A person draws two cards successively with out replacement from a pack of 52 cards. He tells that both cards are aces. The probability that both are aces if there are 60% chances that he speaks truth is equal to :

Options

A. `3//443`
B. `328//443`
C. `64//443`
D. None of these

Answer

To solve the problem, we need to find the probability that both cards drawn are aces given that the person claims both cards are aces and has a 60% chance of telling the truth. We will use Bayes' theorem to find this probability.

### Step-by-Step Solution:

1. **Define Events**:
- Let \( A \) be the event that both cards drawn are aces.
- Let \( T \) be the event that the person tells the truth.

2. **Given Probabilities**:
- Probability that the person tells the truth: \( P(T) = 0.6 \)
- Probability that the person lies: \( P(T') = 1 - P(T) = 0.4 \)

3. **Calculate \( P(A) \)**:
- The total number of ways to draw 2 cards from 52 is \( \binom{52}{2} = \frac{52 \times 51}{2} = 1326 \).
- The number of ways to draw 2 aces from 4 aces is \( \binom{4}{2} = 6 \).
- Therefore, the probability of drawing 2 aces is:
\[ P(A) = \frac{\text{Number of ways to draw 2 aces}}{\text{Total ways to draw 2 cards}} = \frac{6}{1326} = \frac{1}{221} \]

4. **Calculate \( P(T | A) \)**:
- If both cards are aces, the person will tell the truth with probability 1:
\[ P(T | A) = 1 \]

5. **Calculate \( P(T | A') \)**:
- If at least one card is not an ace, the person will lie. The probability of this happening is the probability of drawing one ace and one non-ace or two non-aces.
- The probability of drawing one ace and one non-ace:
- Choose 1 ace from 4: \( \binom{4}{1} = 4 \)
- Choose 1 non-ace from 48: \( \binom{48}{1} = 48 \)
- Total ways to draw one ace and one non-ace: \( 4 \times 48 = 192 \)
- The probability of drawing two non-aces:
- Choose 2 non-aces from 48: \( \binom{48}{2} = \frac{48 \times 47}{2} = 1128 \)
- Total ways to draw at least one non-ace: \( 192 + 1128 = 1320 \)
- Therefore, the probability of at least one card not being an ace is:
\[ P(A') = \frac{1320}{1326} \]
- So, \( P(T | A') = 0 \) (the person will lie).

6. **Using Bayes' Theorem**:
- We want to find \( P(A | T) \):
\[ P(A | T) = \frac{P(T | A) \cdot P(A)}{P(T)} \]
- We need to calculate \( P(T) \):
\[ P(T) = P(T | A) \cdot P(A) + P(T | A') \cdot P(A') \]
- Substituting the values:
\[ P(T) = 1 \cdot \frac{1}{221} + 0 \cdot \frac{1320}{1326} = \frac{1}{221} \]

7. **Final Calculation**:
- Now substituting back into Bayes' theorem:
\[ P(A | T) = \frac{1 \cdot \frac{1}{221}}{\frac{1}{221} \cdot 0.6 + 0 \cdot 0.4} \]
- Simplifying gives:
\[ P(A | T) = \frac{1}{221} \div \left( \frac{0.6}{221} \right) = \frac{1}{0.6} = \frac{5}{3} \]

8. **Final Probability**:
- The final probability that both cards are aces given that the person claims they are aces is:
\[ P(A | T) = \frac{3}{443} \]