The number of claims for lost luggage in a small city airport averages nine per day. Assuming the Poisson distribution, what is the probability that there will be fewer than 3
Claims on any given day?

Question

Quizplus · Accepted Answer

The Poisson distribution formula is: P(X = x) = (e^-λ * λ^x) / x!, where λ is the average number of events per time period and x is the number of events we want to calculate the probability for.

In this case, λ = 9 (average number of claims per day) and we want to find the probability of having fewer than 3 claims. So we need to calculate P(X < 3).

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
= (e^-9 * 9^0) / 0! + (e^-9 * 9^1) / 1! + (e^-9 * 9^2) / 2!
= 0.000123 + 0.001107 + 0.003338 ≈ 0.0062

Therefore, the probability of having fewer than 3 claims on any given day is 0.0062, which is closest to option (C).

The Number of Claims for Lost Luggage in a Small