Solution
= Solution
{c}
The blue count never decreases, so $X_n\geq B_0/(m+n)$. For all sufficiently large $n$, this is at least $B_0/[2(n+1)]$. Whenever the fresh draw satisfies $U_{n+1}\leq B_0/[2(n+1)]$, a blue ball is added. Part (a), with $\lambda=B_0/2$, says that such events occur infinitely often almost surely. Therefore $B_n\to\infty$. Consequently
$$
\boxed{nX_n=\frac n{m+n}B_n\longrightarrow\infty\quad\text{almost surely}.}
$$