Question
What is the probability of choosing a diamond or a queen in a standard deck of cards?(1 point) Responses 1352+451−152 Start Fraction 13 over 52 End Fraction plus Start Fraction 4 over 51 End Fraction minus Start Fraction 1 over 52 End Fraction 1352+452−151 Start Fraction 13 over 52 End Fraction plus Start Fraction 4 over 52 End Fraction minus Start Fraction 1 over 51 End Fraction 1352+452+152 Start Fraction 13 over 52 End Fraction plus Start Fraction 4 over 52 End Fraction plus Start Fraction 1 over 52 End Fraction 1352+452−152
Answers
GPT-4o mini
To find the probability of choosing a diamond or a queen from a standard deck of cards, we can break it down into the following steps:
1. **Identify the Events:**
- There are 13 diamonds in a standard deck (one for each rank).
- There are 4 queens in a standard deck (one for each suit).
- However, one of the queens is also a diamond (the queen of diamonds), which we must account for to avoid double-counting.
2. **Use the Formula for the Union of Two Events:**
\[
P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
\]
where:
- \( A \) = event of drawing a diamond
- \( B \) = event of drawing a queen
- \( P(A) = \text{Probability of drawing a diamond} = \frac{13}{52} \)
- \( P(B) = \text{Probability of drawing a queen} = \frac{4}{52} \)
- \( P(A \text{ and } B) = \text{Probability of drawing the queen of diamonds} = \frac{1}{52} \)
3. **Calculate the Total Probability:**
\[
P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
\]
\[
P(A \text{ or } B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52}
\]
\[
= \frac{13 + 4 - 1}{52} = \frac{16}{52}
\]
\[
= \frac{4}{13}
\]
Thus, the probability of choosing a diamond or a queen is \(\frac{4}{13}\).
None of the response options provided match this answer; however, the correct answer reflects the proper calculation according to basic probability principles. If you need further clarification on the provided options, please let me know!
1. **Identify the Events:**
- There are 13 diamonds in a standard deck (one for each rank).
- There are 4 queens in a standard deck (one for each suit).
- However, one of the queens is also a diamond (the queen of diamonds), which we must account for to avoid double-counting.
2. **Use the Formula for the Union of Two Events:**
\[
P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
\]
where:
- \( A \) = event of drawing a diamond
- \( B \) = event of drawing a queen
- \( P(A) = \text{Probability of drawing a diamond} = \frac{13}{52} \)
- \( P(B) = \text{Probability of drawing a queen} = \frac{4}{52} \)
- \( P(A \text{ and } B) = \text{Probability of drawing the queen of diamonds} = \frac{1}{52} \)
3. **Calculate the Total Probability:**
\[
P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)
\]
\[
P(A \text{ or } B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52}
\]
\[
= \frac{13 + 4 - 1}{52} = \frac{16}{52}
\]
\[
= \frac{4}{13}
\]
Thus, the probability of choosing a diamond or a queen is \(\frac{4}{13}\).
None of the response options provided match this answer; however, the correct answer reflects the proper calculation according to basic probability principles. If you need further clarification on the provided options, please let me know!