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.

Options

A. `11/50`
B. `13/50`
C. `17/50`
D. `19/50`

Explanation

To solve the problem of finding the probability of the lost card being a diamond given that two cards drawn from the remaining cards are diamonds, we can follow these steps:

### Step 1: Define Events
Let:
- \( E_1 \): The event that the lost card is a diamond.
- \( E_2 \): The event that both drawn cards are diamonds.

### Step 2: Calculate Initial Probabilities
Initially, there are 52 cards in total, with 13 diamonds. The probability of the lost card being a diamond is:
\[ P(E_1) = \frac{13}{52} = \frac{1}{4} \]
Thus, the probability of the lost card not being a diamond is:
\[ P(E_1') = 1 - P(E_1) = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 3: Calculate Conditional Probabilities
Now, we need to find the probabilities of drawing two diamonds given each event.

**Case 1: Lost card is a diamond (\( E_1 \))**
If the lost card is a diamond, there are 12 diamonds left in a total of 51 cards. The probability of drawing two diamonds is:
\[ P(E_2 | E_1) = \frac{12}{51} \times \frac{11}{50} \]

**Case 2: Lost card is not a diamond (\( E_1' \))**
If the lost card is not a diamond, there are still 13 diamonds left in a total of 51 cards. The probability of drawing two diamonds is:
\[ P(E_2 | E_1') = \frac{13}{51} \times \frac{12}{50} \]

### Step 4: Apply Bayes' Theorem
We want to find \( P(E_1 | E_2) \):
\[ P(E_1 | E_2) = \frac{P(E_2 | E_1) \cdot P(E_1)}{P(E_2 | E_1) \cdot P(E_1) + P(E_2 | E_1') \cdot P(E_1')} \]

Substituting the values:
\[ P(E_1 | E_2) = \frac{\left(\frac{12}{51} \times \frac{11}{50}\right) \cdot \frac{1}{4}}{\left(\frac{12}{51} \times \frac{11}{50}\right) \cdot \frac{1}{4} + \left(\frac{13}{51} \times \frac{12}{50}\right) \cdot \frac{3}{4}} \]

### Step 5: Simplify the Expression
Calculating the numerator:
\[ \text{Numerator} = \frac{12 \times 11}{51 \times 50} \cdot \frac{1}{4} = \frac{132}{10200} \]

Calculating the denominator:
\[ \text{Denominator} = \frac{12 \times 11}{51 \times 50} \cdot \frac{1}{4} + \frac{13 \times 12}{51 \times 50} \cdot \frac{3}{4} \]
\[ = \frac{132}{10200} + \frac{468}{10200} = \frac{600}{10200} \]

### Step 6: Final Calculation
Now substituting back into Bayes' theorem:
\[ P(E_1 | E_2) = \frac{\frac{132}{10200}}{\frac{600}{10200}} = \frac{132}{600} = \frac{11}{50} \]

### Conclusion
Thus, the probability of the lost card being a diamond is:
\[ \boxed{\frac{11}{50}} \]