A bag contains $5$ black balls and $7$ white balls. Balls are drawn of the bag e organized in a row of $12$ balls. Every configuration has the same probability of happening.
For example, if we set $B = Black$ and $W = White$, one possible configuration is $B \ W \ B \ B \ W \ W \ W \ B \ W \ W \ B \ W$.
Going through this row from left to right, whats is the expected value of the number of black balls that are immediately followed for a white ball?
My attempt: I think I should enumerate tha black balls ($k = 1, 2, 3, 4, 5$) and define $X_k$ as follows:
$X_k =\left\{\begin{array}{rc} 1,&\mbox{if ball $k$ is followed by a white ball}\\ 0, \mbox{otherwise} \end{array}\right. $
I'm having some trouble to evaluate $\mathbb{E}(X_k)$. Once I do that, then $\mathbb{E}(A) = \sum_{k=1}^{5} \mathbb{E}(X_k)$, where $A$ is the event of a black ball being followed by a white ball. And I also know that $\mathbb{E}(X_k) = \mathbb{E}(X_1), \forall k \in \{1,2,3,4,5\}$.
Because $\mathbb{E}(A) = \sum_{k=1}^{5} \mathbb{E}(X_k)$ and $\mathbb{E}(X_k) = \mathbb{E}(X_1), \forall k \in \{1,2,3,4,5\}$, we have $\mathbb{E}(A)=5\mathbb{E}(X_{1})$. $P(X_{1}=1)=\frac{7}{5+7}=\frac{7}{12}$.
Thus $\mathbb{E}(X_{1})=\frac{7}{12}$ and $\mathbb{E}(A)=\frac{35}{12}$.