6 customers enters 3 floored restaurant. Each customer decides on which floor to dinner. Assume customer decisions are independent and each floor is equally likely. Find probability exactly 1 customer dines at 1st floor
First, we need to determine the total number of possible outcomes for the 6 customers to dine on the 3 floors. Since each customer can choose any floor independently, there are $3^6$ total possible outcomes.
Next, we need to determine the number of ways in which exactly 1 customer dines at the 1st floor. There are 6 ways to choose which of the 6 customers will dine at the 1st floor. The remaining customers can choose any of the 2 remaining floors, so there are $2^5$ ways for them to do so.
Therefore, the number of favorable outcomes is $6 \cdot 2^5 = 192$.
Finally, the probability that exactly 1 customer dines at the 1st floor is:
$$\frac{192}{3^6} = \frac{192}{729} \approx 0.2639$$
So, the probability is approximately 0.2639 or 26.39%.