What is the probability that exactly one of the customers dines on the first floor of the restaurant? **Question** : Six customers enter a three-floor restaurant. Each customer decides on which floor to have dinner. Assume that the decisions of different customers are independent, and that for each customer, each floor is equally likely. Find the probability that exactly one customer dines on the first floor. **My solution** :It is as each customers were tossing three possibilities of probabilities 1/3 each, so it is a product of multinomial models. But, can't we say that they can toss the first floor with probability 1/3 versus the second floor + third with probability 2/3? In which case the answer is $$\frac13\Big(\frac23\Big)^5 $$ but somehow, it seems to me that its wrong... **Edit** This is the answer for "the first client is at floor 1", so the answer of the problem is $$ 6\times \frac13\Big(\frac23\Big)^5 $$

I will count the number of possibilties in which exactly one customer dines on the first floor, then count the total number of possibliities (without restriction), then divide the former by the latter.

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There are $6$ choices for which customer is the one that will dine on the first floor. Given this, each of the $5$ remaining customers have $2$ choices for the floor on which they'll dine, for which there are $2^5$ possibilities.

This means that the number of possibilities in which exactly $1$ customer is on the first floor is $$6\cdot 2^5$$

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The total number of possibilities is $3^6$, since each customer has $3$ choices.

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This means that the desired probability is $$\frac{6 \cdot 2^5}{3^6} = \frac{2^6}{3^5} = \boxed{\frac{64}{243}\,}$$