What is the probability of choosing a diamond or a queen in a standard deck of cards?(1 point)

Question

Responses

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
Start Fraction 13 over 52 End Fraction plus Start Fraction 4 over 52 End Fraction minus Start Fraction 1 over 52 End Fraction

1352+451−152

Answer 1

To find the probability of choosing a diamond or a queen from a standard deck of 52 cards, we need to use the principle of inclusion-exclusion.

1. The probability of choosing a diamond (there are 13 diamonds in a deck) is: \[ P(D) = \frac{13}{52} \]

2. The probability of choosing a queen (there are 4 queens in a deck) is: \[ P(Q) = \frac{4}{52} \]

3. However, one of the queens is also a diamond (the queen of diamonds). Therefore, we must subtract the probability of choosing the queen of diamonds, which is: \[ P(D \cap Q) = \frac{1}{52} \]

Now, we apply the inclusion-exclusion principle: \[ P(D \cup Q) = P(D) + P(Q) - P(D \cap Q) \] Substituting in our values: \[ P(D \cup Q) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \] Combining these fractions: \[ P(D \cup Q) = \frac{13 + 4 - 1}{52} = \frac{16}{52} = \frac{4}{13} \]

Therefore, the correct answer in the format you provided is: \[ \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \] This corresponds to the option: 1352 + 452 − 152.