(i) Bag I contains 3 red and 4 black balls while another bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from bag II.
(ii) There are two bags I and II. Bag I contains 4 white and 3 red balls and bag II contains 6 white and 5 red balls. One ball is drawn at random from one of the bags and is found to be red. Find the probability that it was drawn from bag II.

Answer

To solve the given problems step by step, we will use Bayes' theorem, which is a way to find a probability when we know certain other probabilities.

### Part (i)

**Given:**
- Bag I contains 3 red and 4 black balls.
- Bag II contains 5 red and 6 black balls.

**Step 1: Define Events**
- Let \( S_1 \): the event that Bag I is chosen.
- Let \( S_2 \): the event that Bag II is chosen.
- Let \( R \): the event that a red ball is drawn.

**Step 2: Calculate the probabilities of choosing each bag**
- \( P(S_1) = P(S_2) = \frac{1}{2} \) (since one of the two bags is chosen at random)

**Step 3: Calculate the probability of drawing a red ball from each bag**
- From Bag I:
- Total balls = 3 red + 4 black = 7
- \( P(R | S_1) = \frac{3}{7} \)

- From Bag II:
- Total balls = 5 red + 6 black = 11
- \( P(R | S_2) = \frac{5}{11} \)

**Step 4: Use Bayes' Theorem to find \( P(S_2 | R) \)**
\[ P(S_2 | R) = \frac{P(R | S_2) \cdot P(S_2)}{P(R)} \]
Where \( P(R) \) is the total probability of drawing a red ball:
\[ P(R) = P(R | S_1) \cdot P(S_1) + P(R | S_2) \cdot P(S_2) \]
Substituting the values:
\[ P(R) = \left(\frac{3}{7} \cdot \frac{1}{2}\right) + \left(\frac{5}{11} \cdot \frac{1}{2}\right) \]
\[ P(R) = \frac{3}{14} + \frac{5}{22} \]
To add these fractions, we need a common denominator (which is 154):
\[ P(R) = \frac{3 \cdot 11}{154} + \frac{5 \cdot 7}{154} = \frac{33 + 35}{154} = \frac{68}{154} \]

**Step 5: Substitute back into Bayes' Theorem**
\[ P(S_2 | R) = \frac{P(R | S_2) \cdot P(S_2)}{P(R)} = \frac{\left(\frac{5}{11}\right) \cdot \left(\frac{1}{2}\right)}{\frac{68}{154}} \]
\[ = \frac{\frac{5}{22}}{\frac{68}{154}} = \frac{5 \cdot 154}{22 \cdot 68} = \frac{770}{1496} = \frac{35}{68} \]

### Final Answer for Part (i)
The probability that the red ball was drawn from Bag II is \( \frac{35}{68} \).

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### Part (ii)

**Given:**
- Bag I contains 4 white and 3 red balls.
- Bag II contains 6 white and 5 red balls.

**Step 1: Define Events**
- Let \( S_1 \): the event that Bag I is chosen.
- Let \( S_2 \): the event that Bag II is chosen.
- Let \( R \): the event that a red ball is drawn.

**Step 2: Calculate the probabilities of choosing each bag**
- \( P(S_1) = P(S_2) = \frac{1}{2} \)

**Step 3: Calculate the probability of drawing a red ball from each bag**
- From Bag I:
- Total balls = 4 white + 3 red = 7
- \( P(R | S_1) = \frac{3}{7} \)

- From Bag II:
- Total balls = 6 white + 5 red = 11
- \( P(R | S_2) = \frac{5}{11} \)

**Step 4: Use Bayes' Theorem to find \( P(S_2 | R) \)**
\[ P(S_2 | R) = \frac{P(R | S_2) \cdot P(S_2)}{P(R)} \]
Where \( P(R) \) is the total probability of drawing a red ball:
\[ P(R) = P(R | S_1) \cdot P(S_1) + P(R | S_2) \cdot P(S_2) \]
Substituting the values:
\[ P(R) = \left(\frac{3}{7} \cdot \frac{1}{2}\right) + \left(\frac{5}{11} \cdot \frac{1}{2}\right) \]
\[ P(R) = \frac{3}{14} + \frac{5}{22} \]
To add these fractions, we need a common denominator (which is 154):
\[ P(R) = \frac{3 \cdot 11}{154} + \frac{5 \cdot 7}{154} = \frac{33 + 35}{154} = \frac{68}{154} \]

**Step 5: Substitute back into Bayes' Theorem**
\[ P(S_2 | R) = \frac{P(R | S_2) \cdot P(S_2)}{P(R)} = \frac{\left(\frac{5}{11}\right) \cdot \left(\frac{1}{2}\right)}{\frac{68}{154}} \]
\[ = \frac{\frac{5}{22}}{\frac{68}{154}} = \frac{5 \cdot 154}{22 \cdot 68} = \frac{770}{1496} = \frac{35}{68} \]

### Final Answer for Part (ii)
The probability that the red ball was drawn from Bag II is also \( \frac{35}{68} \).

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