when 2 defective items are extracted one by one in a partition consisting of 6 products and checked, Let x be the number of test before finding the last defective items fing probability distribution of X
i realize the possibility are:
D=defective, N=Non defective
X=0 , DD
X=1, NDD
X=2, NNDD
X=3, NNNDD
X=4 , NNNNDD
$\mathbb P(X=2)=\frac{{{2\choose 2}}}{{6\choose 2}}$
$\mathbb P(X=3)=\frac{{{2\choose 2}}{{4\choose 1}}}{{6\choose 2}}$
$\mathbb P(X=4)=\frac{{{2\choose 2}}{{4\choose 2}}}{{6\choose 2}}$
$\mathbb P(X=5)=\frac{{{2\choose 2}}{{4\choose 4}}}{{6\choose 2}}$
is this right?
$P(X=k)$ is the probability that at the $kth$ try you found the second defective, and in the first $k-1$ tries you found only one defective. So $$\begin{align}P(X=k)=\frac{\text{number of ways of choosing 1 defective from k-1 tries}}{\text{number of ways of choosing k-1 objects from six}}&=\frac{k-1\choose1}{6\choose k-1}\end{align}$$