Negative Binomial and Geometric Distributions

Question

So this is the problem I'm dealing with.

An actuary has determined that the number of claims per month can take any number 0, 1, 2, 3,... and follows a negative binomial distribution with mean 3 and variance 12. Calculate the probability that the number of claims is at least three but less than six.

So by using some properties of the negative binomial we can derive that p=0.25, and r=1. It was my understanding that a geometric distribution is just a negative binomial distribution with r=1. Given this I tried using the pmf for geometric distribution and got the wrong answer. Can someone explain to me what is going on here.

Answer 1

Since the number of claims $X\sim \text{Geometric}(\frac 14)$, the desired porbability is $P(X=3)+P(X=4)+P(X=5)=\left(\frac 34\right)^3\frac 14+\left(\frac 34\right)^4\frac 14+\left(\frac 34\right)^5\frac 14=0.2438965$

Answer 2

There are too many ways to parameterize the negative binomial, but using Klugman-Panjer-Wilmott as most US actuaries do, you can write it as:

$$ {r+k-1\choose k}\frac{\beta^k}{\left(1+\beta\right)^{r+k}} $$

where the mean is $r\beta$ and the variance is $r\beta(1+\beta)$.

So if $r\beta = 3$ and $r\beta(1+\beta) = 12$ then $(1+\beta) = 4$ so $\beta = 3$ and $r = 1$. Now solve for $P(k \in {3, 4, 5})$