You are missing the information that you've got from the fact that the first ball is black.
So, if you have drawn out one of the three black balls, the probabilities that it came from urn $U_i$ are
$$P(U_1|B) = 2/3$$
$$P(U_2|B) = 1/3$$
If it's hard to understand, you can think of a similar situation - let's say you have an $U_1'$ with 100 black balls and $U_2'$ with 99 white balls and one black ball. If you take one urn and pull out a ball and it turns out to be black, you can be quite sure that you got $U_1'$. In this case it's the same, just not that obvious.
The total number of combinations is:
$$\dbinom{5+6+7}{5}=\dfrac{18!}{5!\cdot13!}=8568$$
The number of combinations with no red balls is:
$$\dbinom{6+7}{5}=\dfrac{13!}{5!\cdot8!}=1287$$
The number of combinations with no white balls is:
$$\dbinom{5+7}{5}=\dfrac{12!}{5!\cdot7!}=792$$
The number of combinations with no blue balls is:
$$\dbinom{5+6}{5}=\dfrac{11!}{5!\cdot6!}=462$$
The number of combinations with no red balls and no white balls is:
$$\dbinom{7}{5}=\dfrac{7!}{5!\cdot2!}=21$$
The number of combinations with no red balls and no blue balls is:
$$\dbinom{6}{5}=\dfrac{6!}{5!\cdot1!}=6$$
The number of combinations with no white balls and no blue balls is:
$$\dbinom{5}{5}=\dfrac{5!}{5!\cdot0!}=1$$
So the probability of a combination without at least one ball of each color is:
$$\dfrac{1287+792+462-21-6-1}{8568}=\dfrac{2513}{8568}$$
And the probability of a combination with at least one ball of each color is:
$$1-\dfrac{2513}{8568}=\dfrac{6055}{8568}\approx0.7067$$
Best Answer
Yes.
A more rigorous way of writing it would be using conditioning.
Let $W_i$ be the random variable corresponding to drawing of white balls.
$$W_i=\begin{cases}1\text{ If ball i is white}\\0\text{ Otherwise}\end{cases}$$
\begin{align} P(W_2=1)&=\sum_xP(W_2=1\;|\;W_1=x)\times P(W_1=x)\\ &=P(W_2=1\;|\;W_1=0)\times P(W_1=0)+P(W_2=1\;|\;W_1=1)\times P(W_1=1)\\ &=\dfrac{10}{13}\times \dfrac{3}{12}+\dfrac{3}{13}\times \dfrac{2}{12}\\ &=\dfrac{3}{13}\\ &\blacksquare \end{align}