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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.

wkwilliam
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  • Oh, sorry. That was a typo. I meant $7^3/8^3$ – wkwilliam Nov 02 '16 at 12:07
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    Yes this is correct... what is the probability of the first ball not going in that cell? AND the second? AND the third? Answering these gives you another way of coming up with $7^3/8^3$. – JP McCarthy Nov 02 '16 at 12:08
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    It is indeed also the probability that any other SPECIFIC cell is empty. It is smaller, however, from the probability that at least one cell is empty, and even from the probability that exactly one cell is empty. – barak manos Nov 02 '16 at 12:22
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    @barakmanos: Since we only have three balls but eight cells, the other probabilities you mentioned will be $1$ and $0$ respectively. – String Nov 02 '16 at 12:24
  • @String: Haha, ooops, that's right :) – barak manos Nov 02 '16 at 12:26

2 Answers2

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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, $\mathsf{P}(A_1) = \frac{7^3}{8^3}$.

user980341
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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!

UweM.
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