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

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+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+452−152

1 answer

To find the probability of choosing a diamond or a queen from a standard deck of 52 cards, we can use the formula for the probability of either event occurring, which is:

\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

Where:

  • \( P(A) \) is the probability of drawing a diamond.
  • \( P(B) \) is the probability of drawing a queen.
  • \( P(A \cap B) \) is the probability of drawing a card that is both a diamond and a queen.
  1. Probability of drawing a diamond (P(A)): There are 13 diamonds in a deck of 52 cards. \[ P(A) = \frac{13}{52} \]

  2. Probability of drawing a queen (P(B)): There are 4 queens in the deck. \[ P(B) = \frac{4}{52} \]

  3. Probability of drawing the queen of diamonds (P(A ∩ B)): There is only 1 queen of diamonds in the deck. \[ P(A \cap B) = \frac{1}{52} \]

Now we can substitute these values into the formula: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] \[ P(A \cup B) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \] \[ P(A \cup B) = \frac{13 + 4 - 1}{52} \] \[ P(A \cup B) = \frac{16}{52} \]

Simplifying that gives: \[ P(A \cup B) = \frac{4}{13} \]

Now, looking at the options given:

  • \( \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \) corresponds to our calculation.

Thus, the correct response is: \[ \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \]