To find the probability of choosing a diamond or a queen in a standard deck of cards, we need to consider the following:
- Number of diamonds: There are 13 diamonds in a deck.
- Number of queens: There are 4 queens in a deck.
- Number of cards that are both diamonds and queens: There is 1 queen of diamonds.
Using the principle of inclusion-exclusion, the total probability can be calculated as follows:
\[ P(\text{diamond or queen}) = P(\text{diamond}) + P(\text{queen}) - P(\text{diamond and queen}) \]
Calculating each part:
- \(P(\text{diamond}) = \frac{13}{52}\)
- \(P(\text{queen}) = \frac{4}{52}\)
- \(P(\text{diamond and queen}) = \frac{1}{52}\)
Now plug these values into the formula:
\[ P(\text{diamond or queen}) = \frac{13}{52} + \frac{4}{52} - \frac{1}{52} = \frac{13 + 4 - 1}{52} = \frac{16}{52} = \frac{4}{13} \]
Now let's examine the options given:
A. \(\frac{13}{52} + \frac{4}{52} + \frac{1}{52}\)
B. \(\frac{13}{52} + \frac{4}{52} - \frac{1}{52}\)
C. \(\frac{13}{52} + \frac{4}{52} - \frac{1}{51}\)
D. \(\frac{13}{32} + \frac{4}{52} - \frac{1}{52}\)
The correct option is B, which correctly applies the principle of inclusion-exclusion:
\[ \frac{13}{52} + \frac{4}{52} - \frac{1}{52} \]
Thus, the answer is B.
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