Let $X,Y,Z$ be independent random variables taking values $0$ and $1$ each with probability $1/2$. Define the random variable $U$ as $U = X$ if $Z=0$ and $U = Y$ if $Z=1$
Let $A$ be another random variable that takes $0$ with probability $p$ and $1$ with probability $1-p$. Define $C = U+A$. Find the probabilities $P(C=X), P(C=Y)$ and $P(C=Z)$.
I have done the first two and have a confusion with the third. I feel it should be zero.
You can make a table of outcomes. Note that I put $Z$ first since its outcome determines whether $U = X$ or $U = Y$. $$\begin{array}{c|c|c|c|c} Z & X & Y & C = U & C = U + 1 \\ \hline 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 & 2 \\ 0 & 1 & 1 & 1 & 2 \\ 1 & 0 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 2 \\ 1 & 1 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 & 2 \\ \end{array}$$ The fourth column, $C = U$, is taken with probability $p$. The fifth column, $C = U + 1$, is taken with probability $1-p$. Then $\Pr[C = Z]$ is $p$ times the number of rows in which $C = Z$ from the fourth column, plus $1-p$ times the number of rows in which $C = Z$ from the fifth column, divided by $8$, since each triplet $(Z, X, Y)$ has equal probability $1/8$ of occurring. Since there are $4$ rows in the fourth column that match, and $2$ rows in the fifth column that match, the desired probability is $$\Pr[C = Z] = \frac{4p + 2(1-p)}{8} = \frac{p+1}{4}.$$