Case 1
You picked a white ball from the first box.
This happens with a chance of $3/5$. Putting it into the second box, now there are $5$ white and $4$ black balls. Now the chance to pick a white ball is $5/9$. So the total chance in case 1 is $3/5 \cdot 5/9 = 1/3$.
Case 2
You picked a black ball from the first box.
This happens with a chance of $2/5$. Putting it into the second box, now there are $4$ white and $5$ black balls. Now the chance to pick a white ball is $4/9$. So The total chance in case 2 is $2/5 \cdot 4/9 = 8/45$.
Combining the two cases, the chance is $$1/3 + 8/45 = 23/45 \approx 51\%.$$
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,
$$P(A \mid B) = \frac{a^2 + ab + b}{a^2 + 2ab + b^2 + a + b}$$
Best Answer
Case $1$: $2$ white balls are put from first box to second and one ball chosen from the second box is white
Case $2$: $2$ black balls are put from first box to second and one ball chosen from the second box is white
Case $3$: $1$ white ball and $1$ black balls are put from first box to second and one ball chosen from the second box is white
Sum of these three cases will give required probability