Negative binomial distribution.Calculate the probability that exactly 12 packages are inspected. In a shipment of 20 packages, 7 packages are damaged. The packages are randomly inspected, one at a time, without replacement, until the fourth damaged package is discovered.
Calculate the probability that exactly 12 packages are inspected.
I used negative binomial distribution, with $p=$ $\frac{7}{20}$,$r=4$ and $x=8$, where $r$ is the number of successes, $x$ is the number of failures before $r^{th}$ success and $p$ is the probability of success of a particular trial.
$p(x)  = 
\binom {r+x-1}{x}p^r(1-p)^x
$
$p(x=8)  = 
\binom {11}{8}(\frac{7}{20})^4(1-\frac{7}{20})^8=0.079
$
However the answer is 0.119. What have I done wrong?

Take all possibilities of arranging 7 defective packages amongst 20.
The ones with fourth package on position 12 can be calculated by placing 3 packages among the first 11 and 3 among the last 8:
${{11}\choose{3}} * {{8}\choose{3}} / {{20}\choose{7}}$
(11 choose 3) * (8 choose 3)/(20 choose 7) = 0.11919504644
