A flood insurance company determines that [math]N[/math], the number of claims received in a month, is a random variable with [math]\operatorname{P}[N=n] = \frac{2}{3^{n+1}}[/math] for [math]n = 1,2, \ldots [/math]. The numbers of claims received in different months are mutually independent.

Calculate the probability that more than three claims will be received during a consecutive two-month period, given that fewer than two claims were received in the first of the two months.

• 0.0062
• 0.0123
• 0.0139
• 0.0165
• 0.0185

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

Solution: E

Let M and N be the random variables for the number of claims in the first and second month. Then

[[math]] \begin{align*} \operatorname{P}[M + N \gt 3 | M \lt 2] − \operatorname{P}[ M + N ≤ 3 | M \lt 2] &= 1 - \frac{\operatorname{P}[ M + N ≤ 3, M \lt 2]}{\operatorname{P}[ M \lt 2]} \\ &= \operatorname{P}[ M = 0, N = 0] + \operatorname{P}[ M =1, N = 0] + \operatorname{P}[ M = 0, N = 1] + \operatorname{P}[ M =1, N =1] \\ &= 1 - \frac{\operatorname{P}[ M =0, N = 2] + \operatorname{P}[ M = 1, N = 2] + \operatorname{P}[ M =0, N = 3]}{\operatorname{P}[ M =0] \operatorname{P}[ M =1]} \\ &= 1 - \frac{(2 / 3)(2 / 3) + (2 / 9)(2 / 3) + (2 / 3)(2 / 9) + (2 / 9)(2 / 9) + (2 / 3)(2 / 27) + (2 / 9)(2 / 27) + (2 / 3)(2 / 81)}{2 / 3+ 2 / 9} \\ &= 1 - \frac{0.87243}{0.88889} = 0.0185. \end{align*} [[/math]]

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