You are given the following properties of the distribution of the annual number of claims, N:

1. [math]\operatorname{P}(N=k) = p_k, \quad k = 0,1,2,\ldots[/math]
2. [math]p_0 = 0.45 [/math]
3. [math]\frac{p_n}{p_m} = \frac{m!}{n!} [/math] for [math]m \geq 1 [/math] and [math] n \geq 1 [/math]

Calculate the probability that at least two claims occur during a year.

• 0.16
• 0.18
• 0.21
• 0.23
• 0.26

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

Key: D

Members of the ( a, b,1) class have

[[math]]\frac{p_k}{p_{k-1}} = a + \frac{b}{k}[[/math]]

for [math]k = 2,3,4, \ldots, [/math] so point iii) implies that [math]N [/math] is an (a,b,1) distribution with [math]a = 0 [/math] and [math] b = \lambda = 1 [/math]. This, together with point ii), implies the distribution is a zero-modified Poisson distribution with [math]p_0^M = 0.45 [/math].

[[math]] \begin{aligned} P( N \geq 2) = 1 − P( N \leq 1) &= 1 − ( P( N = 0) + P( N = 1)) \\ &= 1 −(0.45 + \frac{1-p_0^M}{1-P(L=0)} P(L =1 ) ), \end{aligned} [[/math]]

where [math]L[/math] has a Poisson distribution with [math]\lambda = 1 [/math] and [math]P(L=0) = P(L=1) e^{-} [/math], so [math]P(N \geq 2) = 1- (0.45 + \frac{0.55}{1-e^{-1}} ) = 0.2299 [/math].

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