Let the transition probability matrix of a Markov chain, $\left(X_{n}\right)_{n \geq 0}$, be given by $$ P=\left(\begin{array}{cccccc} 0 & 0 & 1 / 2 & 1 / 2 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 / 2 & 1 / 2 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{array}\right) $$
Compute $\mathbb{P}\left(X_{2}=6 \mid X_{1} \in\{3,4\}, X_{0}=2\right)$ and $\mathbb{P}\left(X_{2}=6 \mid X_{1} \in\{3,4\}\right)$.
My attempt
Edited:
$\mathbb{P}\left(X_{2}=6 \mid X_{1} \in\{3,4\}, X_{0}=2\right)$
The only path you can take is $ 2 -> 4 -> 6 $ hence the probability is $1$.
Now for
$\mathbb{P}\left(X_{2}=6 \mid X_{1} \in\{3,4\}\right)$ since we don't know what $X_0$ there can be multiple probability as below:
Path 1:
$1 -> 4 -> 6$ with probability $\frac{1}{2}$
Path 2:
$1 -> 3 -> 6$ with probability $\frac{1}{4}$
Path 3:
$2 -> 4 -> 6$ with probability $1$
Am I right?