An electronic scale in an automated filling operation stops the manufacturing line after five overweight packages are detected. Suppose that the probability of an overweight package is 0.05 and each fill is independent. What is the mean number of fills before the line is stopped?
This was my approach:
Let the r.v. $X$ be the boxes with no overweight such that $r=5$ is the number of boxes detected and $p=0.05$ is the probability of detecting those boxes. This implies that $X\sim \text{Negative Binomial}(r=5,p=0.05)$.
$$E(X)=\frac{r(1-p)}{p}=\frac{5(1-0.05)}{0.05}=\frac{5(0.95)}{0.05}=95$$
I am not sure if this is the correct approach?