(i) A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.
(ii) A card from a pack of 52 cards is lost. Form the remaining cards of the pack, two cards are drawn at random and are found to be both clubs. Find the probability of the lost card being of clubs.

Answer

To solve the problem step by step, we will break down each part of the question.

### Part (i): Finding the Probability of the Lost Card Being a Diamond

1. **Define Events**:
- Let \( E_1 \): The event that the lost card is a diamond.
- Let \( E_2 \): The event that the lost card is not a diamond.

2. **Calculate the Total Diamonds**:
- Initially, there are 13 diamonds in a pack of 52 cards.
- If one card is lost, we have 51 cards left. If the lost card is a diamond, there will be 12 diamonds left. If it is not a diamond, there will still be 13 diamonds left.

3. **Calculate the Probability of Drawing 2 Diamonds**:
- **If the lost card is a diamond (Event \( E_1 \))**:
- Probability of drawing the first diamond: \( \frac{12}{51} \)
- Probability of drawing the second diamond: \( \frac{11}{50} \)
- Therefore, the probability of drawing 2 diamonds given \( E_1 \) is:
\[ P(D|E_1) = \frac{12}{51} \times \frac{11}{50} \]

- **If the lost card is not a diamond (Event \( E_2 \))**:
- Probability of drawing the first diamond: \( \frac{13}{51} \)
- Probability of drawing the second diamond: \( \frac{12}{50} \)
- Therefore, the probability of drawing 2 diamonds given \( E_2 \) is:
\[ P(D|E_2) = \frac{13}{51} \times \frac{12}{50} \]

4. **Calculate the Prior Probabilities**:
- The probability of losing a diamond:
\[ P(E_1) = \frac{13}{52} = \frac{1}{4} \]
- The probability of losing a non-diamond:
\[ P(E_2) = 1 - P(E_1) = \frac{3}{4} \]

5. **Apply Bayes' Theorem**:
- We want to find \( P(E_1|D) \):
\[ P(E_1|D) = \frac{P(D|E_1) \cdot P(E_1)}{P(D|E_1) \cdot P(E_1) + P(D|E_2) \cdot P(E_2)} \]

6. **Substitute Values**:
- Substitute the values into the equation:
\[ P(E_1|D) = \frac{\left(\frac{12}{51} \times \frac{11}{50}\right) \cdot \frac{1}{4}}{\left(\frac{12}{51} \times \frac{11}{50}\right) \cdot \frac{1}{4} + \left(\frac{13}{51} \times \frac{12}{50}\right) \cdot \frac{3}{4}} \]

7. **Simplify**:
- The common terms in the numerator and denominator can be canceled out:
\[ P(E_1|D) = \frac{12 \times 11 \times 1}{12 \times 11 + 13 \times 12 \times 3} \]
- This simplifies to:
\[ P(E_1|D) = \frac{132}{132 + 468} = \frac{132}{600} = \frac{11}{50} \]

### Part (ii): Finding the Probability of the Lost Card Being a Club

1. **Define Events**:
- Let \( E_1 \): The event that the lost card is a club.
- Let \( E_2 \): The event that the lost card is not a club.

2. **Calculate the Total Clubs**:
- Initially, there are 13 clubs in a pack of 52 cards.
- If one card is lost, we have 51 cards left. If the lost card is a club, there will be 12 clubs left. If it is not a club, there will still be 13 clubs left.

3. **Calculate the Probability of Drawing 2 Clubs**:
- **If the lost card is a club (Event \( E_1 \))**:
- Probability of drawing the first club: \( \frac{12}{51} \)
- Probability of drawing the second club: \( \frac{11}{50} \)
- Therefore, the probability of drawing 2 clubs given \( E_1 \) is:
\[ P(C|E_1) = \frac{12}{51} \times \frac{11}{50} \]

- **If the lost card is not a club (Event \( E_2 \))**:
- Probability of drawing the first club: \( \frac{13}{51} \)
- Probability of drawing the second club: \( \frac{12}{50} \)
- Therefore, the probability of drawing 2 clubs given \( E_2 \) is:
\[ P(C|E_2) = \frac{13}{51} \times \frac{12}{50} \]

4. **Calculate the Prior Probabilities**:
- The probability of losing a club:
\[ P(E_1) = \frac{13}{52} = \frac{1}{4} \]
- The probability of losing a non-club:
\[ P(E_2) = 1 - P(E_1) = \frac{3}{4} \]

5. **Apply Bayes' Theorem**:
- We want to find \( P(E_1|C) \):
\[ P(E_1|C) = \frac{P(C|E_1) \cdot P(E_1)}{P(C|E_1) \cdot P(E_1) + P(C|E_2) \cdot P(E_2)} \]

6. **Substitute Values**:
- Substitute the values into the equation:
\[ P(E_1|C) = \frac{\left(\frac{12}{51} \times \frac{11}{50}\right) \cdot \frac{1}{4}}{\left(\frac{12}{51} \times \frac{11}{50}\right) \cdot \frac{1}{4} + \left(\frac{13}{51} \times \frac{12}{50}\right) \cdot \frac{3}{4}} \]

7. **Simplify**:
- The common terms in the numerator and denominator can be canceled out:
\[ P(E_1|C) = \frac{12 \times 11 \times 1}{12 \times 11 + 13 \times 12 \times 3} \]
- This simplifies to:
\[ P(E_1|C) = \frac{132}{132 + 468} = \frac{132}{600} = \frac{11}{50} \]

### Final Answers:
- (i) The probability of the lost card being a diamond is \( \frac{11}{50} \).
- (ii) The probability of the lost card being a club is \( \frac{11}{50} \).