I am currently studying this course chapter 5.3. I have this labeled CTMC (Continuous-Time Markov Chains) with three BSCCs (bottom strongly connected components). Let's say we want to calculate the steady-state probability of being in a state with the label {a}.
We know that the states s1, s2 and s5 hold the label {a}. s2 is a transient state, so we are only interested in B1 and B3.
Using the equation below:
We derive one of the solutions is:
My question is how did we get 2/3? In other words, how did we get 2/3 from Pr{reach B1 from s0)?
I confess, it took me too long to admit how much, but I got it:
First of all, that $B_3$ is wrong, it's still $B_1$, then just solve $$x = \frac 1 2 \cdot 1 + \frac 1 2 \cdot\frac 1 2 \cdot x$$
If we are reading the same document (the link you posted is broken), then you can find the same in the next paragraph (click me).
This time, that $B_1$ is wrong, it's still $B_3$, then just solve $$x = \frac 1 2 \cdot \left( \frac 1 2 \cdot x + \frac 1 4 \right)$$