= Solution
An uneventful day leaves one aphid, a one-offspring day leaves the parent plus its offspring, and a two-offspring day leaves three aphids. Death leaves none. Thus the daily descendant counts are $0,1,2,3$ with <probabilities> $(1-q)t$, $q$, $(1-q)r$, $(1-q)s$, respectively. The <mean> and <probability generating function> are
$$
\boxed{\mathbb EX_1=q+(1-q)(2r+3s),\qquad
G(z)=(1-q)t+qz+(1-q)rz^2+(1-q)sz^3.}
$$
Independence across individuals and days makes the daily population a <Galton-Watson process>, with each surviving parent counted as one of the next day's descendants.
The standard <Galton-Watson extinction fixed point> result states that the eventual <extinction probability of a branching process> $e$ from one ancestor is the smallest root in $[0,1]$ of $G(e)=e$. Moreover a nondegenerate offspring law of <mean> at most one has <extinction probability of a branching process> one. Here, since $q<1$, the fixed-point equation reduces to
$$
\boxed{e=t+re^2+se^3,}
$$
which contains no $q$. This proves <laziness invariance of branching extinction>: uneventful days change the timescale, not eventual extinction. The excluded case $q=1$ would keep one aphid forever.
If $2r+3s\le1$, the offspring <mean> is at most one. In this parameter range $t>0$: if $t=0$, then $r+s=1$ would imply $2r+3s\ge2$. Thus the offspring law is not deterministically one, and the extinction criterion applies. \b[Extinction is certain.]
For the specified <probabilities>, the fixed-point equation factors as
$$
2e^3+e^2-5e+2=(e-1)(2e-1)(e+2)=0.
$$
Its roots in $[0,1]$ are $1/2$ and $1$, and the smaller root is the <extinction probability of a branching process>:
$$
\boxed{e=\frac12.}
$$
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