= Solution
{c}
Starting at one, state two can first be reached only after an even number of steps. At every even time before that hit, the chain is again at one. By the <Markov property> and part (i), each successive two-step block hits two with <probability> $4/9$, while a failure returns to one with <probability> $5/9$. Therefore, for the first hitting time $H$,
$$
\boxed{\mathbb P(H=2k)=\left(\frac59\right)^{k-1}\frac49\quad(k\geq1),\qquad\mathbb P(H\text{ odd})=0.}
$$
Equivalently $H/2$ has the <geometric distribution> on the positive integers with success <probability> $4/9$.
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