Markov Chain Exercise. A Markov chain has the transition probability matrix

$$P=
        \begin{bmatrix}
        0.7 &amp; 0.2 &amp; 0.1 \\
        0 &amp; 0.6 &amp; 0.4 \\
        0.5 &amp; 0 &amp; 0.5 \\
        \end{bmatrix}.
$$

Let us number the states $0,1,2$.

Now I have to determine (a) the conditional probability $Pr{\{X_2=1,X_3=1|X_1=0\}}$

(b) If it is known that the process starts in state $X_0=1$, what is the probability $Pr{\{X_0=1,X_1=0,X_2=2\}}$ ?

My Attempt:

(a) $$Pr{\{X_2=1,X_3=1|X_1=0\}}=Pr{\{X_2=1|X_1=0\}}Pr{\{X_3=1|X_1=0\}}=(0.2)Pr{\{X_3=1|X_1=0\}},$$

for calculating $Pr{\{X_3=1|X_1=0\}}$, I need two step transition probability matrix :

$$P^2=
        \begin{bmatrix}
        0.54 &amp; 0.26 &amp; 0.2 \\
        0.2 &amp; 0.36 &amp; 0.44 \\
        0.6 &amp; 0.1 &amp; 0.3 \\
        \end{bmatrix}.
$$

Hence  $$Pr{\{X_2=1,X_3=1|X_1=0\}}=(0.2)Pr{\{X_3=1|X_1=0\}}=(0.2)(0.26)=0.052,$$ which doesn't match with the result. Where am I doing mistake ?

(b)$$Pr{\{X_0=1,X_1=0,X_2=2\}}=Pr{\{X_0=1\}}P_{1,0}P_{0,2},$$
where $P_{0,2}$ denotes transition from state $0$ to state $2$ in one step. 

But my question is  the problem hasn't specified the initial distribution, so I am not getting the value of $Pr{\{X_0=1\}}$, consequently not getting $Pr{\{X_0=1,X_1=0,X_2=2\}}$. But the problem has an answer. How can I proceed ?
(a) Since the process if Markov, we can write

$$
P(X_2 = 1, X_3 = 1 \mid X_1 = 0)
    = P(X_2 = 1 \mid X_1 = 0) \cdot P(X_3 = 1 \mid X_2 = 1)
$$

and you only need the transition matrix itself to answer.

(b) If you're given that $X_0 = 1$, then (as Antitheos also points out) $P(X_0 = 1) = 1$, and you can write

\begin{align}
P(X_0 = 1, X_1 = 0, X_2 = 2)
    &amp; = P(X_1 = 0, X_2 = 2 \mid X_0 = 1) \\
    &amp; = P(X_1 = 0 \mid X_0 = 1) \cdot P(X_2 = 2 \mid X_1 = 0)
\end{align}

and you can proceed as before.
a)
First, by the definition of the term
$$P(X_2=1,X_3=1 |X_1=0) = P(X_3=1|X_2=1,X_1=0)\cdot P( X_2=1|X_1=0)$$
and by Markoc chain definition
$$P(X_3=1|X_2=1,X_1=0)\cdot P( X_2=1|X_1=0) =P(X_3=1|X_2=1)\cdot P( X_2=1|X_1=0) =p_{1,1}\cdot p_{1,0} $$

b)
Read the problem again:


  it is known that the process starts in state $X_0=1$


This means
$$P(X_0=1) =1.$$