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

Question

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.