6 customers enter three floor restaurant. Each customer can decide about which floor to dine-in. Assume decisions are independent and each floor is equally likely. Find probability that exactly 1 customer dines at 1 st floor
Let's denote the floors as follows: floor 1, floor 2, and floor 3.
The total number of ways 6 customers can choose their floors is 3^6 = 729.
To find the probability that exactly 1 customer dines at the 1st floor, we need to calculate the following:
(1 customer at 1st floor) * (5 customers at floors 2 and 3)
There are 6 ways for 1 customer to dine at the 1st floor.
For the remaining 5 customers, they can choose either floor 2 or floor 3. There are 2 choices for each of them.
Therefore, the total number of ways for 1 customer to dine at the 1st floor and 5 customers to dine at floors 2 and 3 is 6 * 2^5 = 192.
Therefore, the probability that exactly 1 customer dines at the 1st floor is:
P(exactly 1 customer at 1st floor) = 192 / 729 ≈ 0.2639
So the probability is approximately 0.2639 or 26.39%.