E. Chapter 3 HW- Probability Topics All Category: Homework (c) Are \( C \) and \( D \) independent events? Why or why not? A \( C \) and \( D \) are independent because \( \mathrm{P}(C \) AND \( D) \neq 0 \). C \( C \) and \( D \) are not independent because \( \mathrm{P}(C \mid D) \neq \mathrm{P}(C) \) are independent because \( \mathrm{P}(C \mid D) \neq \mathrm{P}(C)+\mathrm{P}(D) \). \( \mathrm{D} ~ \) B and \( D \) are not independent because \( \mathrm{P}(C) \neq \mathrm{P}(D) \). (d) Find \( \mathrm{P}(C \) OR \( D) \). \( \mathrm{P}(C \) OR \( D)=C \) (e) Find \( \mathrm{P}(D \mid C) \). \( \mathrm{P}(D \mid C)=1 \)

Sure, let's go through each part of your homework step by step to ensure your understanding is clear. --- ### **Part (c): Are \( C \) and \( D \) independent events? Why or why not?** **Your Answer:** - *C and D are independent because \( \mathrm{P}(C \text{ AND } D) \neq 0 \).* - *C and D are not independent because \( \mathrm{P}(C \mid D) \neq \mathrm{P}(C) \) are independent because \( \mathrm{P}(C \mid D) \neq \mathrm{P}(C)+\mathrm{P}(D) \).* - *B and D are not independent because \( \mathrm{P}(C) \neq \mathrm{P}(D) \).* **Feedback:** There seems to be some confusion in your reasoning. Let's clarify the concept of **independent events**: - **Definition:** Two events, \( C \) and \( D \), are **independent** if and only if: \[ \mathrm{P}(C \text{ AND } D) = \mathrm{P}(C) \times \mathrm{P}(D) \] - **Alternative Criterion:** Independence can also be assessed by checking if: \[ \mathrm{P}(C \mid D) = \mathrm{P}(C) \] or equivalently, \[ \mathrm{P}(D \mid C) = \mathrm{P}(D) \] **Analysis of Your Statements:** 1. **"C and D are independent because \( \mathrm{P}(C \text{ AND } D) \neq 0 \)."** - **Incorrect.** The fact that \( \mathrm{P}(C \text{ AND } D) \neq 0 \) merely indicates that the events can occur together. It doesn't establish independence. What matters is *whether* \( \mathrm{P}(C \text{ AND } D) \) equals \( \mathrm{P}(C) \times \mathrm{P}(D) \). 2. **"C and D are not independent because \( \mathrm{P}(C \mid D) \neq \mathrm{P}(C) \) are independent because \( \mathrm{P}(C \mid D) \neq \mathrm{P}(C)+\mathrm{P}(D) \)."** - **Partially Correct.** If \( \mathrm{P}(C \mid D) \neq \mathrm{P}(C) \), then \( C \) and \( D \) are **not** independent. - **Confusion in Wording:** The phrase "are independent because \( \mathrm{P}(C \mid D) \neq \mathrm{P}(C)+\mathrm{P}(D) \)" is unclear and likely incorrect. Independence relates to multiplication, not addition. 3. **"B and D are not independent because \( \mathrm{P}(C) \neq \mathrm{P}(D) \)."** - **Incorrect.** Independence does **not** require \( \mathrm{P}(C) = \mathrm{P}(D) \). Events can be independent even if their probabilities are different. **Correct Approach:** To determine if \( C \) and \( D \) are independent: - **Calculate \( \mathrm{P}(C \text{ AND } D) \).** - **Compare it to \( \mathrm{P}(C) \times \mathrm{P}(D) \).** If they are equal, the events are independent; otherwise, they are not. --- ### **Part (d): Find \( \mathrm{P}(C \text{ OR } D) \).** **Your Answer:** \[ \mathrm{P}(C \text{ OR } D) = C \] **Feedback:** Your answer appears to be incomplete or possibly mistyped. The probability of the union of two events \( C \) and \( D \) is not simply \( C \). Here's the correct formula: \[ \mathrm{P}(C \text{ OR } D) = \mathrm{P}(C) + \mathrm{P}(D) - \mathrm{P}(C \text{ AND } D) \] **Explanation:** - **Addition Rule for Probability:** \[ \mathrm{P}(C \text{ OR } D) = \mathrm{P}(C) + \mathrm{P}(D) - \mathrm{P}(C \text{ AND } D) \] - **\( \mathrm{P}(C) + \mathrm{P}(D) \):** Adds the probabilities of each event occurring. - **Subtract \( \mathrm{P}(C \text{ AND } D) \):** Removes the double-counted intersection where both events occur. **Example:** If \( \mathrm{P}(C) = 0.3 \), \( \mathrm{P}(D) = 0.4 \), and \( \mathrm{P}(C \text{ AND } D) = 0.12 \), then: \[ \mathrm{P}(C \text{ OR } D) = 0.3 + 0.4 - 0.12 = 0.58 \] --- ### **Part (e): Find \( \mathrm{P}(D \mid C) \).** **Your Answer:** \[ \mathrm{P}(D \mid C) = 1 \] **Feedback:** Your answer states that \( \mathrm{P}(D \mid C) = 1 \), implying that whenever event \( C \) occurs, event \( D \) **always** occurs. This is a strong statement and depends entirely on the context or data provided in your homework problem. Let's clarify the general approach: **Formula for Conditional Probability:** \[ \mathrm{P}(D \mid C) = \frac{\mathrm{P}(C \text{ AND } D)}{\mathrm{P}(C)} \] **Interpretation:** - **\( \mathrm{P}(D \mid C) = 1 \):** This means that \( D \) always occurs when \( C \) occurs. In other words, \( C \) is a subset of \( D \). **Possible Scenarios:** 1. **If \( \mathrm{P}(C \text{ AND } D) = \mathrm{P}(C) \):** - Then, \( \mathrm{P}(D \mid C) = 1 \). - This implies that every occurrence of \( C \) is accompanied by \( D \). 2. **If \( \mathrm{P}(C \text{ AND } D) < \mathrm{P}(C) \):** - Then, \( \mathrm{P}(D \mid C) < 1 \). - This means that \( D \) does **not** always occur when \( C \) occurs. **Recommendation:** - **Check Your Data:** Ensure you have the correct values for \( \mathrm{P}(C \text{ AND } D) \) and \( \mathrm{P}(C) \) to compute \( \mathrm{P}(D \mid C) \) accurately. - **Avoid Assumptions:** Unless the problem explicitly states that \( D \) always occurs when \( C \) does, you cannot assume \( \mathrm{P}(D \mid C) = 1 \). **Example:** If \( \mathrm{P}(C \text{ AND } D) = 0.2 \) and \( \mathrm{P}(C) = 0.5 \), then: \[ \mathrm{P}(D \mid C) = \frac{0.2}{0.5} = 0.4 \] --- ### **General Tips:** 1. **Review Definitions:** - Understanding the fundamental definitions of independence, conditional probability, and probability rules is crucial. 2. **Use Formulas Carefully:** - Ensure you're applying the correct formulas to the given information. 3. **Check for Typos:** - Clear and accurate notation helps in avoiding confusion, both for you and anyone assisting you. 4. **Work with Concrete Numbers:** - If possible, apply your formulas to specific numbers to test your understanding. If you have specific probability values or additional context for events \( C \) and \( D \), feel free to share them, and I can provide more detailed assistance!