Probability when cutting the stick twice 
Given a stick of length $l$. We cut the stick twice. Let $X$ be the random variable defined by the length of the stick after the first cut, and $Y$ be the random variable defined by the length of the stick after the second cut. What is the probability for $Y>\frac{l}{4}$?

First, the PDF of $X$ is uniform, thus $f_{X}(x)=\frac{1}{l}$ (since $0\le X\le l$). $Y$ is uniform in $[0,X]$, then $f_{Y|X}(y|x)=\frac{1}{x}$. Then the probability of $Y>\frac{l}{4}$ is $$\int_{l/4}^{x}\frac{1}{x}dy=1-\frac{l}{4x}$$
On the other hand, we have $$f_{X,Y}=f_Xf_{Y|X}=\frac{1}{lx}$$ 
We write the probability of $Y>\frac{l}{4}$ as follow: $$P\left(\frac{l}{4}<Y<x|\frac{l}{4}\le X \le l\right)=\frac{P(l/4<Y<x,l/4\le X \le l)}{P(l/4\le X\le l)}=\frac{\int_{l/4}^{x}\int_{l/4}^{l} \frac{1}{lx}dxdy}{\frac{1}{l}}$$
which leads to a different result from $1-\frac{l}{4x}$.
My question is which of the solution is correct, and what's wrong with the incorrect solution?
Thanks in advance.
 A: Both solutions are incorrect. Without loss of generality we might assume $l=1$. The first solution computes $P(Y > 1/4 | X = x)$ and is furthermore not valid if $x < 1/4$.
The second solution has a very confusing notation (the $x$ should be $X$?) but I think you try to compute $P(Y > 1/4 | X > 1/4)$, however the integration order is swapped. You could integrate with respect to x for the outer integral with limits $1/4$ to $1$ and y for the inner integral with limits from $1/4$ to $x$. Or you could use the order you have but then outer integral (for y) should have limits from $1/4$ to 1 and inner integral (for x) should have limits from $y$ to $1$.
The probability in the denominator should be $P(X > 1/4) = 3/4$ but is irrelevant since they didn't ask for a conditional probability.
A: Let the random variables $U$ and $V$ be uniformly distributed in $[0,1]$. We have to compute the probability $P$ that $UV\geq{1\over4}$. This probability is equal to the area above the hyperbola $uv={1\over4}$ in the square $[0,1]^2$, and is given by
$$P=\int_{1/4}^1\left(1-{1\over 4 u}\right)\>du=\left(u-{1\over4}\log u\right)\Biggr|_{1/4}^1={3\over4}-\log\sqrt{2}\doteq0.4034\ .$$
