I am struggling with the question below and would appreciate any help.
Let $Z$ be a random variable with a continuous distribution having density $$ f_{Z}(z)=\left\{\begin{array}{ll} 2 z & \text { if } 0 \leq z \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ Let $X$ and $Y$ be two random variables, independent given $Z=z$ satisfying $$ \begin{array}{l} \mathbf{P}(X=1 \mid Z=z)=z=1-\mathbf{P}(X=0 \mid Z=z) \end{array} $$ and $$ \begin{array}{l} \mathbf{P}(Y=1 \mid Z=z)=z=1-\mathbf{P}(Y=0 \mid Z=z) \end{array} $$ Find the conditional probability $$ \mathbf{P}(Z \leq 1 / 2 \mid X=1 \text { and } Y=1). $$
I tried to solve it below: \begin{align*} &\mathbf{P}(Z \leq 1 / 2 \mid X=1 \text { and } Y=1)\\&=\frac{\mathbf{P}(X=1,Y=1|Z\leq1/2)\mathbf{P}(Z\leq1/2)}{\mathbf{P}(X=1,Y=1|Z\leq1/2)\mathbf{P}(Z\leq1/2)+\mathbf{P}(X=1,Y=1|Z>1/2)\mathbf{P}(Z>1/2)}\\ &=\frac{\mathbf{P}(X=1|Z\leq1/2)\mathbf{P}(Y=1|Z\leq1/2)\mathbf{P}(Z\leq1/2)}{\mathbf{P}(X=1|Z\leq1/2)\mathbf{P}(Y=1|Z\leq1/2)\mathbf{P}(Z\leq1/2)+\mathbf{P}(X=1|Z>1/2)\mathbf{P}(Y=1|Z>1/2)\mathbf{P}(Z>1/2)}\\ &=\frac{\mathbf{P}(X=1|Z\leq1/2)\mathbf{P}(Y=1|Z\leq1/2)1/4}{\mathbf{P}(X=1|Z\leq1/2)\mathbf{P}(Y=1|Z\leq1/2)1/4+\mathbf{P}(X=1|Z>1/2)\mathbf{P}(Y=1|Z>1/2)3/4}\\ \end{align*} I know $X$ and $Y$ are Bernoulli random variables given $Z$ but I don't know how to proceed from here. I think the right answer is 0.125 based on my simulation.
I think I know how to solve it now, $$ \begin{array}{l} P(Z \leq 1 / 2 \mid X=1 \text { and } Y=1) &=\dfrac{P(X=1, Y=1\mid Z \leq 1 / 2 )P(Z \leq 1 / 2 )}{P(X=1,Y=1)}\\ &=\dfrac{\int_0^{1/2}z\cdot z\cdot 2z dz}{\int_0^{1}z\cdot z\cdot 2z dz}\\ &=1/16. \end{array} $$