Using Baye's theorem:
$$Pr=\frac{\frac{1}{2}\times\frac{6}{21}}{(\frac{1}{2}\times\frac{6}{21})+(\frac{7}{12}\times\frac{12}{21})+(\frac{2}{3}\times\frac{3}{21})}=\frac{3}{12}=\frac{1}{4}$$
in which the first term at denominator is for the case that both ball are black, the second term is for the case that one of them is white, and the other is black, and the last term is for the case that both balls are white.
According to the Baye's theorem, the probability that both ball are black given that the output ball is white is equal to a ratio, that nominator is equal to the probability of the intended event given the two balls are black and the denominator is equal to the summation of the probabilities of the desired event (that output is white) given all the situations, one by one
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
I would find all the possible first draws (4B 0W), (3B 1W), (2B 2W), (1B 3W) and give each of them a probability of happening. Then, from each of those nodes, find the probability that a white is drawn from them. Multiply the two probabilities together in each case. Then make a fraction $$\frac{\text{probability of drawing a white from 1B 3W}}{\text{sum of all probabilities that led to drawing a white ball}}$$