Finding a conditional probability from joint p.d.f

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Hamiltonian
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Homework Statement
If the following joint p.d.f. can be considered for the random variables X, Y, and Z:
$$f(x,y,z) = \begin{cases} 2 & for & 0<x<y<1\ \&\ 0<z<1 \\ 0 & otherwise\end{cases}$$

Evaluate ##\mathbb{P}(2X > Y |1 < 4Z < 3).##
Relevant Equations
$$f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)} {f_{Y}(y)}$$
using the equation mentioned under Relevant Equations I can get, $$\mathbb{P}(2X > Y |1 < 4Z < 3) = \frac{\mathbb{P}(2X>Y, 1<4z<3)}{\mathbb{P}(1<4z<3)}$$ I can find the denominator by finding the marginal probability distribution, ##f_{Z}(z)## and then integrating that with bounds 0 to 1. But I am a little confused as to the limits of integration I need to use to find ##f_{Z}(z)## and then there's still the question of what I need to do to find the numerator.
$$f_{Z}(z) = \int_{?}^{?}\int_{?}^{?} f(x,y,z) dx dy$$

Additionally, I wonder if this approach is completely flawed and whether there is a better way to approach this problem.
 
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The approach is not flawed, but there is an easier way.
Are X and Y independent of Z?
If so how can we simplify the target expression ##\mathbb{P}(2X > Y |1 < 4Z < 3)##?

Regarding limits for integration, the starting point is ##-\infty## to ##+\infty##. But usually you can narrow that down by identifying the region over which the integrand is nonzero. If the region is rectangular, with sides aligned with coordinate axes, your limits will be simple constants. Otherwise your limits for the inner integral will depend on the values of the integration variable of the outer integral
 
andrewkirk said:
The approach is not flawed, but there is an easier way.
Are X and Y independent of Z?
If so how can we simplify the target expression ##\mathbb{P}(2X > Y |1 < 4Z < 3)##?
X and Y are independent of Z but are dependent on each other. So is ##\mathbb{P}(2X > Y |1 < 4Z < 3) = \mathbb{P}(2X>Y)##