by integrating $Y$, i.e, $$ \hat{p} = P(Y\in [3,5]) E_Y[P(\max\{XY-1,0\} > 0 | Y \in [3, 5])] $$ which gives $$ \hat{p} = \frac{1}{5}\int_{3}^5 (1-1/Y)\frac{dy}{10}. $$
However this seems strange to me as the conditional probability should satisfy $$ P(A\cap B) = P(A|B)P(B) $$ without integration. I know that the conditional expectation satisfies $$ E_Y[E[XY|Y=y]] = E_Y[yE[X|Y=y]] = E[XY]. $$ and it seems that it is highly related to the answer for my question. However, I am sure how exactly these are related.
Any comments/suggestions/answers will be very appreciated. Thank you.
It seems all these conditioning/deconditioning/reconditioning steps only serve to muddy the waters for you.
To compute the probability of an event depending on $(X,Y)$, all that is needed is the joint distribution of $(X,Y)$, and we have it, hence we can simply write $$p=P(XY>1,3\leqslant Y\leqslant 5)=\iint_{\mathbb R^2}\mathbf 1_{xy>1,3\leqslant y\leqslant5}\,f_{X,Y}(x,y)\,dxdy$$ Using $$f_{X,Y}(x,y)=\frac1{20}\,\mathbf 1_{-1<x<1,-5<y<5}$$ this yields $$20p=\int_3^5\int_{1/y}^1dxdy=\int_3^5\left(1-\frac1y\right)dy=\left.(y-\log y)\right|_{y=3}^{y=5}$$ Thus, finally, $$p=\frac1{20}\left(2-\log\frac53\right)$$