= Solution
The fitted <progressive illness-death model> has three allowed arrows: $1\to2$ with estimated <transition intensity> $0.1849$, $1\to3$ with $0.01935$, and $2\to3$ with $0.06143$, all in years$^{-1}$. State 3 is an <absorbing state>, and state 2 has no return arrow.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-30-state-transitions.png]
{title=Progressive three-state model with estimated annual transition intensities}
{height=360}
Once in state 2, the <holding time> has an <exponential distribution> with rate $q_{23}$, so the <mean holding time from a transition intensity matrix> is
$$
\boxed{\widehat E(T_2)=\frac1{0.06143}=16.28\ \text{years}.}
$$
Apply a <confidence interval for an inverse rate> to the printed rate interval $(0.03552,0.1063)$: since inversion reverses order, the approximate 95% interval for the mean is
$$
\boxed{\left(\frac1{0.1063},\frac1{0.03552}\right)=(9.41,28.15)\ \text{years}.}
$$
The <time-homogeneous Markov property> makes the future depend only on the current state; the <exponential distribution> also has the <memoryless property>. For a person currently in state 2,
$$
\boxed{p_{23}(2)=1-e^{-2(0.06143)}=0.11561.}
$$
Equivalently, $p_{22}(2)=p_{22}(1)^2\simeq0.9404163^2$. Thus the fitted two-year death <probability> is approximately \b[11.6%].
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