Find the marginal density for [tex]Y[/tex] by integrating the joint PDF over all possible values of [tex]x[/tex]:
[tex]f_Y(y)=\displaystyle\int_{-\infty}^\infty f_{X,Y}(x,y)\,\mathrm dx=\int_0^y2\,\mathrm dx[/tex]
[tex]\implies f_Y(y)=\begin{cases}2y&\text{for }0<y<1\\0&\text{otherwise}\end{cases}[/tex]
Then the density of [tex]X[/tex] conditioned on [tex]Y[/tex] is
[tex]f_{X|Y}(x\mid y)=\dfrac{f_{X,Y}(x,y)}{f_Y(y)}=\frac1y[/tex]
for [tex]0<x<y[/tex] and undefined elsewhere.
Thus
[tex]P(0.25<X<0.50\mid Y=0.9)=\displaystyle\int_{0.25}^{0.50}f_{X|Y}(x\mid y=0.9)\,\mathrm dx=\int_{0.25}^{0.50}\frac{\mathrm dx}{0.9}\approx\boxed{0.28}[/tex]
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