2-state Continuous-time Markov Chain: probability of at least 1 state change given in state B The process stays in state A for exponentially distributed amount of time with mean 2 hours and then moves to state B. The process stays in state B with mean of 3 hours. 
So the question asks what the probability would be for at least one state change to occur in one hour given that it starts in B. So my approach was as follows:
$$P(X_1 \geq 1|\lambda_B = \tfrac{1}{3}) = 1 - P(X_1 = 0|\lambda_B = \tfrac{1}{3})\\ = 1 - \frac{\tfrac{1}{3}^0e^{-\tfrac{1}{3}}}{0!} = 0.283  $$
I used a Poisson process as I only need to count that it doesn't transition and get the probability that at least one transition occurs from that value then. I am not sure whether this is the correct procedure though. I am having trouble unifying this with the Transition Function $P(t)$.
If we let generator matrix be as follows:
$$ Q = \left[ \begin{matrix}-0.5 & 0.5\\ \tfrac{1}{3} & - \tfrac{1}{3} \end{matrix} \right] $$
Then we get transition function for the first hour to be:
$$P(1) =  \left[\begin{matrix}0.660758925104247 & 0.339241074895753\\0.226160716597169 & 0.773839283402831\end{matrix}\right]
 $$
This doesn't correspond with the above calculation. But, in all honesty I do not actually know how to  interpret the transition function i.e. I am not sure what $P(1)$ actually represents. Could you please help explain the meaning behind it and help me to figure out which calculation is the correct approach. I want to ask first before attempting to verify my question via simulation. Oh, I  noticed $(P(1)_{12} + P(1)_{12})/2 = 0.2827 $ but I have very little justification for doing this.
 A: The transition matrix, being an exponential of the generator matrix, takes into account all possible transitions during that hour. Thus, while you calculated the probability $\mathrm e^{-\frac13}$ of staying in state $B$ for the whole hour, the bottom right element of the transition matrix is greater than this because it includes not only this probability but also the probabilities for any number of back-and-forth transitions from $B$ to $A$ and back. Similarly, the bottom left entry is not your probability $1-\mathrm e^{-\frac13}$ because this is just the probability of making a transition from $B$ to $A$ at some point during the hour, whereas the transition matrix contains the probability of ending up at $A$ at the end of the hour (after starting the hour at $B$), which is the probability of transitioning to $A$ and then staying there, plus the probability of transitioning to $A$ and then going from $A$ to $B$ and back any number of times. The matrix entry is less than your probability because the dominating term in the difference is the probability of transitioning to $A$ and back to $B$ and then staying there until the end of the hour.
Let me know if you have trouble figuring out which of these probabilities you actually need in order to solve your problem and I can say more on that.
