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. 
 A: 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, $\mathsf{P}(A_1) = \frac{7^3}{8^3}$.
A: Is there an equal probability for each cell to receive a ball? In this case, you can treat this problem using the Binomial distribution. There are two possibilities: either a ball comes into the 1st cell or not. The probability for a ball to get into the 1st cell is 1/8, the probability not to get into this cell is 7/8. We have 3 trials in total. Let X be a random variable, the number of balls in the 1st cell. Then, $P(X=k)=\binom{N}{k}\cdot p^k (1-p)^{N-k}$. You are looking for the probability that $k=0$ for $N=3$ with $p=1/8$. We get $$P(X=0)=\binom{3}{0}\cdot \left(\frac{1}{8} \right)^0 \left( \frac{7}{8}\right)^3 = \left(\frac{7}{8}\right)^3$$ It is the probability to hit the other cells three times in a row. So your answer is correct!
