A fair die is rolled. If a face 1 turns up, a ball is drawn from Bag A. If face 2 or 3 turns up a ball is drawn from Bag B. If face 4 or 5 or 6 turns up, a ball is drawn from Bag C. A contain 3 red and 2 white balls, Bag B contains 3 red and 4 white balls and Bag C contains 4 red and 5 white balls. The die is rolled, a Bag is picked up and a ball is drawn. If the drawn ball is red, what is the probability that it is drawn from Bag B?

Answer

To solve the problem, we will use Bayes' theorem to find the probability that a red ball drawn is from Bag B, given that a red ball is drawn. Let's break down the solution step by step.

### Step 1: Define Events
Let:
- \( E_1 \): Event that a ball is drawn from Bag A.
- \( E_2 \): Event that a ball is drawn from Bag B.
- \( E_3 \): Event that a ball is drawn from Bag C.
- \( A \): Event that a red ball is drawn.

### Step 2: Calculate Probabilities of Selecting Each Bag
- The probability of rolling a 1 (and thus selecting Bag A) is:
\[ P(E_1) = \frac{1}{6} \]
- The probability of rolling a 2 or 3 (and thus selecting Bag B) is:
\[ P(E_2) = \frac{2}{6} = \frac{1}{3} \]
- The probability of rolling a 4, 5, or 6 (and thus selecting Bag C) is:
\[ P(E_3) = \frac{3}{6} = \frac{1}{2} \]

### Step 3: Calculate the Probability of Drawing a Red Ball from Each Bag
- For Bag A (3 red, 2 white):
\[ P(A | E_1) = \frac{3}{5} \]
- For Bag B (3 red, 4 white):
\[ P(A | E_2) = \frac{3}{7} \]
- For Bag C (4 red, 5 white):
\[ P(A | E_3) = \frac{4}{9} \]

### Step 4: Use Bayes' Theorem
We want to find \( P(E_2 | A) \), which can be calculated using Bayes' theorem:
\[ P(E_2 | A) = \frac{P(A | E_2) \cdot P(E_2)}{P(A)} \]
Where \( P(A) \) is the total probability of drawing a red ball:
\[ P(A) = P(A | E_1) \cdot P(E_1) + P(A | E_2) \cdot P(E_2) + P(A | E_3) \cdot P(E_3) \]

### Step 5: Calculate \( P(A) \)
Substituting the values:
\[ P(A) = \left(\frac{3}{5} \cdot \frac{1}{6}\right) + \left(\frac{3}{7} \cdot \frac{1}{3}\right) + \left(\frac{4}{9} \cdot \frac{1}{2}\right) \]
Calculating each term:
- From Bag A:
\[ \frac{3}{5} \cdot \frac{1}{6} = \frac{3}{30} = \frac{1}{10} \]
- From Bag B:
\[ \frac{3}{7} \cdot \frac{1}{3} = \frac{3}{21} = \frac{1}{7} \]
- From Bag C:
\[ \frac{4}{9} \cdot \frac{1}{2} = \frac{4}{18} = \frac{2}{9} \]

Now, we need a common denominator to add these fractions. The least common multiple of 10, 7, and 9 is 630. Converting each fraction:
- \( \frac{1}{10} = \frac{63}{630} \)
- \( \frac{1}{7} = \frac{90}{630} \)
- \( \frac{2}{9} = \frac{140}{630} \)

Adding these together:
\[ P(A) = \frac{63 + 90 + 140}{630} = \frac{293}{630} \]

### Step 6: Calculate \( P(E_2 | A) \)
Now substitute back into Bayes' theorem:
\[ P(E_2 | A) = \frac{P(A | E_2) \cdot P(E_2)}{P(A)} = \frac{\left(\frac{3}{7}\right) \cdot \left(\frac{1}{3}\right)}{\frac{293}{630}} \]
Calculating the numerator:
\[ \frac{3}{7} \cdot \frac{1}{3} = \frac{1}{7} \]
Thus:
\[ P(E_2 | A) = \frac{\frac{1}{7}}{\frac{293}{630}} = \frac{630}{7 \cdot 293} = \frac{90}{293} \]

### Final Answer
The probability that the drawn red ball is from Bag B is:
\[ \boxed{\frac{90}{293}} \]