Consider a green ball and a yellow distributed in two urnas.Em each step, a ball is selected at random, then if that ball is green it changes of urn with probability $1/4$, and if the ball is yellow it goes to the urn containing the green ball with probability $2/3$, and goes to another urn with probability $1/3$. Calculate the approximate probability that after 140320 steps each one contains one ball.
What I think,
Let $X_t$ a Markov Chain denotes the number of balls in Urn 1 at time $t$ then the possible values of $X_t$ is $S=(0,1,2)$. Then we need to find the stationary distribution of the chain, and the probabilidade that after 140320 steps each urn contains one ball is $$\pi(1)$$
but the problem is being calculate the probability for mounting the chain $$P(X_{t+1}=2/X_t=0)=P(X_{t+1}=0|X_t=2)=0$$ $$P(X_{t+1}=1|X_t=0)=7/24$$ $$P(X_{t+1}=1|X_t=2)=7/24$$ $$P(X_{t+1}=0|X_t=1)=11/24$$ $$P(X_{t+1}=2|X_t=1)=11/24$$ then I found this transition matrix $$\begin{bmatrix}17/24&7/24&0\\11/24&2/24&11/24\\0&7/24&17/24\end{bmatrix}$$
is this right?