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 customer, the hotel chain will have no more than three mishandled bags?
To solve this problem, we need to use the Poisson probability formula:
P(X=k) = (e^(-λ) * λ^k) / k!
Where:
- P(X=k) is the probability of getting k mishandled bags
- e is the base of the natural logarithm (approximately equal to 2.71828)
- λ is the average number of mishandled bags per 1,000 customers (4.06 in this case)
- k is the number of mishandled bags we are interested in (0, 1, 2, or 3 in this case)
Let's calculate the probabilities for each possible value of k and then sum them up:
P(X=0) = (e^(-4.06) * 4.06^0) / 0! = e^(-4.06) ≈ 0.017
P(X=1) = (e^(-4.06) * 4.06^1) / 1! = e^(-4.06) * 4.06 / 1 ≈ 0.069
P(X=2) = (e^(-4.06) * 4.06^2) / 2! = e^(-4.06) * 4.06^2 / 2 ≈ 0.141
P(X=3) = (e^(-4.06) * 4.06^3) / 3! = e^(-4.06) * 4.06^3 / 6 ≈ 0.183
Therefore, the probability that in the next 1,000 customers the hotel chain will have no more than three mishandled bags is the sum of these probabilities:
P(X<=3) = P(X=0) + P(X=1) + P(X=2) + P(X=3) ≈ 0.410
So, the probability that in the next 1,000 customers the hotel chain will have no more than three mishandled bags is approximately 0.410.