Given a matrix with probability of moving from one state to another.
$$\begin{bmatrix} 0&1&0&0&0&1 \\ 4&0&0&3&2&0 \\ 0&0&0&0&0&0 \\ 0&0&0&0&0&0 \\ 0&0&0&0&0&0 \\ 0&0&0&0&0&0 \end{bmatrix}$$
I have to calculate the probability of reaching state s3 s4 s5 from s0. the answer to them are $3\over14$, $1\over7$, $9\over{14}$ respectively.
While my answers are s3, s4, s5 are $2\over9$, $1\over6$, $1\over9$.
Am I missing anything? Also How would I go with this when there are multiple paths?
$$\begin{bmatrix} 0&1&0&0&1&1 \\ 4&0&0&3&2&1 \\ 0&0&0&0&0&0 \\ 0&0&0&0&0&0 \\ 0&0&0&0&0&0 \\ 0&0&0&0&0&0 \end{bmatrix}$$
Elaborating on my comment -
Explanation of the matrix:
If you index the matrix starting from $0$, then the element at $M_{ij}$ gives the number of ways of getting from state $S_i$ to $S_j$.
Take the second row, $i=1$, which is $[4,0,0,3,2,0]$, for example. This tells us that there are $4$ ways of getting to $S_0$, $3$ to $S_3$ and $2$ to $S_4$.
The probability of $A$ given $B$ is
$$P(A|B)={\textrm{the number of ways of getting to $A$ from $B$}\over\textrm{the total number of ways of getting anywhere from $B$}}$$
The probabilities you get, therefore, are $P(S_0|S_1)=\frac{4}{9}$, $P(S_3|S_1)=\frac{1}{3}$, $P(S_4|S_1)=\frac{2}{9}$.
With a general matrix, $M$, let the probability of eventually reaching state $b$ from state $a$ be written as $P(S_a \rightarrow S_b)$.
Then
$$P(S_a \rightarrow S_b)=\sum_i P(S_i|S_a)P(S_i \rightarrow S_b)$$
Using this, you can iteratively calculate the probabilities (this would get harder to do with more complicated matrices).
Example calculation:
Say you want to know $P(S_0 \rightarrow S_3)$.
The only way of getting to $S_3$ is from $S_1$. If you land back at $S_0$ from $S_1$, you're back where the original problem started. Landing in states 2, 4 and 5 would get you trapped, so $P(S_{2,4,5} \rightarrow S_3)=0$. Hence you get:
$$P(S_0 \rightarrow S_3)=P(S_1|S_0)\left(P(S_0|S_1)P(S_0 \rightarrow S_3)+P(S_3|S_1)\right)$$
which gives the numbers I provided in the comment.