Probability that the first cell is empty There are three distinctive balls to distribute to 8 cells. Each cell can hold multiple balls. I'm trying to figure out the probability $P(A)$ that, after distribution, the first cell is empty.
My thoughts: In total, there are $8^3$ possibilities to distribute the three distinctive balls to the cells, and there are $7^3$ possibilities to distribute the balls to all cells but the first.
So $P(A) = 7^3/8^3.$
Is this correct? I'm confused since this could also be the probability of any one cell being 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, $\mathsf{P}(A_1) = \frac{7^3}{8^3}$.
