Does Binomial variables independence implies Bernoulli variables independence $X$, $Y$ are independent variables with Binomial distribution. $X={\sum_{i=1}^nX_i}$, $Y={\sum_{i=1}^nY_i}$.
$X_i$, ($1\le i\le n$) are independent Bernoulli variables.
Same applies for $Y_i$
Is the set of $X_i$ and $Y_i$ independent?

Surprisingly, the answer is no.
Consider the case $n=2$ with probability space $\{0,1\}^4$ and
$X_1, X_2$ the first two coordinate functions and $Y_1, Y_2$ the second two.
The probabilities of the $16$ different configurations are
$$\begin{array}[cccc]{}
x_1 & x_2 & y_1 & y_2 & p(x_1,x_2,y_1,y_2)\cr
0 & 0 & 0 & 0 &1/16\cr
0 & 0 & 0 & 1 &2/16\cr
0 & 0 & 1 & 0 &0\cr
0 & 0 & 1 & 1 &1/16\cr
0 & 1 & 0 & 0 &2/16\cr
0 & 1 & 0 & 1 &2/16\cr
0 & 1 & 1 & 0 &0\cr
0 & 1 & 1 & 1 &0\cr
1 & 0 & 0 & 0 &0\cr
1 & 0 & 0 & 1 &0\cr
1 & 0 & 1 & 0 &2/16\cr
1 & 0 & 1 & 1 &2/16\cr
1 & 1 & 0 & 0 &1/16\cr
1 & 1 & 0 & 1 &0\cr
1 & 1 & 1 & 0 &2/16\cr
1 & 1 & 1 & 1 &1/16\cr
\end{array}$$
Then unless I've miscalculated $X_1, X_2$ are independent Bernoulli(1/2), $Y_1, Y_2$ are independent Bernoulli(1/2), $X_1+X_2$ and $Y_1+Y_2$ are independent, but e.g. $X_1$ and $Y_1$ are dependent: $P(X_1 = 1, Y_1 = 1) = 7/16$. 
