Conditional expectation and variance

Question

$$f_Y(y)\in U(0,1)$$ $$f_{X\mid Y=y}(y)\in \Gamma(3,y)$$ $$0\lt X\lt \infty,\quad 0\lt Y \lt 1 $$

Question: find $EX$ and $\operatorname{var}X$

My attempt:

$$EX=E(E(X\mid Y))=E(3Y)=\frac{3}{2}$$ $$\operatorname{var}X=E(X^2)-E(X)^2=E(9y^2)-\frac{9}{4}=9\int_0^1\mkern -8mu y^2 f_Y(y)\,dy-\frac{9}{4}=9*\frac{1}{3}-\frac{9}{4}=\frac{3}{4}$$

Solution in the book: $EX=\frac{3}{2}$ and $\operatorname{var}X=\frac{7}{4}$

I can't seems to find where my calculation went wrong, any feedback is appreciated!

Answer 1

Thanks for the hints.

$$\operatorname{var}X=E(\operatorname{var}(X\mid Y))+\operatorname{var}(E(X\mid Y))=E(3Y^2)+\operatorname{var}(3Y)= 3\int_0^1y^2f_Y(y)dy+9\operatorname{var}Y=1+\frac{9}{12}=\frac{7}{4}$$