Bag A contains 2 white, 1 black and 3 red balls, Bag B contains 3 white, 2 black and 4 red balls and Bag C contains 4 white, 3 black and 2 red balls. One Bag is chosen at random and 2 balls are drawn at random from that Bag. If the randomly drawn balls happen to be red and black, what is the probability that both balls come from Bag B ?

Answer

To solve the problem step by step, we will use the concept of conditional probability and the law of total probability.

### Step 1: Define the events
Let:
- \( A_1 \): Event that Bag A is chosen.
- \( A_2 \): Event that Bag B is chosen.
- \( A_3 \): Event that Bag C is chosen.
- \( E \): Event that one red and one black ball are drawn.

### Step 2: Calculate the probabilities of choosing each bag
Since one bag is chosen at random, the probabilities are:
\[ P(A_1) = P(A_2) = P(A_3) = \frac{1}{3} \]

### Step 3: Calculate \( P(E | A_i) \) for each bag
Now we need to calculate the probability of drawing one red and one black ball from each bag.

#### For Bag A:
- Total balls = 2 white + 1 black + 3 red = 6
- The number of ways to choose 1 red and 1 black:
\[ P(E | A_1) = \frac{\text{Ways to choose 1 red and 1 black}}{\text{Total ways to choose 2 balls}} = \frac{3C1 \cdot 1C1}{6C2} = \frac{3 \cdot 1}{15} = \frac{3}{15} = \frac{1}{5} \]

#### For Bag B:
- Total balls = 3 white + 2 black + 4 red = 9
- The number of ways to choose 1 red and 1 black:
\[ P(E | A_2) = \frac{4C1 \cdot 2C1}{9C2} = \frac{4 \cdot 2}{36} = \frac{8}{36} = \frac{2}{9} \]

#### For Bag C:
- Total balls = 4 white + 3 black + 2 red = 9
- The number of ways to choose 1 red and 1 black:
\[ P(E | A_3) = \frac{2C1 \cdot 3C1}{9C2} = \frac{2 \cdot 3}{36} = \frac{6}{36} = \frac{1}{6} \]

### Step 4: Use the law of total probability to find \( P(E) \)
\[ P(E) = P(E | A_1) P(A_1) + P(E | A_2) P(A_2) + P(E | A_3) P(A_3) \]
\[ P(E) = \left(\frac{1}{5} \cdot \frac{1}{3}\right) + \left(\frac{2}{9} \cdot \frac{1}{3}\right) + \left(\frac{1}{6} \cdot \frac{1}{3}\right) \]
\[ = \frac{1}{15} + \frac{2}{27} + \frac{1}{18} \]

To add these fractions, we need a common denominator. The least common multiple of 15, 27, and 18 is 270.

Converting each fraction:
\[ \frac{1}{15} = \frac{18}{270}, \quad \frac{2}{27} = \frac{20}{270}, \quad \frac{1}{18} = \frac{15}{270} \]
\[ P(E) = \frac{18 + 20 + 15}{270} = \frac{53}{270} \]

### Step 5: Calculate \( P(A_2 | E) \)
Using Bayes' theorem:
\[ P(A_2 | E) = \frac{P(E | A_2) P(A_2)}{P(E)} \]
\[ = \frac{\left(\frac{2}{9}\right) \left(\frac{1}{3}\right)}{\frac{53}{270}} = \frac{\frac{2}{27}}{\frac{53}{270}} = \frac{2 \cdot 270}{27 \cdot 53} = \frac{20}{53} \]

### Final Answer
Thus, the probability that both balls come from Bag B given that one red and one black ball are drawn is:
\[ \boxed{\frac{20}{53}} \]