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

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.

• Why do you think you'd have a negative number or a zero? Commented Oct 9, 2019 at 19:47
• Because (from my limited understanding), I assumed my set up would be: 1-[P(X=0)+P(X=1)+P(X=2)...] and so forth up until 7. In the negative binomial, the initial combination is (x-1) and (k-1) and if X equals 0 or 1, then it would be negative or zero. Commented Oct 9, 2019 at 19:51
• You can model the number of failures using a negative binomial, then the total number of trials is just the number of failures + 3. The NB distribution can be used either way, but you have to define $Y$ differently, as you have discovered; the way you have written it in the response to my question is the "# of failures" way. Commented Oct 9, 2019 at 20:13
• Alternatively, can't I use the binomial distribution where I look for P(8 or more mice are required) = the probability that in 7 mice where less than or equal to 2 mice contracted the disease? Commented Oct 9, 2019 at 20:36
• Look at the different formulations of the negative binomial here: en.wikipedia.org/wiki/Negative_binomial_distribution (compare the main "number of failures" one with the "number of trials" one in the table). Commented Oct 10, 2019 at 0:59

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