Let's denote the events as follows :

Let $$ A_1, A_2, $$ and $$ A_3 $$ be the events that urns A, B, and C are chosen, respectively.

Let $$ B $$ be the event that a black ball is drawn.

We need to find the probability that the chosen urn is A given that a black ball is drawn, which is $$ P(A_1|B) $$.

Using Bayes' theorem, we have :

$$ P(A_1|B) = \frac{P(B|A_1)P(A_1)}{P(B)} $$

First, we calculate each term individually :

1. The probability of choosing any urn, since they are chosen at random :

$$ P(A_1) = P(A_2) = P(A_3) = \frac{1}{3} $$

2. The probability of drawing a black ball from each urn :

Urn A: $$ P(B|A_1) = \frac{5}{12} $$

Urn B: $$ P(B|A_2) = \frac{7}{12} $$

Urn C: $$ P(B|A_3) = \frac{6}{12} = \frac{1}{2} $$

3. The total probability of drawing a black ball, $$ P(B) $$ :

$$ P(B) = P(B|A_1)P(A_1) + P(B|A_2)P(A_2) + P(B|A_3)P(A_3) $$

$$ P(B) = \frac{5}{12} \cdot \frac{1}{3} + \frac{7}{12} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} $$

$$ P(B) = \frac{5}{36} + \frac{7}{36} + \frac{6}{36} $$

$$ P(B) = \frac{18}{36} = \frac{1}{2} $$

Now, substitute these into Bayes' theorem :

$$ P(A_1|B)= \frac{P(B|A_1)P(A_1)}{P(B)} $$

$$ P(A_1|B) = \frac{\frac{5}{12} \cdot \frac{1}{3}}{\frac{1}{2}} $$

$$ P(A_1|B) = \frac{\frac{5}{36}}{\frac{1}{2}} $$

$$ P(A_1|B) = \frac{5}{18} $$

Therefore, the probability that the black ball drawn is from urn A is option C :

$$ \frac{5}{18} $$