A continuous time Markov chain {$X(t); \ t \ \geq0$} has the state space $S_X = \ ${$0,1,2$} and the following transition matrix : $$\begin{bmatrix}-8 & 4 & 4 \\ 3 & -5 & 2 \\ 0 & 2 & -2\end{bmatrix}$$ The chain starts in state $2$, so $X(0) = 2$.
We must compute the probability of A : "the chain stays in state 2 at least until time $t=1$".
I tried to compute it as : $$P(A) = 1 - P(\text{chain goes to state $1$ in the time interval} \ (0,1]) - P(\text{chain goes to state $0$ in the time interval} \ (0,1])$$
But I found that $$P(\text{chain goes to state $1$ in the time interval} \ (0,1]) = \frac{2}{0+2} = 1$$ and $$P(\text{chain goes to state $0$ in the time interval} \ (0,1]) = \frac{0}{0+2} = 0$$
which gives the result $$P(A) = 1 - 1 - 0 = 0$$
that seems a bit weird, can someone explain to me what I did wrong ?