1st Yr Probability: question about marginal and joint pdfs for $3$ uniform continuous independent random variables Background
I'm trying to improve my understanding of the relationship between marginal and joint pdfs for calculating specific probabilities.
The Problem
$X$, $Y$, $Z$ are independent and uniformly distributed $(0,1)$.
What is $P(X>YZ)$?
My question
The book solution is below, but I'm wondering if I can solve this with the marginal distribution of $X$ alone.
The marginal pdf of $X$ is $f_X(x) = 1$
In theory with $f_X(x)$ I can calculate any probability for $X$. I believe that is the whole point of having a pdf for a random variable.
Therefore:$$P(X>YZ) = \int_{YZ}^{1}dx$$
$$=1-YZ$$
But this definitely isn't the right answer (which as you see below is $3/4$).
The book solution makes complete sense to me.
My question is why can't we get the answer from the marginal pdf of $X$? Shouldn't a marginal pdf for a RV answer all probability statements for that RV?
Thanks for your help and patience!

Book Solution

 A: To understand why you also need to integrate with respect to $Y$ and $Z$,
then think about this sentence:

Evaluating a probability is very similar to evaluating the area/volume
  of a set.

To be more precise, suppose that you want to find the probability that $$(X,Y,Z) \in V,$$
where $V$ is a particular set in $\mathbb{R}^3$.
In you case, 
$$V = \{(X,Y,Z) : X > YZ, X \in [0,1], Y \in [0,1], Z \in [0,1]\}.$$ 
Then, the probability associated is:
$$P((X,Y,Z) \in V) = \iiint_V f_{X,Y,Z}\,dx\,dy\,dz,$$
where $f_{X,Y,Z}$ is the joint distribution of $X$, $Y$ and $Z$.
In your specific case, we know that $f_{X,Y,Z} = 1$ for all $(X,Y,Z) \in Q$, where $Q = [0, 1] \times [0, 1] \times [0,1].$
Then, the probability is
$$P((X,Y,Z) \in V) = \iiint_V \,dx\,dy\,dz.$$
Notice that, $\iiint_V \,dx\,dy\,dz$ corresponds to the volume of $V$. We can say that the volume of $V$ corresponds to the number of favorable cases.
As a final (and obvious) remark, notice that:
$$P((X,Y,Z) \in Q) = \iiint_Q \,dx\,dy\,dz$$
is exactly the volume of the cube $Q$. In particular, the probability that $(X,Y,Z) \in Q$, is $1$, and corresponds to the number of possible cases.
Concluding, you need to integrate with respect to all the variables, since calculating a probability is like evaluating the volume of a set.
