Questions: Experiment A deck contains 10 cards numbered 1 through 10. A card is randomly selected and returned, then another random selection is made. Charlie randomly selects a sock from a drawer containing black socks and brown socks, and puts it on. Then another random selection is made from the remaining socks. A family has two children. Event A: The first selection is a 6. Event B: The second selection is a 5. Event A: The first selection is black. Event B: The second selection is brown. Event A: The older child has red hair. Event B: Both children have red hair. Independent Dependent

Solution Steps

To determine whether the events are independent or dependent, we need to check if the occurrence of one event affects the probability of the other event.

1. For the card selection:

   • Event A: The first selection is a 6.
   • Event B: The second selection is a 5. Since the card is returned to the deck after the first selection, the probability of the second selection is not affected by the first selection. Therefore, these events are independent.

2. For the sock selection:

   • Event A: The first selection is black.
   • Event B: The second selection is brown. Since the sock is not returned to the drawer after the first selection, the probability of the second selection is affected by the first selection. Therefore, these events are dependent.

3. For the children:

   • Event A: The older child has red hair.
   • Event B: Both children have red hair. The probability of both children having red hair is affected by the older child having red hair. Therefore, these events are dependent.

For the card selection, we define:

• Event \( A \): The first selection is a 6.
• Event \( B \): The second selection is a 5.

Since the card is returned to the deck after the first selection, the probabilities are: \[ P(A) = \frac{1}{10}, \quad P(B) = \frac{1}{10} \] The joint probability is: \[ P(A \cap B) = P(A) \cdot P(B) = \frac{1}{10} \cdot \frac{1}{10} = \frac{1}{100} \] Since \( P(A \cap B) = P(A) \cdot P(B) \), the events are independent.

For the sock selection, we define:

• Event \( A \): The first selection is black.
• Event \( B \): The second selection is brown.

Assuming there are 5 black socks and 5 brown socks, the probabilities are: \[ P(A) = \frac{5}{10} = \frac{1}{2} \] After selecting a black sock, there are now 9 socks left (4 black and 5 brown): \[ P(B|A) = \frac{5}{9} \] The joint probability is: \[ P(A \cap B) = P(A) \cdot P(B|A) = \frac{1}{2} \cdot \frac{5}{9} = \frac{5}{18} \] Since \( P(A \cap B) \neq P(A) \cdot P(B) \), the events are dependent.

For the children, we define:

• Event \( A \): The older child has red hair.
• Event \( B \): Both children have red hair.

Assuming the probability of the older child having red hair is: \[ P(A) = \frac{1}{2} \] If the older child has red hair, the probability that both children have red hair is also: \[ P(B|A) = \frac{1}{2} \] The joint probability is: \[ P(A \cap B) = P(A) \cdot P(B|A) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \] Since \( P(A \cap B) \neq P(A) \cdot P(B) \), the events are dependent.

Final Answer