= 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> gives
$$
h_1=\frac14+\frac34h_2,\qquad h_2=\frac12h_1+\frac12h_3,\qquad h_3=\frac34h_2.
$$
Substitution yields $h_2=1/2$, $h_1=5/8$ and $h_3=3/8$. The initial state has one red ball, so \b[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>.
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