Probability that the first cell is empty

Define $\Omega:= \{C_1,\ldots, C_8\}^3$. Each element $\omega \in \Omega$ has the form $$ \omega = (C_i,C_j,C_k) $$ and tells you that the first ball is in cell $C_i$, the second is in $C_j$ and the third ball is in cell $C_k$. It is also allowed that for instacne $C_i = C_j$. Now consider the event that no ball is in the cell $C_n$ with fixed index $n \in \{1, \ldots, 8\}$, i.e. $$ A_n := \{ \omega \in \Omega \colon C_n \notin \omega\}= \big(\{C_1,\ldots, C_8\} \setminus \{C_n\}\big)^3. $$ Then $|\Omega| = 8^3$, $|A_n|=7^3$ and so $$ \mathsf{P}(A_n) = \frac{|A_n|}{|\Omega|} = \frac{7^3}{8^3}. $$ Especially, $\mathsf{P}(A_1) = \frac{7^3}{8^3}$.