Three white and three black balls are distributed in two urns in such a way that each contains three balls. We say that the system is in state i, i = 0, 1, 2, 3, if the first urn contains i white balls. At each step, we draw one ball from each urn and place the ball drawn from the first urn into the second, and conversely with the ball from the second urn. Let $X_n$ denote the state of the system after the nth step. Explain why ${X_n, n = 0, 1, 2, . . .}$ is a Markov chain and calculate its transition probability matrix.

Solution: Author computed transition probability matrix as follows:$$\begin{bmatrix} 0 & 1 & 0 & 0 \\ \frac19 & \frac49 & \frac49 & 0 \\ 0 & \frac49 & \frac49 & \frac19 \\ 0 & 0 & 1 & 0 \end{bmatrix}$$

Is above transition probability matrix correct? How is that computed?

If the first urn contains 1 white ball (state 1), how can it enter into state 2 ( 2 white balls into first urn) with probability $\frac49?$

If the first urn contains 2 white balls (state 2), how can it enter into state 3 (3 white balls into first urn) with probability $\frac19?$

Note: At each step, we draw one ball from each urn and place the ball drawn from the first urn into the second urn and place the ball drawn from the second urn into first urn.

Would any member of "Cross Validated" stack exchange answer my questions?

Three white and three black balls are distributed in two urns in such a way that each contains three balls. We say that the system is in state i, i = 0, 1, 2, 3, if the first urn contains i white balls. At each step, we draw one ball from each urn and place the ball drawn from the first urn into the second, and conversely with the ball from the second urn. Let $X_n$ denote the state of the system after the nth step. Explain why ${X_n, n = 0, 1, 2, . . .}$ is a Markov chain and calculate its transition probability matrix.

Solution: Author computed transition probability matrix as follows:$$\begin{bmatrix} 0 & 1 & 0 & 0 \\ \frac19 & \frac49 & \frac49 & 0 \\ 0 & \frac49 & \frac49 & \frac19 \\ 0 & 0 & 1 & 0 \end{bmatrix}$$

After making some careful thinking, I came to the conclusion that author's computations for the above matrix is correct.