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 draw in black, then the probability that it is drawn from urn A is:

Options

A. `5/(16)`
B. `7/(18)`
C. `4/(17)`
D. `5/(18)`

Answer

To solve the problem, we will use Bayes' theorem, which helps us find the probability of an event given that another event has occurred.

### Step-by-Step Solution:

1. **Define the Events**:
- Let \( A \) be the event that the ball is drawn from urn A.
- Let \( B \) be the event that a black ball is drawn.

2. **Calculate the Total Number of Balls in Each Urn**:
- Urn A: 7 red + 5 black = 12 balls
- Urn B: 5 red + 7 black = 12 balls
- Urn C: 6 red + 6 black = 12 balls

3. **Find the Probability of Drawing a Black Ball from Each Urn**:
- Probability of drawing a black ball from urn A:
\[ P(B|A) = \frac{5}{12} \]
- Probability of drawing a black ball from urn B:
\[ P(B|B) = \frac{7}{12} \]
- Probability of drawing a black ball from urn C:
\[ P(B|C) = \frac{6}{12} = \frac{1}{2} \]

4. **Calculate the Prior Probabilities of Selecting Each Urn**:
Since the urn is selected at random:
\[ P(A) = P(B) = P(C) = \frac{1}{3} \]

5. **Calculate the Total Probability of Drawing a Black Ball**:
Using the law of total probability:
\[ P(B) = P(B|A) \cdot P(A) + P(B|B) \cdot P(B) + P(B|C) \cdot P(C) \]
Substituting the values:
\[ P(B) = \left(\frac{5}{12} \cdot \frac{1}{3}\right) + \left(\frac{7}{12} \cdot \frac{1}{3}\right) + \left(\frac{1}{2} \cdot \frac{1}{3}\right) \]
\[ = \frac{5}{36} + \frac{7}{36} + \frac{6}{36} = \frac{18}{36} = \frac{1}{2} \]

6. **Apply Bayes' Theorem to Find \( P(A|B) \)**:
We want to find the probability that the ball was drawn from urn A given that a black ball was drawn:
\[ P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)} \]
Substituting the known values:
\[ P(A|B) = \frac{\left(\frac{5}{12}\right) \cdot \left(\frac{1}{3}\right)}{\frac{1}{2}} = \frac{\frac{5}{36}}{\frac{1}{2}} = \frac{5}{36} \cdot \frac{2}{1} = \frac{10}{36} = \frac{5}{18} \]

### Final Answer:
The probability that the black ball drawn is from urn A is \( \frac{5}{18} \).