The conditional probability and the expected value for a sequence of independent variables $X_1, X_2, \dots $ is sequences of independent random variables with a value in $\{ 0,1\}$ s.t $p(X_i=1)=p,~~ \text{for}~~ i \geq 1 $ where $p \in (0,1)$. Assume now we have another sequence $(Y_i )_{i \geq 1}$ of independent random variables with a value in $\{ 0,1\}$ such that $ p (Y_i = 1) = q~$ and also  $ q  \in (0, 1).$
Knowing that $U_n = \sum_{i=1}^n X_i$, $V_n = \sum_{i=1}^n X_i Y_i$ and $N = \inf\{n \geq 0, V_{n+1} = 1\}$, I would like to find $E(U_n)$ and the probability $p(U_n = i / N=n )$.
I just see let and assume, I don't know from where to start so any kind of help would be appreciated.
From the comments below, $U_n$ distirbute Binomial $B(n,p)$, the same distribution for $V_n$ as $B(n,pq)$ and $N$ distributes geometric$(pq)$.
How could I use all of this to find the conditional probability?
\begin{eqnarray}
p(U_n = i / N=n ) &=& \frac{p(U_n = i, N=n)}{p(N=n)} ~~\text{but}~~p(U_n = i, N=n)=unknown \nonumber \\
&=& \frac{p(N=n/ U_n = i)p(U_n=i)}{p(N=n)}~~ \text{how to find}~~ p(N=n/ U_n = i)p(U_n=i)
\end{eqnarray}
 A: Since $X_i$'s independent Bernoulli with parameter $p$, $U_i\sim Binom(i,p)$, hence $\mathbb E[U_i] = ip$. 
Concerning $N$ we have
$$ P(N=n) = P(V_1=0,\ldots,V_n =0, V_{n+1} = 1) = P\bigg(\bigcap_{i=1}^n (\{X_i = 0\}\cup \{Y_i = 0\}) \cap \{X_{n+1}=1, Y_{n+1}=1\} \bigg) = (1-pq)^n (pq) $$
Considering the event $\{U_n=i,N=n\}$:
the favorable outcomes are $ ((x_1,y_1),(x_2,y_2),\ldots,(x_n,y_n), (x_{n+1},y_{n+1}))$ s.t. $\sum_{j=1}^n x_j = i$ and $ \prod_{j=1}^n x_jy_j = 0$ and $y_{n+1}=1=x_{n+1}$. The probability of such a fixed outcome is $p^j(1-q)^{i} (1-p)^{n-i} pq$, since there need to be $i$ indices with $x_j=1$ and the corresponding $y_j$'s have to be zero, the remaining $n-i$ (not counting the very last one, where $x_{n+1}=y_{n+1}=1$) pairs can be anything but $(1,1)$ and $(1,0)$. The number of such outcomes is $ \binom{n}i $, since there is $\binom{n}i$ ways to chose those $i$ indices with $x_j=1$ and $y_j=0$. So
$$ P\bigg( U_n = i, N=n \bigg) = \binom{n}i p^i(1-q)^i (1-p)^{n-i}pq $$
Update: this means that 
$$U_n | N=n \sim Binom\bigg(n, \frac{p(1-q)}{1-pq}\bigg)$$
