I'm reading the solution from my lecture note:
In my understanding, $$\begin{cases}P\left(\mu_{t}=m_{2} | \mu_{t-1}=m_{1}\right)+P\left(\mu_{t}=m_{1} | \mu_{t-1}=m_{1}\right) = P(\mu_{t-1}=m_{1}) \\ P\left(\mu_{t}=m_{1} | \mu_{t-1}=m_{2}\right) + P\left(\mu_{t}=m_{2} | \mu_{t-1}=m_{2}\right) = P(\mu_{t-1}=m_{2}) \end{cases}$$
Then $$\begin{cases}P\left(\mu_{t}=m_{2} | \mu_{t-1}=m_{1}\right)+p = P(\mu_{t-1}=m_{1}) \\ P\left(\mu_{t}=m_{1} | \mu_{t-1}=m_{2}\right) + q = P(\mu_{t-1}=m_{2}) \end{cases}$$
Finally, $$\begin{cases}P\left(\mu_{t}=m_{2} | \mu_{t-1}=m_{1}\right)= P(\mu_{t-1}=m_{1}) -p\\ P\left(\mu_{t}=m_{1} | \mu_{t-1}=m_{2}\right) = P(\mu_{t-1}=m_{2})-q \end{cases}$$
As such, the author is wrong in asserting $$\begin{cases} {P\left(\mu_{t}=m_{2} | \mu_{t-1}=m_{1}\right)=1-p} \\ {P\left(\mu_{t}=m_{1} | \mu_{t-1}=m_{2}\right)=1-q} \end{cases}$$
Could you please confirm if my observation is correct? Thank you so much!
The author is correct. This $$P\left(\mu_{t}=m_{2} | \mu_{t-1}=m_{1}\right)+P\left(\mu_{t}=m_{1} | \mu_{t-1}=m_{1}\right) = P(\mu_{t-1}=m_{1})$$ is incorrect. For any events $A$ and $B$, $$P(A|B) + P(A^c|B) = 1,$$ where $A^c$ denotes the complement of $A$. This is because no matter what happened (i.e., no matter what $B$ is), with certainty either $A$ occurs or $A$ does not occur.
In your case, either $\mu_t = m_1$ or $\mu_t = m_2$; nothing else can occur. You can verify this with Bayes' rule, $P(A|B) = P(A\cap B) / P(B)$, and setting $A = \{\mu_t = m_1\}, B = \{\mu_{t-1} = m_1\}.$