 Solution

The blue count never decreases, so $X_n\ge B_0/(m+n)$. For all sufficiently large $n$, this is at least $B_0/[2(n+1)]$. Whenever the fresh draw satisfies $U_{n+1}\le B_0/[2(n+1)]$, 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