Given the transition probability matrix
$P=\begin{Vmatrix} 0.9 & 0.08 & 0.02\\ 0.85 & 0.10 & 0.05\\ 0 & 0 & 1 \end{Vmatrix},\qquad P^{(2)}=P^2=\begin{Vmatrix} 0.878 & 0.08 & 0.042\\ 0.85 & 0.078 & 0.072\\ 0 & 0 & 1 \end{Vmatrix} $
calculate $P(X_3\not=2,X_1\not=1\mid X_0=0)$.
Here is what I have done so far:
$P(X_3\not=2,X_1\not=1\mid X_0=0)$
$= P(X_3\not=2 \mid X_1\not=1, X_0=0) \cdot P(X_1\not=1\mid X_0=0)$
$= (P_{00}^2+P_{01}^2+P_{20}^2+P_{21}^2)\cdot(P_{00}+P_{02})$
$=(0.878 + 0.08 + 0 + 0)\cdot (0.90 + 0.02)\approx 0.881 $.
My though process was to sum up the probability of the possible routes between $X_0,X_1$ and $X_1,X_3$. The problem is that this seems like a very manual way of doing it, and it includes $P_{02}$, which makes $X_3\not=2$ impossible since state 2 is absorbing.
Is there a rule or formula I'm forgetting, or have I done it correctly?
Since state $2$ is absorbing, it's clear that both $\{X_1=2\}$ and $\{X_2=2\}$ imply $\{X_3=2\}$. So conditioned on $\{X_0=0\}$, to have $\{X_3\ne 2,X_1\ne 1\}$ we must have $\{X_1=0\}$, which occurs with probability $0.9$. Conditioning on $\{X_1=0\}$, we compute \begin{align} \mathbb P(X_3\ne2\mid X_1=0) &= \mathbb P(X_3=0\mid X_1=0) +\mathbb P(X_3=1\mid X_1=0)\\ &= P^2_{00} + P^2_{01}\\ &= 0.878 + 0.08\\ &= 0.958, \end{align} and hence \begin{align} \mathbb P(X_3\ne 2,X_1\ne1\mid X_0=0) &= \mathbb P(X_3\ne2\mid X_1=0)\cdot\mathbb P(X_1=0\mid X_0=0)\\ &= 0.958\cdot0.9\\ &= 0.8622. \end{align} Your work is correct aside from including the transition from state $0$ to state $2$.