I did the question below using both Binomial and Negative Binomial distribution and got very different answers. Being 100% positive that I applied the formulae correctly, I would like to know where my mistake is, is it in the assumption or variables values?
Question: Given that the probability of drink and drive is 3%, what is the probability that the police will find exactly 4 cases of drinking and driving in testing 200 drivers?
My attempt:
$X= B(200,0.03)$ where 200 is the # of trials, and 0.03 is the success rate.
$Y = NB(4,0.03)$ where 4 is the # of success, and 0.03 is the rate of success.
\begin{align} \Pr(X = 4) &= \binom{200}{4}0.03^40.97^{200-4}\\ &= 0.1338 \\ \end{align}
\begin{align} \Pr(Y = 200) &= \binom{199}{3}0.03^40.97^{200-4}\\ &= 0.002676 \\ \end{align}
If you used negative binomial, and calculated $\Pr(Y=200)$, it means that the fourth drink and drive person had to be tested the last (200th person), and you denied any other option. That is why the probability is so small, compared to the binomial one.
Use binomial instead, $\Pr(X=4)$ means that there are 4 person who drink and drive, regardless whether the last person to be tested is drink and drive person or not