 Solution

One excursion from zero escapes forever on the positive branch with probability $a_{+}=\frac{1}{2}(1-\frac{1}{2})=1/4$, and on the negative branch with probability $a_{-}=\frac{1}{2}(1-\frac{1}{3})=1/3$. The remaining probability is the return probability $r=5/12$. By the last-excursion direction law at a transient junction, the eventual positive escape probability isEquivalently, if $p$ denotes the desired probability, the Strong Markov property gives $p=1/4+(5/12)p$. The negative escape probability is $4/7$, and their sum is one, as independently established in part (b).