= Solution
State $2$ can be reached from state $1$ only at an even time. In the two-step sampled chain, each step while still at state $1$ reaches state $2$ with probability $4/9$, and otherwise stays at state $1$ with probability $5/9$. Thus, if $T$ is the original hitting time,
$$
\boxed{P(T=2r)=\left(\frac59\right)^{r-1}\frac49,\quad r\geq1,
\qquad P(T\text{ odd})=0.}
$$
Equivalently $T=2G$ where $G$ has the <geometric distribution> on positive integers with parameter $4/9$.
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