My attempt:
$\mathbb{P}(X\_1,...,X\_4|X\_1+...+X\_4=2)=\frac{\mathbb{P}(X\_1,...,X\_4,X\_1+...+X\_4=2)}{\mathbb{P}(X\_1+...+X\_4=2)}$

$\mathbb{P}(X\_1+...+X\_4)\sim Binom(4,p)\Leftrightarrow \mathbb{P}(X\_1+...+X\_4=2)= {4 \choose 2}p^2(1-p)^2$

And I dont know how to find the following.
$\mathbb{P}(X\_1=x\_1,...,X\_4=x\_4,X\_1+...+X\_4=2)$

edit:

By drawing an event table of $X\_1,...,X\_4$ against $\sum X\_i$ we see that for
all the ${4 \choose 2}$ cases of two $\ 0s$ and two $1s$, each case have a probability of $p^2(1-p)^2$. Which is $\frac{1}{{4 \choose 2}}\mathbb{P}(\sum X\_i=2)$.

Is this the answer for $\mathbb{P}(X\_1=x\_1,...,X\_4=x\_4|X\_1+...+X\_4=2)$?