= Consecutive-clearance inspection model
{title2=$I=[1+(r-1)(1-p)^K]^{-1}$}
In the <consecutive-clearance inspection model>, a run of $K$ consecutive good items is required to leave full inspection. With $q=1-p$, $0<p<1$, its <Markov chain> has transitions $P_{i0}=p$, $P_{i,i+1}=q$ for $i<K$, and $P_{K0}=p/r$, $P_{KK}=1-p/r$. Thus every defect in the clearance phase resets the count, rather than preserving the cumulative total.
Writing $D=1+(r-1)q^K$, the <stationary distribution> is
$$
\pi_i=\frac{pq^i}{D}\quad(0\leq i<K),\qquad\pi_K=\frac{rq^K}{D}.
$$
To prove this, balance at $i=1,\ldots,K-1$ gives $\pi_i=q^i\pi_0$, while balance at $K$ gives $(p/r)\pi_K=q^K\pi_0$. Summing the geometric series and normalizing gives $\pi_0=p/D$; balance at zero follows as well. The inspected fraction is $1-(1-1/r)\pi_K=1/D$. The fraction of all production that is both defective and uninspected is $p(r-1)q^K/D$; the fraction among defective items alone is $(r-1)q^K/D$. All these long-run limits follow from the <Markov chain ergodic theorem> and the independent inspection coins. The distinction from cumulative clearance matters already at $K=2$.
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