Find $P(X/Y \leq t)$, $P(XY \leq t)$, and use it to find $P(XY/Z \leq t)$ Let X, Y, Z be independent uniform (0,1) random variables.  
We know that the uniform (0, 1) pdf will be f(x|a,b)=1/(b-a).  I'm not really seeing how you would deal with this.  For instance, how would you be able to bring Z into the equation if you're given two different $P(X/Y \leq t)$ and $P(XY \leq t)$?  It doesn't seem like it makes a whole lot of sense.  
 A: Because $(X , Y)$ is a uniformly distributed vector on the unit square, therefore $P(X/Y\leq t)$ is equal to the proportion of that square above the line $y = x/t$.
Since $(X, Y)$ is such, $P(XY\leq t)$ is the proportion of the unit square below the curve $y = t/x$.
Can you find formula for these proportionate areas? (Hint: piecewise functions)
Likewise evaluate the proportionate volume covered by $xy/z\leq t$ inside the unit cube.
A: Don't think about it from the perspective of the pdf, but think about it probabilistically. Knowing that $X$, $Y$, and $Z$ are independent allows you to multiply probabilities. In addition, since $X$, $Y$, and $Z$ are between 0 and 1, you know that the product of any two of these must be less than either on their own. For example, $XY \leq Y$ and $XY \leq X$. If $XY \leq t$, then $X$ must be less than $t$. Similarly, $X/Y \geq X$ is true but $X/Y \geq Y$ need not be.
Hope this helps!
A: This is Ex. 4.51 from Casella and Berger's Statistical Inference. The solution provided online does not seem correct. Here is my take. Since the integration gets a bit messy, I am not sure if mine is correct either. Someone interested in this may check it


A: \begin{align}
P\left(\frac XY \le t\right) = P\left(X\le tY\right)&=\int_0^1\int_0^{\min(ty, 1)} \mathrm dx \mathrm dy\\
&= \int_{0}^1 \left(ty\mathbf 1_{ty < 1} + \mathbf 1_{ty\ge 1}\right)\mathrm dy\\
&= t\int_{0}^1y\mathbf 1_{y< \frac1t}\mathrm dy + \int_0^1 \mathbf 1_{y\ge \frac1t}\mathrm dy\\
&= \begin{cases}
\frac12 t&\text{if $0\le t < 1$}\\
1-\frac1{2t} &\text{if $t\ge 1$}
\end{cases}
\end{align}
Do the same thing for others.
