A person drawn a card from a pack of playing cards replaces it and shuffles the pack .He continues doing this until he draws a spade .The chance that he will fail in first two times is

Options

A. `(9)/(64)`
B. `(1)/(64)`
C. `(1)/(16)`
D. `(9)/(16)`

Answer

To solve the problem step by step, we need to calculate the probability of failing to draw a spade in the first two attempts when drawing cards from a standard deck of playing cards.

### Step-by-Step Solution:

1. **Understand the total number of cards**:
A standard deck of playing cards has 52 cards, which includes 4 suits: hearts, diamonds, clubs, and spades. Each suit has 13 cards.

**Hint**: Remember that there are 4 suits in a deck of cards, each containing 13 cards.

2. **Determine the probability of drawing a spade**:
Since there are 13 spades in a deck of 52 cards, the probability of drawing a spade (P(spade)) is:
\[ P(\text{spade}) = \frac{13}{52} = \frac{1}{4} \]

**Hint**: The probability of an event is calculated as the number of favorable outcomes divided by the total number of outcomes.

3. **Calculate the probability of not drawing a spade**:
The probability of not drawing a spade (P(not spade)) is:
\[ P(\text{not spade}) = 1 - P(\text{spade}) = 1 - \frac{1}{4} = \frac{3}{4} \]

**Hint**: The complement of an event (not drawing a spade) can be found by subtracting the probability of the event from 1.

4. **Calculate the probability of failing to draw a spade in the first two attempts**:
Since the draws are independent (the card is replaced and the deck is shuffled each time), the probability of failing to draw a spade in the first two attempts is:
\[ P(\text{fail first two times}) = P(\text{not spade}) \times P(\text{not spade}) = \left(\frac{3}{4}\right) \times \left(\frac{3}{4}\right) = \frac{9}{16} \]

**Hint**: For independent events, the probability of both events occurring is the product of their individual probabilities.

5. **Conclusion**:
The probability that the person will fail to draw a spade in the first two attempts is \(\frac{9}{16}\).

**Final Answer**: The chance that he will fail in the first two times is \(\frac{9}{16}\).