= Solution
{c}
At the empty <Markov chain state>, only the constant-rate production <chemical reaction> is possible. Thus the <first-passage time> $\tau_1$ has an <exponential distribution> of rate $2$:
$$
\mathbb P(\tau_1>t)=e^{-2t},\qquad t\geq0.
$$
Its <expected value> can be obtained by integrating this <survival function>, and its second <moment> by integrating $2t$ times the <survival function>:
$$
\mathbb E\tau_1=\int_0^\infty e^{-2t}\,dt=\frac12,
\qquad
\mathbb E\tau_1^2=2\int_0^\infty te^{-2t}\,dt=\frac12.
$$
Consequently the requested <expected value> and <variance> are
$$
\boxed{\mathbb E\tau_1=\frac12,\qquad \operatorname{Var}(\tau_1)=\frac14.}
$$
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