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 4.06 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?
0.0254
0.0687
0.0256
0.0172
0.0025
The probability of having no mishandled bags in a Poisson distribution with 4.06 mishandled bags per 1000 customers is given by the formula:
P(X = 0) = (e^(-λ) * λ^0) / 0!
Where λ is the average number of mishandled bags per 1000 customers.
In this case, λ = 4.06. Thus, the probability can be calculated as:
P(X = 0) = (e^(-4.06) * 4.06^0) / 0!
P(X = 0) = (e^(-4.06) * 1) / 1
P(X = 0) = e^(-4.06)
P(X = 0) = 0.0172
Therefore, the probability that in the next 1,000 customers, the hotel chain will have no mishandled bags is 0.0172 or 1.72%.
Hence, option "0.0172" is the correct answer.