In each of the months June, July, and August, the number of accidents occurring in that month is modeled by a Poisson random variable with mean 1. In each of the other 9 months of the year, the number of accidents occurring is modeled by a Poisson random variable with mean 0.5. Assume that these 12 random variables are mutually independent.

Calculate the probability that exactly two accidents occur in July through November.

• 0.084
• 0.185
• 0.251
• 0.257
• 0.271

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: B

The months in question have 1, 1, 0.5, 0.5, and 0.5 respectively for their means. The sum of independent Poisson random variables is also Poisson, with the parameters added. So the total number of accidents is Poisson with mean 3.5. The probability of two accidents is

[[math]] \frac{e^{-3.5}3.5^2}{2!} = 0.185. [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.