6 customers enter 3 floored 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 1 st floor
Let's denote the customers as C1, C2, C3, C4, C5, C6.
For exactly 1 customer to dine on the 1st floor, we need one customer to choose the first floor and the other 5 customers to choose any of the other two floors.
The total number of ways this can happen is $\binom{6}{1} \times \binom{5}{5} = 6 \times 1 = 6$.
The total number of ways for each customer to choose a floor is $3^6 = 729$.
Therefore, the probability that exactly 1 customer dines on the 1st floor is $6/729 = 2/243$ or approximately 0.0082.