The first box has $a$ white and $b$ black, the second box has $b$ white and $a$ black. One ball is either moved from the first box to the second (with probability $1/2$) or from the second box to the first (also probability $1/2$). You then select a ball from the box to which a ball was added. You are looking for the probability that the box you select from is the first box, given that it is white.
Let $A$ be the event that the box you select from is the first box. Since both boxes are equally likely to be the box that is added to, $P(A) = 1/2$.
Let $B$ be the event that the ball you select is white. By symmetry, $P(B) = 1/2$.
You're looking for $P(A \mid B$), so you apply Bayes' theorem: $$P(A \mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}$$
Since $P(A) = P(B) = 1/2$, this means $$P(A \mid B) = P(B \mid A)$$
$P(B \mid A)$, the probability that the ball is white, given that you choose from the first box, is considerably easier to calculate. The fact that you choose from the first box means that a ball was taken from the second box and put into the first.
The probability $P(W)$ that a white ball was taken from the second box and added to the first is $$P(W) = \frac{b}{a+b}$$
The probability $P(B)$ that a black ball was taken from the second box and added to the first is $$P(B) = 1 - P(W) = \frac{a}{a+b}$$
Now, we write $P(B \mid A)$ in terms of $P(W)$, $P(B)$, $a$, and $b$:
$$P(B \mid A) = P(W) \cdot \frac{a+1}{a+b+1} + P(B) \cdot \frac{a}{a+b+1}$$
The terms multiplied by $P(W)$ and $P(B)$ are the probabilities of getting a white ball, given a white-ball-transfer and a black-ball-transfer, respectively. In each case, this is just the number of white balls in the first box, divided by the total number of balls in the first box.
$$P(B \mid A) = \frac{b}{a+b} \cdot \frac{a+1}{a+b+1} + \frac{a}{a+b} \cdot \frac{a}{a+b+1}$$
$$P(B \mid A) = \frac{a^2 + ab + b}{a^2 + 2ab + b^2 + a + b}$$
Finally,
$$P(A \mid B) = \frac{a^2 + ab + b}{a^2 + 2ab + b^2 + a + b}$$
There are three blue balls, all equally likely to be the first ball chosen. In two cases, the second ball chosen will also be blue, and in one case, it will be red.
EDIT What you say in your objections is true, but it has nothing to do with the case. Instead of balls, let us says that we have one urn with two gold coins, one with a gold coin and a silver coin, and one with two silver coins. The coins are the same size and cannot be distinguished by touch. You draw a coin at random from a randomly chosen urn, and it's gold. What is the probability that the other coin in the urn you have chosen is gold? According to your logic, it $1/2.$
Now I tell you that the gold coin in the urn with the silver coin is dated $1900,$ and the two gold coins in the same urn are dated $1901$ and $1902.$ You draw a gold coin, and place it on the table, and tails is showing, so that date cannot be seen. Now, what is the probability that the other coin in the urn is gold? This is the question being asked. The answer is that the other coin is gold if when we flip the coin we've drawn, the date is $1901$ or $1902$ but not if it's $1900.$
Do you really want to maintain that it's equally likely to be $1900$ as it is to be $1901$ or $1902?$
Best Answer
Either through experimentation or theory. Here you can evaluate from the model of drawing balls from urns without bias.
Here you have $F$, the event of obtaining exactly two white balls when drawing one ball from each of the three urns, and $E_A, E_B, E_C$, the respective events of obtaining a white ball when drawing from the indicated urn. Note: these events will be independent, and you can easily calculate $\mathsf P(E_A)$, $\mathsf P(E_B)$ and $\mathsf P(E_C)$
Then $F= (E_A\cap E_B\cap E_C^\complement)\cup(E_A\cap E_B^\complement\cap E_C)\cup(E_A^\complement\cap E_B\cap E_C)$
You seek $\mathsf P(E_A\mid F)$, and do know :
$$\begin{align}\mathsf P(E_A\mid F) &=\dfrac{\mathsf P(E_A\cap F)}{ \mathsf P(F)}\\[2ex]&=\dfrac{\mathsf P\big(E_A\cap((E_B\cap E_C^\complement)\cup(E_B^\complement\cap E_C))\big)}{\mathsf P(F)}\end{align}$$
I shall leave the rest to you.