Given are two urns, urn $U_1$ contains $3$ black and $2$ white balls, and urn $U_2$ contains $2$ black and $3$ white balls. A fair coin is flipped to decide which urn we should draw from. We draw $2$ balls from the selected urn with replacement (putting each ball back after we draw it). The question then asks:
$(i)$ What is the probability that the second ball we draw is black, if the first one is also black.
$(ii)$ What is the probability that the second ball is black, if urn $U_1$ is selected and the first ball is black.
$(iii)$ What is the probability that urn $U_1$ was selected, if the first ball is black.
$(iv)$ Given two events $A$ and $B$:
$A:$ $U_1$ is selected and the first ball is black.
$B:$ The second ball is black.
Are $A$ and $B$ independent?
I'm stuck at point $(iii)$. My idea was to define two events for $(i)$:
$A_1:$ The first ball is black.
$A_2: $ The second ball is black.
Since the question does not specify which urn to draw from, then $P(A_1) = P(A_2) = 0.5$ (as our sample space has $10$ balls, $5$ of them are black). Using the conditional probability definition and since $A_1$ and $A_2$ are independant:
$P(A_2 \mid A_1) = \frac{P(A_2 \cap A_1)}{P(A_1)} = \frac{P(A) \cdot P(B)}{P(A_1)} = \frac{0.5 \cdot 0.5}{0.5} = 0.5$.
$(ii)$ Here my approach was to define event $A_3 : $ "Urn $U_1$ is selected" with $P(A_3) = 0.5 $ (since it's a fair coin) and recalculate $P(A_1)$ as our sample space got smaller:
$P(A_1) = \frac{3}{5}$ $\Rightarrow P(A_3 \mid A_1) = 0.5$.
How do I proceed with $(iii)$? If I used the same approach, what would $P(A_3 \cap A_1)$ be? Is my work correct at all? Any help would be much appreciated.