Consider the general two-state chain where p and q are not both 0. Let $T$ be the first reutrn time to state 1, for the chain started in 1. (a) Show that $P(T\ge n) = p(1-q)^{n-2}$, for $n\ge 2$. $$ P= \begin{pmatrix} 1-p & p \\ q & 1-q \end{pmatrix} $$
the $p$ term is self evidence. Going from state 1 to state 2 occurs with probability $p$.
However, what about the return probability $q$?
It is my belief that it should be,
$P(T\ge n) = p(1-q)^{n-2}q$, for $n\ge 2$.
Is this a typo or am I completely misunderstanding my probabilities?
The quantity you have written is the probability that $T=n$ exactly. For the event $\{T\geq n\}$, all that is required is that at the first $n-1$ steps you are not at $1$, which happens with probability $p(1-q)^{n-2}$ (jump to $2$ first, then stay there the next $n-2$ steps).