= Solution
This is an <immigration--death process>, equivalently an <M-M-infinity queue>: admission occurs at rate $\alpha$ regardless of occupancy, and each of $n$ animals is discharged independently at rate $\delta$. Its state-transition diagram through four animals is
$$
0\underset{\delta}{\overset{\alpha}{\rightleftarrows}}1\underset{2\delta}{\overset{\alpha}{\rightleftarrows}}2\underset{3\delta}{\overset{\alpha}{\rightleftarrows}}3\underset{4\delta}{\overset{\alpha}{\rightleftarrows}}4\underset{5\delta}{\overset{\alpha}{\rightleftarrows}}\cdots.
$$
Every upper arrow points right and every lower arrow points left. Thus $q_{n,n+1}=\alpha$ and $q_{n,n-1}=n\delta$ for $n\geq1$, with no discharge from zero. Four is only the limit of the requested sketch, not an absorbing state or a capacity boundary.
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