How can I tell whether this is a negative binomial distribution or binomial distribution?

Question

A scientist inoculates several mice, one at a time, with a disease until he finds 3 that have contracted the disease. If the probability of contracting the disease is 1/6, what is the probability that 8 or more mice are required?

At first, my initial instinct was that this is an example of a negative binomial, where X = the total number of trials to achieve the rth success (being 3 mice that have contracted the disease). I then realized if I had set up the problem as such, where $P(Y\geq 8)$, or $1-P(Y<8)$, I would have a negative and a zero value in my set up, which would not be possible.

This is the point where I find myself rather stuck at how to proceed.

Answer 1

you can do it like this:

$$P(Y \ge 8| succ == 3) = P(Y == 7| succ == 0) + P(Y == 7| succ == 1) + P(Y == 7| succ == 2) = \binom{7}{0}(\frac{5}{6})^7 + \binom{7}{1}\frac{1}{6}(\frac{5}{6})^6 + \binom{7}{2}(\frac{1}{6})^2(\frac{5}{6})^5$$

it's possible to model it with negative binomial distribution but you would end up with sumation over some condition anyway

Answer 2

Considering the definition of each is the best way to tell. For binomial, you have independent, constant number of trials and the variable (what is random) is the number of successes. Just based on this definition, the situation you have already does not fit since in your situation, the number of trials is not constant and is in fact, the variable.

When we look at NB, one of its formulations (there are several), fits your situation, where we count the number of trials before some condition is met (3 successes). So NB does fit the distribution of your random variable.