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To find a stationary distribution for a Markov chain, one solves the system of linear equations given by $\pi P = \pi$, subject to the constraint that the probabilities in $\pi$ sum to 1.

Conditions: $P$ is the transition matrix of the Markov chain.; $\pi$ is a row vector of probabilities.

To find a stationary distribution for a Markov chain, one solves the system of linear equations given by $\pi P = \pi$, subject to the constraint that the probabilities in $\pi$ sum to 1.

Conditions: $P$ is the transition matrix of the Markov chain.; $\pi$ is a row vector of probabilities.

By substituting $\pi(b)=7/4 \pi(a)$ into $\pi(a)+\pi(b)=1$, we get $\pi(a) + 7/4 \pi(a) = 1$. Combining the terms yields $11/4 \pi(a) = 1$, and solving for $\pi(a)$ gives $\pi(a) = 4/11$.

Conditions: The balance equation gives $4\pi(b) = 7\pi(a)$, which rearranges to $\pi(b) = 7/4 \pi(a)$.; The normalization equation is $\pi(a) + \pi(b) = 1$.

By substituting $\pi(b)=7/4 \pi(a)$ into $\pi(a)+\pi(b)=1$, we get $\pi(a) + 7/4 \pi(a) = 1$. Combining the terms yields $11/4 \pi(a) = 1$, and solving for $\pi(a)$ gives $\pi(a) = 4/11$.

Conditions: The balance equation gives $4\pi(b) = 7\pi(a)$, which rearranges to $\pi(b) = 7/4 \pi(a)$.; The normalization equation is $\pi(a) + \pi(b) = 1$.