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

Question

Responses

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+452−151
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

Answer 1

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

1. Count the total diamonds: There are 13 diamonds in a deck.
2. Count the total queens: There are 4 queens in a deck.
3. Count the overlap: There is 1 queen of diamonds.

Using the inclusion-exclusion principle:

\[ P(\text{diamond or queen}) = P(\text{diamond}) + P(\text{queen}) - P(\text{diamond and queen}) \]

Plugging in the values:

\[ P(\text{diamond or queen}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \]

Simplifying this gives:

\[ = \frac{13 + 4 - 1}{52} = \frac{16}{52} = \frac{4}{13} \]

Among the given options, the correct expression that captures this probability is:

\[ \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \]

So the first option, which is:

\[ \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \]

is the correct answer.