I'm trying to solve an excercise that states:
"A random number X is drawn at (0,1). Next, another random number Y with uniform distribution is drawn at (0, X). The probability that Y is greater than 0.5 is: (answer: 0.15)"
In theory I have two uniform distributed variables $X≈U(0, 1)$ and $Y≈(0, X)$ and I am given the conditional distribution: given $X=x$, $Y$ is uniformly distributed between $0$ and $x$.
That is, for any $x\in(0,\,1)$ and any $y\in(0,\,x)$, the conditional density function of $Y$ should be $f_{Y|X}(y\mid x)=\frac{1}{x}$ but i don't really know how to go on from here, since its conditionals, it seems like using the total law of probability would be a choice:
$P\left(Y\gt \tfrac12\right)=1 - P\left(Y\leq \tfrac12\right)=1 -\int_0^1 P\left(Y\leq \tfrac12\mid X=x\right)f_X(x)\,\mathrm dx.$
$f_X(x)$ should be $1$ I think but I have no clue on how to calculate $P\left(Y\leq \tfrac12\mid X=x\right)$
Solutions are welcome if possible.
HINT: they said to you that $Y|X\sim \operatorname{Unif}(0,X)$, this means that
$$ \Pr [Y\leqslant c|X=x]=\begin{cases} \frac{c}{x},& \text{ when }c\in(0,x)\\ 1,&\text{ when }c\geqslant x\\ 0,&\text{ otherwise } \end{cases} $$
Now observe that
$$ \Pr [Y>1/2]=1-\Pr [Y\leqslant 1/2]=1-\int_{\mathbb{R}}\Pr [Y\leqslant 1/2|X=x]P_X(dx) $$
where $P_X$ is the distribution of $X$.