Imagine that you are randomly placing students into 4 different groups. You "draw" (or randomly place) graduate student 1 first, and you put him or her in group x. You then have to place the remaining 15 students into the 4 groups (where group x has 3 remaining spots, and the other groups have 4 remaining spots).
What is the probability that any single student (i.e. graduate student 2) will also be in group x? Since there are 3 spots left in group x and a total of 15 students to choose from, $ P = \frac{3}{15} $. However, you are interested in the probability that any single student (i.e. graduate student 2) will not be in group x, which amounts to $ P(A_1) = 1 - P = 1 - \frac{3}{15} = \frac{15-3}{15} = \frac{12}{15} $
The order of the draws is not important (i.e. whether you choose student 4 or student 8 first doesn't change anything), but it often helps to think about these processes in such a way.
You can use a similar approach/method to get $ P(A_2|A_1) $ and $ P(A_3|A_1\cap A_2) $...
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
Just use the definition of Conditional Probability, and the fact that $A_1\subseteq A_1\cup A_2\cup A_3$.
$$\mathsf P(A_1\mid A_1\cup A_2\cup A_3)=\dfrac{\mathsf P(A_1)}{\mathsf P(A_1\cup A_2\cup A_3)}=\dfrac{0.200}{0.488}$$