I'll continue on the answer by Ross Millikan, and give an exact solution (I'll just replace $R$ and $W$ by lower case $r$ and $w$, which intimidates me less). So for $r,w\in\mathbb N$, not both zero, let $P(r,w)$ denote the probability of leaving a white ball as last one when starting with $r$ red and $w$ white balls. One obviously has $P(r,0)=0$ for any $r>0$ and $P(0,w)=1$ for any $w>0$; moreover by the argument Ross gives one has the recurrence relation
$$
P(r,w) = \frac{r^2}{(r+w)^2}P(r-1,w) + \frac{w^2+2rw}{(r+w)^2}P(r,w-1) \quad\text{for all } r,w>0,
$$
since the first step reduces the problem defining $P(r,w)$ either to the one defining $P(r-1,w)$ or to the one defining $P(r,w-1)$, with the indicated factors as probablilities for the first step.
Now this recurrence has the (surprisingly simple) solution
$$
P(r,w)=\frac{w}{(r+1)(r+w)}\quad\text{for }(r,w)\in\mathbb N^2\setminus\{(0,0)\}.
$$
The proof is by a simple induction on $r+w$, expanding the recurrence relation, factoring out $\frac w{(r+w)^2(r+w-1)(r+1)}$ which leaves $r(r+1)+(w+2r)(w-1)$ as other factor, and rewriting that factor as $(r+w)(r+w-1)$ permits some cancelling and arriving at the desired result.
I did not guess the formula just like that of course. Rather I calculated a number of values $P(1,w)$ with exact rational arithmetic, noticing substantial simplifications and easily guessing $P(1,w)=\frac{w}{2(w+1)}$. I proceeded similarly for $P(2,w)$, again with significant simplfications, after which I guessed the general form.
For the concrete problem one gets $P(10,90)=\frac9{110}\approx0.08181818$, a chance of a bit less than $1$ in $12$ to be left with a white ball. In fact one sees that throwing in any number of white balls (with $10$ red balls) initially never even raises the chance to $1$ in $11$. And if there are any red balls at all initially, it is always more likely to be left with one of them than with a white ball!
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 $$1/3 + 8/45 = 23/45 \approx 51\%.$$
Best Answer
No need to use Bayes Rule/conditional probabilities.
Pick the second ball first (when you pick a pair of balls, you can swap balls/change their order and preserve probabilities). You are left with 3 whites out of total 10 balls to draw a white ball from. The needed probability is 3/10