All the spades are taken out from a pack of cards. From these cards; cards are drawn one by one without replacement till the ace of spades comes. The probability that the ace comes in the 4th draw is

Options

A. `(1)/(13)`
B. `(12)/(13)`
C. `(4)/(13)`
D. None of these

Answer

To find the probability that the ace of spades comes in the 4th draw when drawing cards one by one without replacement from a set of spades, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Scenario**:
- We have a total of 13 spade cards (from a standard deck of cards).
- We want to find the probability that the ace of spades is drawn specifically on the 4th draw.

2. **Conditions for the Ace of Spades to be in the 4th Draw**:
- For the ace of spades to be drawn on the 4th draw, the first three draws must not be the ace of spades. This means we can draw any of the other 12 spade cards in the first three draws.

3. **Calculating the Probability**:
- The probability that the first card drawn is not the ace of spades is \( \frac{12}{13} \) (since there are 12 cards that are not the ace).
- The probability that the second card drawn is also not the ace of spades is \( \frac{11}{12} \) (since one card has already been drawn and it was not the ace).
- The probability that the third card drawn is not the ace of spades is \( \frac{10}{11} \) (since two cards have already been drawn and neither was the ace).

4. **Probability of the Ace of Spades on the 4th Draw**:
- The probability that the 4th card drawn is the ace of spades is \( \frac{1}{10} \) (since there is only one ace of spades left among the remaining 10 cards).

5. **Combining the Probabilities**:
- The overall probability that the ace of spades is drawn on the 4th draw can be calculated by multiplying the probabilities of the first three cards not being the ace and the 4th card being the ace:
\[ P(\text{Ace on 4th draw}) = P(\text{1st not Ace}) \times P(\text{2nd not Ace}) \times P(\text{3rd not Ace}) \times P(\text{4th is Ace}) \]
\[ = \left(\frac{12}{13}\right) \times \left(\frac{11}{12}\right) \times \left(\frac{10}{11}\right) \times \left(\frac{1}{10}\right) \]

6. **Simplifying the Expression**:
- When we multiply these fractions, we can see that the \(12\), \(11\), and \(10\) in the numerators and denominators will cancel out:
\[ = \frac{12 \times 11 \times 10 \times 1}{13 \times 12 \times 11 \times 10} = \frac{1}{13} \]

### Final Answer:
The probability that the ace of spades comes in the 4th draw is \( \frac{1}{13} \).