Let $U_i$ be the $i$th urn just like you described, and let $A =\{\text{Picked a white ball}\},B_i = \{\text{Pick $U_i$}\}.$ Then the event "all 3 balls in the selected urn are white" is the same as event $B_4$.
Thus, if we use conditioning, the probability we are interested in is
\begin{align*}
P(B_4|A)&=\frac{P(B_4A)}{P(A)}\\
&=\frac{P(A|B_4)P(B_4)}{P(A)}\\
&=\frac{P(A|B_4)P(B_4)}{P(A|B_1)P(B_1)+P(A|B_2)P(B_2)+P(A|B_3)P(B_3)+P(A|B_4)P(B_4)}\\
&=\frac{1\cdot \frac{1}{4}}{0(1/4)+(1/3)(1/4)+(2/3)(1/4)+1(1/4)}\\
&=\frac{1}{2}.
\end{align*}
Hint: We can model this by arranging the ten balls in a line, so that the first four are those moved into the second urn then removed in that order. Given that the first ball is white, what is the probability that the second ball is too?
That is the short method.
The long method is:
Let $E_x$ be the event of $x$ white balls among the four balls drawn from five white and five black, for $x\in\{0,1,2,3,4\}$. This corresponds to a hypergeometric distribution.
$$\mathsf P(E_x) ~=~ \dfrac{\dbinom 5 x\dbinom 5 {4-x}}{\dbinom {10}4}\cdot\mathbf 1_{x\in\{0,1,2,3,4\}}$$
The events $E_0,E_1,E_2,E_3,E_4$ are thus not equally probable.
Let $A, B$ be the event of drawing white balls in two consecutive draws from among those four.
$$\mathsf P(B\mid A)=\dfrac{\sum_{x=2}^4 \mathsf P(E_x)~\mathsf P(A\cap B\mid E_x)}{\sum_{x=1}^4 \mathsf P(E_x)~\mathsf P(A\mid E_x)}=\dfrac{\sum_{x=2}^4 \binom 5x\binom 5{4-x}\binom x 2/\binom 4 2}{\sum_{x=1}^4 \binom 5x\binom 5{4-x}\binom x 1/\binom 4 1}$$
Best Answer
Correct.
Application of law of total probability is what they call this: $$P(E)=P(A)P(E\mid A)+P(B)P(E\mid B)+P(C)P(E\mid C)$$ where $A,B,C$ are mutually exclusive and covering.