Question

The number of claims per month paid by an insurance company is modelled by a random variable N with p.m.f satisfying the relation

p(n + 1) =

1/3p(n), n = 0,1,2,...

where p(n) is the probability that n claims are filed during a given month

(a) Find p(0).

(b) Calculate the probability of at least one claim during a particular month given that there have been at most four claims during the month. 

Answer 1

$a:\\p\left( n+1 \right) =\frac{1}{3}p\left( n \right) \Rightarrow p\left( n \right) =\frac{p\left( 0 \right)}{3^n}\\\sum_{n=0}^{\infty}{p\left( n \right) =1}\Rightarrow p\left( 0 \right) \sum_{n=0}^{\infty}{\left( \frac{1}{3} \right) ^n}=1\Rightarrow p\left( 0 \right) =\frac{1}{\frac{1}{1-\frac{1}{3}}}=\frac{2}{3}\\b:\\P\left( X\geqslant 1|X\leqslant 4 \right) =\frac{P\left( 1\leqslant X\leqslant 4 \right)}{P\left( X\leqslant 4 \right)}=\frac{\sum_{n=1}^4{\frac{2}{3}\cdot \frac{1}{3^n}}}{\sum_{n=0}^4{\frac{2}{3}\cdot \frac{1}{3^n}}}=\frac{\frac{2}{3}\left( \frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4} \right)}{\frac{2}{3}\left( 1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4} \right)}=0.330579$