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 $$1/3 + 8/45 = 23/45 \approx 51\%.$$
I am not sure what you mean by "vectors". Here is how I would solve the problem:
Number the boxes 1 through n. Let $B_r$ be the event that box $r$ winds up with exactly one lone black ball. This occurs by placing a black ball in that particular box and then randomly distributing all other balls. There are $j$ balls to place in box $B_r$. Then, for every other of the $k-1$ balls, there are $n-1$ choices to place them. So, that is: $j(n-1)^{k-1}$ different possible ways of placing them.
Next, consider $|B_r \cap B_s|$ for $r\neq s$. We place two black balls into the two boxes. Then, we randomly distribute the remaining balls. So, that is $j(j-1)(n-2)^{k-2}$ ways of placing them.
For three different boxes with one black ball each, there are $j(j-1)(j-2)(n-3)^{k-3}$ ways of placing them.
Etc.
Next, apply Inclusion/Exclusion. Start with all possibilities, subtract off where at least one box has a single black ball. Add back where at least two boxes have exactly one black ball. Subtract where at least three boxes have one lone black ball, etc.
You wind up with something like this:
$$\sum_{i=0}^j(-1)^i \dbinom{n}{i}(j)_i(n-i)^{k-i}$$
Where $(j)_i$ is the falling factorial: $(j)_i = j(j-1)\cdots (j-i+1) = \dfrac{j!}{(j-i)!}$
Finally, divide by the total number of ways of distributing the balls: $n^k$.
Edit: I mixed up the $n$'s and the $k$'s. I think I fixed them all, but you may want to double check my work. I have to go for a bit.
Best Answer
Yes, your calculation is somewhat off.
Tip: Evaluate the following
Then use Bayes' Rule: $$\mathsf P(W_1\mid B_2)=\dfrac{\mathsf P(W_1)\mathsf P(B_2\mid W_1)}{\mathsf P(W_1)\mathsf P(B_2\mid W_1)+\mathsf P(W_1^\complement)\mathsf P(B_2\mid W_1^\complement)}$$