Suppose that we have started with $w$ white balls and $b$ black balls. Then
\begin{align*}
\mathbb{P}(w_n = w+k)
&= \binom{n}{k} \frac{w(w+1)\dots(w+k-1)b(b+1)\dots(b+n-k-1)}{(w+b)(w+b+1)\dots(w+b+n-1)} \\
&= \frac{1}{B(w, b)} \binom{n}{k} \frac{\Gamma(w+k)\Gamma(b+n-k)}{\Gamma(w+b+n)} \\
&= \frac{1}{B(w, b)} \frac{k^{w-1} (n-k)^{b-1}}{n^{w+b-1}} \frac{E_k(w)E_{n-k}(b)}{E_n(b+w)} ,
\end{align*}
where $\Gamma(\cdot)$ is the gamma function, $B(\alpha, \beta) = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$ is the beta function, and
$$ E_n(z) := \frac{\Gamma(n+z)}{n!n^{z-1}}. $$
Note that $E_n(z) \to 1$ as $n\to\infty$. So, if we write $p_k = k/n$, then the m.g.f. of $X_n$ is explicitly given by
\begin{align*}
\mathbb{E}[e^{\lambda X_n}]
= \frac{1}{B(w, b)} \sum_{k=0}^{n} \exp\biggl( \lambda \frac{p_k + w/n}{1 + (w+b)/n} \biggr) p_k^{w-1}(1 - p_k)^{b-1} \frac{1}{n} \cdot \frac{E_k(w)E_{n-k}(b)}{E_n(b+w)}.
\end{align*}
Letting $n \to \infty$, this converges to
\begin{align*}
\mathbb{E}[e^{\lambda X_{\infty}}]
= \frac{1}{B(w, b)} \int_{0}^{1} e^{\lambda p} p^{w-1}(1 - p)^{b-1} \, \mathrm{d}p.
\end{align*}
From this, we read out that the distribution of $X_{\infty}$ has the density
$$ f(p) = \frac{1}{B(w, b)} p^{w-1}(1 - p)^{b-1} \mathbf{1}_{(0,1)}(p), $$
proving that the limit distribution is $\operatorname{Beta}(w, b)$.
If events A and B are conditionally independent over event C (and its complement), then that and the Law of Total Probability states: $$\mathsf P(A, B)=\mathsf P(A\mid C)\mathsf P(B\mid C)\mathsf P(C)+\mathsf P(A\mid C^\complement)\mathsf P(B\mid C^\complement)\mathsf P(C^\complement)$$Use this.
(i) What is the probability that the second ball we draw is black, if the first one is also black.
Using $A_1, A_2$ for "the first/second ball is black" and $U_1,U_2$ for the complementary events of selecting urn 1 or 2 respectively. Since we are drawing with replacement from the same urn, $A_1,A_2$ are conditionally independent given whichever urn is drawn and have identical conditional probabilities for a given urn.$$\mathsf P(A_1\mid U_1)=\mathsf P(A_2\mid U_1)=3/5\\\mathsf P(A_1\mid U_2)=\mathsf P(A_2\mid U_2)=2/5\\\mathsf P(U_1)=\mathsf P(U_2)=1/2$$
So using Bayes' Rule with the above.
$$\begin{align}\mathsf P(A_2\mid A_1)&=\dfrac{\mathsf P(A_1,A_2)}{\mathsf P(A_1)}\\[2ex]&=\dfrac{\mathsf P(A_1\mid U_1)\mathsf P(A_2\mid U_1)\mathsf P(U_1)+\mathsf P(A_1\mid U_2)\mathsf P(A_2\mid U_2)\mathsf P(U_2)}{\mathsf P(A_1\mid U_1)\mathsf P(U_1)+\mathsf P(A_1\mid U_2)\mathsf P(U_2)}\\[1ex]&=13/25\end{align}$$
(ii) What is the probability that the second ball is black, if urn
$U_1$ is selected and the first ball is black.
You seek $\mathsf P(A_2\mid A_1, U_1)$ which you can find by using Bayes' Rule as above. However, here's a hint: "$A_1,A_2$ are conditionally independent give $U_1$".
What is the probability that urn $U_1$ was selected, if the first ball is black.
Again, $\mathsf P(U_1\mid A_1)$ can be found using Bayes' Rule. However, notice that because there are five balls in each urn, then they are all equally likely to be the first ball drawn, and three of the five black balls are in urn 1.
Given two events $A$ and $B$:
Well, $A=A_1\cap U_1$ and $B=A_2$ so you seek to see if $\mathsf P(A_1,A_2,U_1)$ equals $\mathsf P(A_1,U_1)\mathsf P(A_2)$ or not.
Do that.
Best Answer
Try using Markov's inequality. Realistically speaking, your probability should be very, very close to $0$. But this inequality provides an (rather weak) upper bound that is just slightly less than $\frac{4}{9}$, which I believe is what you are looking for. You should also use the Optional Sampling Theorem or the Martingale Property along the way.