Conditional probability of correlated Bernoulli +Gaussian. If $X_1$ bea  Bernulli random variables with parameter $p$ (i.e., $P[X_1=0]=p$) such that
\begin{align}
X_2=X_1 \oplus B,
\end{align}
where  $B$ is Bernulli with parameter $q$ and independent of $X_1$  Then, $X_2$ is Bernulli with $r=qp+(1-q)(1-p)$.  
Now let 
\begin{align}
V_1= X_1+Z_1,\\
V_2=X_2+Z_2,
\end{align}
where $Z_1$ and $Z_2$ are strandard normal and independent.  
Can someone show (or outline the method) of how to compute pdf's $f_{V_1,V_2|X_1}, f_{V_1,V_2}$?
I was able to find the marginals
\begin{align}
f_{V_1}(v)&= \frac{1}{\sqrt{2 \pi}}E[ e^{-\frac{(v-X_1)^2}{2}} ]=  (1-p)\frac{1}{\sqrt{2 \pi}} e^{-\frac{(v-1)^2}{2}} + p\frac{1}{\sqrt{2 \pi}} e^{-\frac{(v)^2}{2}}\\
f_{V_2}(v)&= \frac{1}{\sqrt{2 \pi}}E[ e^{-\frac{(v-X_2)^2}{2}} ]=  (1-r)\frac{1}{\sqrt{2 \pi}} e^{-\frac{(v-1)^2}{2}} + r\frac{1}{\sqrt{2 \pi}} e^{-\frac{(v)^2}{2}}\\
\end{align}
 A: I assume $Z_1, Z_2$ are independent. ($X_1,X_2$ clearly aren't).
It's not immediately clear whether $V_1,V_2$ are jointly (even marginally) gaussian. What should be clear is this: if we are given $X_1$, then $V_1,V_2$ are independent.
Indeed, given $X_1=0$ : $V_1 = Z_1 \equiv N(0,1)$ and $V_2 = Z_2 + B \equiv (1-q) N(0,1) + q N(1,1)$
And given $X_1=1$ : $V_1 = Z_1 +1 \equiv N(1,1)$ and $V_2 = Z_2 + 1 - B \equiv (1-q) N(1,1) + q N(0,1)$
This can be written as $f_{V_1|X_1}(v_1)=N(v_1;X_1,1)$ and $f_{V_2|X_1}(v_2)=(1-q) N(v_2;X_1,1) + q N(v_2;1-X_1,1)$
Because they are conditionally independent
$$f_{V_1,V_2|X_1} = f_{V_1|X_1}f_{V_2|X_1}$$
From this you can get 
$$f_{V_1,V_2}  = \sum_{X_1} f_{V_1,V_2|X_1} P(X1)$$

Update:
Can $f_{V_1,V_2|X_2}$ can be factored? It's true that $B$ is independent of $X_1$, not of $X_2$. But, again, consider both cases:
In the case $f_{V_1,V_2|X_2=0} $, we have $V_2=Z_2$ and $V_1=Z_1 +B$, so $V_1,V_2$ are (conditioned to $X_2=0$) independent.
In the case $f_{V_1,V_2|X_2=1} $, we have $V_2=Z_2+1$ and $V_1=Z_1 +1-B$, so again $V_1,V_2$ are (conditioned to $X_2=1$) independent.
Hence, if I'm not mistaken, $f_{V_1,V_2|X_2}=f_{V_1|X_2}f_{V_2|X_2}$ 
