Chapter 3 HW- Probability Topics Current score: AI. Category: Homework 68.95/100 points d. Which of the following explains what " \( P(A \mid B) \) " represents? A \( P(A \mid B) \) is the probability that the sum of rolling two dice is at most ten given the fact that we have a one or three on the first die, followed by an odd number on the second die. B \( P(A \mid B) \) is the probability of rolling a one or three on the first die, followed by an odd number on the second die given that the sum of these two rolls is at least ten. C \( P(A \mid B) \) is the probability of rolling a one or three on the first die, followed by an odd number on the second die given that the sum of these two rolls is at most ten. D None of the above Find \( P(A \mid B) \). \[ P(A \mid B)= \] Enter your answer e. Are \( A \) and \( B \) mutually exclusive events? It is possible to have a one or a three on the first die rolling followed by an odd number on the second die when the sum of these two numbers is A at most ten. Numerically, we have \( P(A \) AND \( B)=\frac{1}{6} \), so the events are not mutually exclusive. It is impossible to have a one or a three on the first die rolling followed by an odd number on the second die when the sum of these two numbers \( B \) is at most ten. Numerically, we have \( P(A \) AND \( B)=0 \), so the events are mutually exclusive.

Sure, let's work through **Problem d** and **Problem e** step by step. --- ## **Problem d** **Question:** Which of the following explains what \( P(A \mid B) \) represents? **Options:** - **A.** \( P(A \mid B) \) is the probability that the sum of rolling two dice is at most ten given the fact that we have a one or three on the first die, followed by an odd number on the second die. - **B.** \( P(A \mid B) \) is the probability of rolling a one or three on the first die, followed by an odd number on the second die given that the sum of these two rolls is at least ten. - **C.** \( P(A \mid B) \) is the probability of rolling a one or three on the first die, followed by an odd number on the second die given that the sum of these two rolls is at most ten. - **D.** None of the above. **Understanding \( P(A \mid B) \):** The notation \( P(A \mid B) \) denotes the **probability of event \( A \)** occurring **given that event \( B \)** has occurred. Let's define the events based on the options: - **Event \( A \):** The sum of rolling two dice is at most ten. - **Event \( B \):** Rolling a one or three on the first die, followed by an odd number on the second die. **Interpreting the Options:** - **Option A** correctly states that \( P(A \mid B) \) is the probability that the sum is at most ten **given** that \( B \) has occurred. - **Option B** incorrectly reverses the conditions, making it \( P(B \mid A) \). - **Option C** misstates \( A \) and \( B \) within the conditional probability. - **Option D** is incorrect because **Option A** is accurate. **Conclusion:** **Option A** is the correct interpretation of \( P(A \mid B) \). --- **Calculating \( P(A \mid B) \):** The formula for conditional probability is: \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \] **Step 1: Calculate \( P(B) \)** - Rolling a one or three on the first die: \( 2/6 = 1/3 \) - Rolling an odd number on the second die: \( 3/6 = 1/2 \) - Since the two dice are independent: \[ P(B) = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6} \] **Step 2: Calculate \( P(A \cap B) \)** - \( A \cap B \) means both: - The sum is at most ten. - The first die is a one or three, and the second die is odd. First, list all possible outcomes for \( B \): - First die: 1 or 3 - Second die: 1, 3, or 5 - Total favorable outcomes for \( B \): \( 2 \times 3 = 6 \) Now, check which of these have a sum at most ten: - \( (1,1) \): \( 1 + 1 = 2 \) ✓ - \( (1,3) \): \( 1 + 3 = 4 \) ✓ - \( (1,5) \): \( 1 + 5 = 6 \) ✓ - \( (3,1) \): \( 3 + 1 = 4 \) ✓ - \( (3,3) \): \( 3 + 3 = 6 \) ✓ - \( (3,5) \): \( 3 + 5 = 8 \) ✓ All outcomes in \( B \) also satisfy \( A \) (since the maximum sum here is 8, which is less than 10). \[ P(A \cap B) = P(B) = \frac{1}{6} \] **Step 3: Compute \( P(A \mid B) \)** \[ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{\frac{1}{6}}{\frac{1}{6}} = 1 \] **Answer:** \[ P(A \mid B) = 1 \] This means that if you have a one or three on the first die followed by an odd number on the second die, the sum of the two dice will **always** be at most ten. --- ## **Problem e** **Question:** Are \( A \) and \( B \) mutually exclusive events? **Options Explanation:** 1. **First Statement:** - It is possible to have a one or a three on the first die followed by an odd number on the second die **when the sum of these two numbers is at most ten**. - Numerically, \( P(A \text{ AND } B) = \frac{1}{6} \), so the events are **not** mutually exclusive. 2. **Second Statement:** - It is **impossible** to have a one or a three on the first die followed by an odd number on the second die **when the sum of these two numbers is at most ten**. - Numerically, \( P(A \text{ AND } B) = 0 \), so the events are **mutually exclusive**. **Determining Mutual Exclusivity:** Two events \( A \) and \( B \) are **mutually exclusive** if they **cannot occur simultaneously**, i.e., \( P(A \cap B) = 0 \). From **Problem d**, we found: \[ P(A \cap B) = \frac{1}{6} \neq 0 \] **Conclusion:** Since \( P(A \cap B) = \frac{1}{6} \), which is not zero, \( A \) and \( B \) are **not mutually exclusive**. **Answer:** - **\( A \) and \( B \) are not mutually exclusive events** because there is a non-zero probability (\( \frac{1}{6} \)) that both events occur together. --- ## **Summary** - **Problem d:** - **Correct Option:** A - **\( P(A \mid B) = 1 \)** - **Problem e:** - **\( A \) and \( B \) are not mutually exclusive events.** Feel free to reach out if you need further clarification or assistance!