A card from a pack of 52 playing cards is lost. From the remaining cards of the pack three cards are drawn at random (without replacement) and found to be all spades. Find the probability of the lost card being a spade.

Answer

To solve the problem of finding the probability that the lost card is a spade given that three cards drawn from the remaining cards are all spades, we can use Bayes' theorem.

### Step-by-Step Solution:

1. **Define Events**:
- Let \( E_1 \): the event that the lost card is a spade.
- Let \( E_2 \): the event that the lost card is not a spade.
- Let \( A \): the event that three cards drawn are all spades.

2. **Calculate the Prior Probabilities**:
- The total number of cards is 52, with 13 spades.
- The probability that the lost card is a spade:
\[ P(E_1) = \frac{13}{52} = \frac{1}{4} \]
- The probability that the lost card is not a spade:
\[ P(E_2) = 1 - P(E_1) = 1 - \frac{1}{4} = \frac{3}{4} \]

3. **Calculate the Probability of Drawing Three Spades**:
- If the lost card is a spade (\( E_1 \)), there are 12 spades left in the remaining 51 cards. The probability of drawing 3 spades from these 12 is:
\[ P(A|E_1) = \frac{\binom{12}{3}}{\binom{51}{3}} \]
- If the lost card is not a spade (\( E_2 \)), there are still 13 spades left in the remaining 51 cards. The probability of drawing 3 spades from these 13 is:
\[ P(A|E_2) = \frac{\binom{13}{3}}{\binom{51}{3}} \]

4. **Calculate the Total Probability of Drawing Three Spades**:
- Using the law of total probability:
\[ P(A) = P(A|E_1)P(E_1) + P(A|E_2)P(E_2) \]
- Substitute the values:
\[ P(A) = \left(\frac{\binom{12}{3}}{\binom{51}{3}} \cdot \frac{1}{4}\right) + \left(\frac{\binom{13}{3}}{\binom{51}{3}} \cdot \frac{3}{4}\right) \]

5. **Use Bayes' Theorem to Find \( P(E_1|A) \)**:
- Bayes' theorem states:
\[ P(E_1|A) = \frac{P(A|E_1)P(E_1)}{P(A)} \]
- Substitute the values:
\[ P(E_1|A) = \frac{\left(\frac{\binom{12}{3}}{\binom{51}{3}} \cdot \frac{1}{4}\right)}{P(A)} \]

6. **Final Calculation**:
- After calculating the combinations:
\[ \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \]
\[ \binom{13}{3} = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} = 286 \]
\[ \binom{51}{3} = \frac{51 \times 50 \times 49}{3 \times 2 \times 1} = 23426 \]
- Substitute these values back into the equation to find \( P(E_1|A) \).