Question

A standard deck of cards has 52 card. The 52 cards are divided into 4 units: diamonds, clubs, hearts, and spades. The diamond and the hearts are red cards and the clubs and the spades are black. See the picture below for an example. For the following problems, consider the following scenario: Brant is doing a card trick and asks you to pick a card from a standard deck. Let event A be that you choose an ace and let event B be that you choose a red card.
Question 1
Determine the following . Enter your answer as a fraction(3 points)
P (A ∩ B)

P(A) =
P (B) =
Question 2
True or False: In the given situation, events A and B are dependent.(1 point)
Responses
True
True
False

Answer 1

Let's determine the probabilities of events A and B.

**Question 1: Calculate the following probabilities.**

1. **P(A ∩ B)**: This is the probability of choosing a card that is both an ace and a red card.

- There are 4 aces in a deck: the ace of diamonds, ace of hearts, ace of clubs, and ace of spades.
- Among these, the red aces are the ace of diamonds and the ace of hearts. Thus, there are 2 favorable outcomes.

\[
P(A \cap B) = \frac{\text{Number of red aces}}{\text{Total number of cards}} = \frac{2}{52} = \frac{1}{26}
\]

2. **P(A)**: This is the probability of choosing an ace.

- As mentioned earlier, there are 4 aces in total.

\[
P(A) = \frac{\text{Number of aces}}{\text{Total number of cards}} = \frac{4}{52} = \frac{1}{13}
\]

3. **P(B)**: This is the probability of choosing a red card.

- There are 26 red cards in total (13 diamonds and 13 hearts).

\[
P(B) = \frac{\text{Number of red cards}}{\text{Total number of cards}} = \frac{26}{52} = \frac{1}{2}
\]

**Summary of the probabilities:**
- \( P(A \cap B) = \frac{1}{26} \)
- \( P(A) = \frac{1}{13} \)
- \( P(B) = \frac{1}{2} \)

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**Question 2: True or False: In the given situation, events A and B are dependent.**

To determine if events A and B are independent, we check if:

\[
P(A \cap B) = P(A) \cdot P(B)
\]

Calculating \( P(A) \cdot P(B) \):

\[
P(A) \cdot P(B) = \left(\frac{1}{13}\right) \cdot \left(\frac{1}{2}\right) = \frac{1}{26}
\]

Since \( P(A \cap B) = \frac{1}{26} \) is equal to \( P(A) \cdot P(B) = \frac{1}{26} \), the events A and B are independent.

Thus, the answer to Question 2 is **False**.

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**Final Answers:**

1. \( P(A \cap B) = \frac{1}{26} \), \( P(A) = \frac{1}{13} \), \( P(B) = \frac{1}{2} \)
2. False