$\mathbb{P}(Z=X)=\alpha$ and $\mathbb{P}(Z=Y)=1-\alpha$, i.e. $Z$ is a mixture of two random variables $X$ and $Y$. I am stuck in the application of the law of total probability to show that $F_Z(z)=\alpha F_X(z)+(1-\alpha)F_Y(z)$:
$$F_Z(z)=\mathbb{P}(Z\leq z)=\mathbb{P}(Z\leq z|Z=X)\mathbb{P}(Z=X)+\mathbb{P}(Z\leq z|Z=Y)\mathbb{P}(Z=Y)=\alpha\mathbb{P}(Z\leq z|Z=X)+(1-\alpha)\mathbb{P}(Z\leq z|Z=Y)=\alpha\mathbb{P}(X\leq z|Z=X)+(1-\alpha)\mathbb{P}(Y\leq z|Z=Y)$$.
Dropping the conditioning is obviously not allowed, and writing the definition of conditional probability makes it look weirder. I feel using the "substitution law" is trickier than what I wrote here.