A major hotel chain keeps a record of the number of mishandled bags per 1 000 customers. In a recent year, the hotel chain had 2.16 mishandled bags per 1 000 customers. Assume that the number of mishandled bags has a Poisson distribution. What is the probability that in the next 1 000 customers, the hotel chain will have no mishandled bags?

Answer 1

To calculate the probability that in the next 1,000 customers there will be no mishandled bags, we can use the Poisson probability formula:

P(X = k) = (e^(-λ) * λ^k) / k!

Where:
- P(X = k) is the probability that exactly k mishandled bags occur
- λ is the average number of mishandled bags per 1,000 customers
- k is the number of mishandled bags (in this case, k = 0)

Given that λ = 2.16 mishandled bags per 1,000 customers, we need to first convert this to the average number of mishandled bags per customer:

λ = 2.16 / 1000 = 0.00216 mishandled bags per customer

Now we can plug in the values to the formula:

P(X = 0) = (e^(-0.00216) * 0.00216^0) / 0!
P(X = 0) = (e^(-0.00216) * 1) / 1
P(X = 0) = e^(-0.00216)

Using a calculator, we find that e^(-0.00216) ≈ 0.997845288

Therefore, the probability that in the next 1,000 customers, the hotel chain will have no mishandled bags is approximately 0.9978 or 99.78%.