 Solution

Let $R$ count red balls and let $h_r$ be the probability of eventual absorption at $R=0$ starting from $R=r$. The birth-death chain moves to $r-1$ with probability $r/4$ and to $r+1$ with probability $(4-r)/4$. Its absorbing states have $h_0=1$, $h_4=0$. First-step analysis givesSubstitution yields $h_2=1/2$, $h_1=5/8$ and $h_3=3/8$. The initial state has one red ball, so the probability that all balls end white is $5/8$.

Absorption occurs almost surely: from each interior state there is probability at least $1/4$ of absorption within the next two draws. The chance of surviving $2m$ draws is therefore at most $(3/4{)}^m$. This also justifies the finite expected times used in part (ii). The process is the absorbing four-ball Ehrenfest chain.