Question

What is the probability of choosing a diamond or a queen in a standard deck of cards?

A. 13/52 + 4/52 + 1/52
B. 13/52 + 4/51 - 1/52
C. 13/52 + 4/52 - 1/51
D. 13 /32 + 4/52 - 1/52

Answer 1

To find the probability of choosing a diamond or a queen in a standard deck of cards, we need to consider the following:

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 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**.