Three urns A, B and C contain 7 red, 5 black; 5 red, 7 black and 6 red, 6 black balls, respectively. One of the urn is selected at random and a ball is drawn from it. If the ball drawn is black, then the probability that it is drawn from urn A is:

We can use the Bayes Theorem:

$$E_1$$: Selecting Urn A

$$E_2$$: Selecting Urn B

$$E_3$$: Selecting Urn C

Because the urn is chosen completely at random, the probability of selecting any single urn is equal:

$$P(E_1) = P(E_2) = P(E_3) = \dfrac{1}{3}$$

Let $$B$$ be the event of drawing a black ball. We have to find the probability of drawing a black ball from each specific urn.

Each urn contains exactly 12 balls in total (7+5 = 12, 5+7 = 12, 6+6 = 12).

Urn A has 5 black balls: $$P(B|E_1) = \dfrac{5}{12}$$

Urn B has 7 black balls: $$P(B|E_2) = \dfrac{7}{12}$$

Urn C has 6 black balls: $$P(B|E_3) = \dfrac{6}{12}$$

 Bayes Theorem: $$P(E_1|B) = \dfrac{P(E_1) \cdot P(B|E_1)}{P(E_1) \cdot P(B|E_1) + P(E_2) \cdot P(B|E_2) + P(E_3) \cdot P(B|E_3)}$$

$$P(E_1|B) = \dfrac{\left(\dfrac{1}{3}\right) \cdot \left(\dfrac{5}{12}\right)}{\left(\dfrac{1}{3}\right) \cdot \left(\dfrac{5}{12}\right) + \left(\dfrac{1}{3}\right) \cdot \left(\dfrac{7}{12}\right) + \left(\dfrac{1}{3}\right) \cdot \left(\dfrac{6}{12}\right)}$$

$$P(E_1|B) = \dfrac{5}{5 + 7 + 6}$$

$$P(E_1|B) = \dfrac{5}{18}$$