Question

What is the probability of randomly selecting a card from a standard 52 card deck and having the card be a face card or an odd numbered card? Enter a fraction or decimal to the nearest hunderdth.

Answer 1

In a standard 52-card deck, there are 12 face cards (3 face cards: King, Queen, Jack per suit, and 4 suits) and 20 odd-numbered cards (1, 3, 5, 7, 9 for each suit).

1. **Counting face cards**:
- There are 3 face cards (King, Queen, Jack) in each of the 4 suits.
- Total face cards = \(3 \times 4 = 12\).

2. **Counting odd-numbered cards**:
- The odd-numbered cards are 1 (Ace), 3, 5, 7, and 9, giving us 5 odd numbers in each of the 4 suits.
- Total odd-numbered cards = \(5 \times 4 = 20\).

3. **Finding the intersection**:
- There are no cards that are both face cards and odd-numbered (the face cards are 11, 12, and 13 in value).

4. **Applying the principle of inclusion-exclusion**:
\[
\text{Total unique cards} = \text{face cards} + \text{odd-numbered cards} - \text{intersection}
\]
\[
\text{Total unique cards} = 12 + 20 - 0 = 32
\]

5. **Calculating the probability**:
- The probability of selecting a card that is either a face card or an odd-numbered card is the number of favorable outcomes divided by the total number of outcomes:
\[
P(\text{face card or odd card}) = \frac{\text{Total unique cards}}{\text{Total cards}} = \frac{32}{52}
\]

6. **Simplifying the fraction**:
\[
\frac{32}{52} = \frac{8}{13} \approx 0.6153846153846154
\]

Rounding to the nearest hundredth:
\[
\text{Probability} \approx 0.62
\]

Thus, the final answer is \( \boxed{0.62} \).