A question on probability theory with 3 urns with return Each of $3$ urns contains twenty balls. First urn contains ten white balls, second urn contains six white balls and third urn contains two white balls. All other balls are black.
One ball is drawn from the random urn with return in the same urn. The ball's color is white.
What is the probability that the second ball drawn from the same urn is white?
I think, this is $\frac{1}{9} \cdot \frac{2}{20} + \frac{3}{9} \cdot \frac{6}{20} + \frac{5}{9} \cdot \frac{10}{20}$ by Bayes'theorem and Law of total probability, but can't be sure.
Thanks for any help.

I am unsure of where your math came from, and I am unfamiliar with the "Law of total probability".  The following is how I would compute the probability.
Let $p(k)$ represent the probability that the 1st ball came from urn $k : k \in \{1,2,3\}.$
Then, the chance that the new drawing will also be a white ball, since the sampling is done with replacement is
$$\left[p(1) \times \frac{10}{20}\right]
~+~ \left[p(2) \times \frac{6}{20}\right]
~+~ \left[p(3) \times \frac{2}{20}\right].$$
Let $D$ (i.e. denominator) =
$$\left[\frac{1}{3} \times \frac{10}{20}\right]
~+~ \left[\frac{1}{3} \times \frac{6}{20}\right]
~+~ \left[\frac{1}{3} \times \frac{2}{20}\right].$$
Then:
$$p(1) = \frac{\frac{1}{3} \times \frac{10}{20}}{D}.$$
$$p(2) = \frac{\frac{1}{3} \times \frac{6}{20}}{D}.$$
$$p(3) = \frac{\frac{1}{3} \times \frac{2}{20}}{D}.$$
