Answer

Let \(a_{i}\) denote the abosorption probabilites into state \(4\) starting from \(i\)

\[ \begin{aligned} a_{5} &= 0, a{4} = 1 \newline a_{i} &= \sum_{j} a_{j}p_{ij}\newline a_{2} &= a_{1}p_{21} + a_{4}p_{24}\newline a_{3} &= a_{1}p_{31} + a_{2}p_{32} + a_{5}p_{35}\newline a_{1} &= a_{2}p_{12} + a_{3}p_{13} \end{aligned} \]

Solving, \(a_{1} = \frac{9}{14}, a_{2} = \frac{5}{7}\) and \(a_{3} = \frac{15}{28}\)

Let \(\mu_{i}\) denote the expected time till absorption starting from \(i\), then

\[ \begin{aligned} \mu_{4} &= 0 \newline \mu_{1} &= 1 + \mu_{2}p_{12} + \mu_{3}p_{13} \newline \mu_{2} &= 1 + \mu_{1}p_{21} + \mu_{4}p_{24} \newline \mu_{3} &= 1 + \mu_{1}p_{31} + \mu_{2}p_{32} \end{aligned} \]

Solving, \(\mu_{1} = \frac{55}{4}, \mu_{2} = 12\) and \(\mu_{3} = \frac{111}{8}\)