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

1 answer

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

  1. Number of Diamonds: There are 13 diamonds in a deck.
  2. Number of Queens: There are 4 queens in a deck.
  3. Number of Queen of Diamonds: There is 1 queen that is also a diamond.

Using inclusion-exclusion, the formula to find the probability \( P(D \cup Q) \) of choosing a diamond (D) or a queen (Q) is:

\[ P(D \cup Q) = P(D) + P(Q) - P(D \cap Q) \]

Where:

  • \( P(D) = \frac{13}{52} \) (probability of choosing a diamond)
  • \( P(Q) = \frac{4}{52} \) (probability of choosing a queen)
  • \( P(D \cap Q) = \frac{1}{52} \) (probability of choosing the queen of diamonds)

Now, substituting these values into the equation gives us:

\[ P(D \cup Q) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \] \[ P(D \cup Q) = \frac{13 + 4 - 1}{52} = \frac{16}{52} = \frac{4}{13} \]

When analyzing your options, we find that the correct formulation aligns with this reasoning:

The response that fits correctly should be:

1352 + 452 - 152

This corresponds to the calculation we performed (\( \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \)).