Constructing the transition matrix for device failure There are three machines. Let the probability that an operable machine fails on any
given day be $0.1$, independently of the other machines. Only one machine can be repaired on the same day (so it is available for the next). 
Machine $1$ feeds both machines $2$ and $3$. Thus, if machine $1$ fails, there is no production that day. If $1$ is down, it is always repaired first. If both $2$ and $3$ are down, and $1$ is up, then $2$ is repaired. Machines that are down remain down for the day. Machines that are up today may fail the next day, regardless if production is available or not. But a repaired machine is certain to operate the next day.
How would I construct the transition matrix for this problem? I have figured that there are $8$ state combinations:
$(1)$ - $1$ is down
$(2)$ - $2$ is down
$(3)$ - $3$ is down
$(4)$ - $1$,$2$ are down
$(5)$ - $1$,$3$ are down
$(6)$ - $2$,$3$ are down
$(7)$ - $1$,$2$,$3$ are down
$(8)$ - Everything works.
But I can't figure out how they would relate to each other (which states are transient and which ones are recurrent/absorbing). Any help would be greatly appreciated.
 A: Clearly all states are recurrent ; there is always a non-zero probability of going from any of the states $1$-$7$ (the broken states) to state $8$ (the repaired state) in at most $3$ steps (repairing all the machines one by one without them breaking down) and at this step, we are in state $8$, and from state $1$ you can go to any other state in one step with non-zero probability. Therefore, starting somewhere, you might end up in any state four steps later with non-zero probability.
You are right, there are $\sum_{k=0}^3 \begin{pmatrix} 3 \\ k \end{pmatrix} = 2^3 = 8$ possible states (and you wrote them all down, that's even better). So you need to compute the probability of going from one state to another. 
Say you are in a complicated state to compute ; suppose you are in state $(5)$. Machines $1$ and $3$ are broken, so in the next state, for sure we will have $1$ repaired and $3$ broken. But $2$ will be working with probability $0.9$, and will be broken with probability $0.1$. So from state $(5)$ you go to state $(3)$ with probability $0.9$, and to state $(6)$ with probability $0.1$.
You compute all the others this way. This gives you the matrix : 
$$
\rho = 
\begin{bmatrix} 
0     & 0.09  & 0.09  & 0     & 0     & 0.01  & 0     & 0.81  \\
0.09  & 0     & 0.09  & 0     & 0.01  & 0     & 0     & 0.81  \\
0.09  & 0.09  & 0     & 0.01  & 0     & 0     & 0     & 0.81  \\
0     & 0.9   & 0     & 0     & 0     & 0.1   & 0     & 0     \\
0     & 0     & 0.9   & 0     & 0     & 0.1   & 0     & 0     \\
0     & 0     & 0.9   & 0     & 0.1   & 0     & 0     & 0     \\
0     & 0     & 0     & 0     & 0     & 1     & 0     & 0     \\
0.081 & 0.081 & 0.081 & 0.009 & 0.009 & 0.009 & 0.001 & 0.729 \\
\end{bmatrix}
$$
where the $ij^{\text{th}}$ entry is the probability from going to state $i$ to state $j$, given the ordering above. (If you don't like my argument that all states are recurrent, compute the fourth power of this matrix with a computer and see that $\rho(i,i) > 0$ for $1 \le i \le 8$ for a more computational proof.)
Hope that helps,
