A card is lost from a pack of 52 playing cards. From remainder of the pack of a card is drawn and is found to be a spade. The probability that the misssing card is spade, is

Options

A. `(2)/(17)`
B. `(3)/(17)`
C. `(4)/(17)`
D. `(5)/(17)`

Answer

To solve the problem, we will use Bayes' theorem to find the probability that the missing card is a spade given that a card drawn from the remaining cards is a spade.

### Step-by-Step Solution:

1. **Define Events**:
- Let \( E_1 \): the event that the missing card is a spade.
- Let \( E_2 \): the event that the missing card is not a spade.
- Let \( A \): the event that a card drawn from the remaining cards is a spade.

2. **Find Probabilities**:
- The total number of cards in a standard deck is 52.
- The number of spades in the deck is 13.
- Therefore, the probability that the missing card is a spade:
\[ P(E_1) = \frac{13}{52} = \frac{1}{4} \]
- The probability that the missing card is not a spade:
\[ P(E_2) = \frac{39}{52} = \frac{3}{4} \]

3. **Calculate Conditional Probabilities**:
- If the missing card is a spade (\( E_1 \)), then there are 12 spades left in the remaining 51 cards. Thus, the probability of drawing a spade:
\[ P(A | E_1) = \frac{12}{51} \]
- If the missing card is not a spade (\( E_2 \)), then all 13 spades are still in the remaining 51 cards. Thus, the probability of drawing a spade:
\[ P(A | E_2) = \frac{13}{51} \]

4. **Apply Bayes' Theorem**:
- We want to find \( P(E_1 | A) \):
\[ P(E_1 | A) = \frac{P(A | E_1) \cdot P(E_1)}{P(A | E_1) \cdot P(E_1) + P(A | E_2) \cdot P(E_2)} \]
- Substitute the values:
\[ P(E_1 | A) = \frac{\left(\frac{12}{51}\right) \cdot \left(\frac{13}{52}\right)}{\left(\frac{12}{51}\right) \cdot \left(\frac{13}{52}\right) + \left(\frac{13}{51}\right) \cdot \left(\frac{39}{52}\right)} \]

5. **Simplify the Expression**:
- The numerator becomes:
\[ \frac{12 \cdot 13}{51 \cdot 52} \]
- The denominator becomes:
\[ \frac{12 \cdot 13}{51 \cdot 52} + \frac{13 \cdot 39}{51 \cdot 52} = \frac{12 \cdot 13 + 13 \cdot 39}{51 \cdot 52} = \frac{13(12 + 39)}{51 \cdot 52} \]
- Thus, we have:
\[ P(E_1 | A) = \frac{\frac{12 \cdot 13}{51 \cdot 52}}{\frac{13(12 + 39)}{51 \cdot 52}} = \frac{12}{12 + 39} = \frac{12}{51} \]

6. **Final Calculation**:
- Simplifying \( \frac{12}{51} \):
\[ \frac{12}{51} = \frac{4}{17} \]

### Conclusion:
The probability that the missing card is a spade is \( \frac{4}{17} \).