I am self-learning probability theory from William Feller's Introduction to Probability theory and its applications. I would like to ask for some help in deriving the correct solution to the below very interesting problem.
Problem IX.12
Suppose that items with a probability $p$ of being acceptable are subject to inspection in such a way, that the probability of an item being inspected is $p'$. We have four classes, namely, "acceptable and inspected", "acceptable but not inspected" and so forth with probabilities $pp'$, $pq'$, $p'q$ and $qq'$. We are concerned with double Bernoulli trials. Let $N$ be the number of items passing the inspection desk (both inspected and uninspected) before the first defective is found, and let $K$ be the (undiscovered) number of defectives among them. Find the joint distributions of $N$ and $K$, and the marginal distributions.
Solution (My Attempt).
We have,
\begin{array}{c|cc} & \text{Acceptable} & \text{Defective}\\ \hline \text{Inspected} & pp' & qp'\\ \text{Undiscovered} & pq' & qq'\\ \end{array}
The first defective item is found at trial number $(n+1)$, if the preceding $n$ items were either acceptable or uninspected.
$P\{\text{Acceptable} \cup \text{Uninspected}\} = P\{\text{Defective},\text{Inspected}\}^C = 1 - qp'$
$N$ is the waiting time to the first defective. $N$ follows a geometric distribution.
$P\{N = n\} = (1 - qp')^n qp' \tag{1}$
$K$ is the number of undiscovered defectives among these $n$ trials. So, given that we waited for time $n$ to find the first defective, the probability that the number of defectives equals $k$ is given by,
$P\{K = k \vert N = n\} = {n \choose k} (qq')^k p^{n-k}$
Thus, the joint distribution
\begin{align*} P\{K = k, N = N\} &= P \{K = k \vert N =n \} \cdot P \{N = n\}\\ &= {n \choose k} (qq')^k p^{n-k} (1 - qp')^n qp' \end{align*}
However, the textbook states the expression for the joint distribution as,
$$ P\{K = k, N = n\} ={n \choose k}(qq')^k p^{n-k} qp' $$
Also, how to sum over all $n$ to derive an expression for the marginal distribution of $N$?
For the joint distribution note that
$$P(n,k)=P(\text{k defects over n trials})P(\text{k not detected})P(\text{defect detected at n+1})$$
$$={n \choose k}p^{n-k}q^k \times (q')^k\times q p' $$
$$={n \choose k}p^{n-k}\times (qq')^k\times q p' $$