 Solution

Assume $K\ge 1$ is an integer, $0<p<1$, and put $q=1-p$. There is a discrepancy between the printed state definition and the formula requested later. We first follow the stated cumulative count, which retains previous good items when a defective item appears during phase A. The transition probabilities arewith all others zero. The balance equations imply $q{\pi }_i=q{\pi }_{i-1}$ for $1\le i<K$ and $(p/r){\pi }_K=q{\pi }_{K-1}$. ThusFor comparison, if clearance requires $K$ consecutive good items, each defect in phase A resets the state to zero. Replace its transitions by $P_{i,0}=p$, $P_{i,i+1}=q$, combining contributions if destinations coincide. Phase B retains its two transitions. For this consecutive-clearance inspection model, balance yields ${\pi }_i=q^i{\pi }_0$ and $(p/r){\pi }_K=q^K{\pi }_0$. With $D=1+(r-1)q^K$, normalization givesThis second model is the one giving the printed formula in part (iii). It is a different transition rule, rather than an algebraic simplification of the cumulative model.