A scientist inoculates several mice, one at a time, with a disease germ until he finds 2 mice that have attacked by the disease. If the probability of getting attack of the disease is 1/6, what is the probability that 8 mice are required

To answer this question, we shall apply the Negative Binomial distribution.

Let contracting the disease be considered a "success". The number of success $r=2$ is what we are interested in.

Let $n=8$ be the sample size of the sample until we get 2 successes. The probability of contracting the disease is $p={1\over6}$. The form of the Negative Binomial distribution we apply is given as,

$p(X=n)=\binom{n-1}{r-1}p^r(1-p)^{n-r}$.

Therefore,

$p(X=8)=\binom{7}{1}({1\over6})^2({5\over6})^6=7\times{1\over36}\times{15625\over46656}=0.0651$

Therefore, the probability that 8 mice are required is 0.0651