A fair die is rolled. If 1 or 2 turns up bag A is picked, if 3 turns up bag B is picked, if 4 or 5 or 6 turns up bag C is picked. A die is rolled, a bag is picked and a ball is drawn. What is the probability that the ball drawn is red given bag A contains 4 red and 5 white balls, bag B contains 3 red and 2 white balls and bag C contains 3 red and 4 white balls? If the ball drawn found to be red then find the probability that 3 turned up on the die?

Answer

To solve the problem step by step, we will first calculate the probabilities of picking each bag and then the probabilities of drawing a red ball from each bag. Finally, we will use these probabilities to find the required answers.

### Step 1: Determine the probabilities of picking each bag

1. **Probability of picking Bag A (P(A))**:
- Bag A is picked if the die shows 1 or 2.
- There are 2 favorable outcomes (1, 2) out of 6 possible outcomes.
- \( P(A) = \frac{2}{6} = \frac{1}{3} \)

2. **Probability of picking Bag B (P(B))**:
- Bag B is picked if the die shows 3.
- There is 1 favorable outcome (3) out of 6 possible outcomes.
- \( P(B) = \frac{1}{6} \)

3. **Probability of picking Bag C (P(C))**:
- Bag C is picked if the die shows 4, 5, or 6.
- There are 3 favorable outcomes (4, 5, 6) out of 6 possible outcomes.
- \( P(C) = \frac{3}{6} = \frac{1}{2} \)

### Step 2: Determine the probabilities of drawing a red ball from each bag

1. **Probability of drawing a red ball from Bag A (P(R|A))**:
- Bag A contains 4 red and 5 white balls, a total of 9 balls.
- \( P(R|A) = \frac{4}{9} \)

2. **Probability of drawing a red ball from Bag B (P(R|B))**:
- Bag B contains 3 red and 2 white balls, a total of 5 balls.
- \( P(R|B) = \frac{3}{5} \)

3. **Probability of drawing a red ball from Bag C (P(R|C))**:
- Bag C contains 3 red and 4 white balls, a total of 7 balls.
- \( P(R|C) = \frac{3}{7} \)

### Step 3: Calculate the total probability of drawing a red ball (P(R))

Using the law of total probability:
\[ P(R) = P(A) \cdot P(R|A) + P(B) \cdot P(R|B) + P(C) \cdot P(R|C) \]

Substituting the values:
\[ P(R) = \left(\frac{1}{3} \cdot \frac{4}{9}\right) + \left(\frac{1}{6} \cdot \frac{3}{5}\right) + \left(\frac{1}{2} \cdot \frac{3}{7}\right) \]

Calculating each term:
1. \( P(A) \cdot P(R|A) = \frac{1}{3} \cdot \frac{4}{9} = \frac{4}{27} \)
2. \( P(B) \cdot P(R|B) = \frac{1}{6} \cdot \frac{3}{5} = \frac{3}{30} = \frac{1}{10} \)
3. \( P(C) \cdot P(R|C) = \frac{1}{2} \cdot \frac{3}{7} = \frac{3}{14} \)

Finding a common denominator (which is 270):
- Convert \( \frac{4}{27} = \frac{40}{270} \)
- Convert \( \frac{1}{10} = \frac{27}{270} \)
- Convert \( \frac{3}{14} = \frac{57}{270} \)

Now, summing these:
\[ P(R) = \frac{40}{270} + \frac{27}{270} + \frac{57}{270} = \frac{124}{270} = \frac{62}{135} \]

### Step 4: Find the probability that the die showed 3 given that a red ball was drawn (P(B|R))

Using Bayes' theorem:
\[ P(B|R) = \frac{P(B) \cdot P(R|B)}{P(R)} \]

Substituting the values:
\[ P(B|R) = \frac{\left(\frac{1}{6}\right) \cdot \left(\frac{3}{5}\right)}{\frac{62}{135}} = \frac{\frac{3}{30}}{\frac{62}{135}} = \frac{3}{30} \cdot \frac{135}{62} = \frac{3 \cdot 135}{30 \cdot 62} = \frac{405}{1860} = \frac{27}{124} \]

### Final Answers
1. The probability that the ball drawn is red is \( \frac{62}{135} \).
2. The probability that 3 turned up on the die given that the ball drawn is red is \( \frac{27}{124} \).