91Ó°ÊÓ

The joint p.d.f. of X, Y, and Z is given by,

\(f\left( {x,y,z} \right) = \left\{ {\begin{array}{*{20}{c}}2&{for\,0 < x < y < 1\,\,and\,\,0 < z < 1,}\\0&{otherwise.}\end{array}} \right.\)

Since,\(f\left( {x,y,z} \right)\)can be factored in the form,\(g\left( {x,y} \right)h\left( z \right)\)it follows that Z is independent of the random variables X and Y.

Hence, the required conditional probability is the same as the conditional probability\(\Pr \left( {3X > Y} \right)\).

It follows from the marginal p.d.f.\(h\left( z \right)\)is constant for\(0 < z < 1\).

Thus, this constant must be 1 and the marginal joint p.d.f. of X and Y must be,

\(g\left( {x,y} \right) = 2\,;\,0 < x < y < 1.\)

Therefore,

\(\begin{aligned}{}\Pr \left( {3X > Y\left| {1 < 4Z < 2} \right.} \right) &= \Pr \left( {3X > Y} \right)\\ &= \Pr \left( {X > \frac{Y}{3}} \right)\\ &= \int_0^1 {\int_{\frac{y}{3}}^y {g\left( {x,y} \right)dxdy} } \\ &= 2\int_0^1 {\int_{\frac{y}{3}}^y {dxdy} } \\ &= 2\int_0^1 {\left[ x \right]_{\frac{y}{3}}^y} dy\\ &= 2\int_0^1 {\left[ {y - \frac{y}{3}} \right]} dy\\ &= 2\int_0^1 {\frac{{2y}}{3}} dy\\ &= \frac{4}{3}\left[ {\frac{{{y^2}}}{2}} \right]_0^1\\ &= \frac{2}{3}\left( {1 - 0} \right)\\ &= \frac{2}{3}\\ &= 0.666667\end{aligned}\)

Therefore, the probability is 0.666667