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 $$ D)
eq 0 $$. C $$ C $$ and $$ D $$ are not independent because $$ \mathrm{P}(C \mid D)
eq \mathrm{P}(C) $$ are independent because $$ \mathrm{P}(C \mid D)
eq \mathrm{P}(C)+\mathrm{P}(D) $$. $$ \mathrm{D} ~ $$ B and $$ D $$ are not independent because $$ \mathrm{P}(C)
eq \mathrm{P}(D) $$. (d) Find $$ \mathrm{P}(C $$ OR $$ D) $$. $$ \mathrm{P}(C $$ OR $$ D)=C $$ (e) Find $$ \mathrm{P}(D \mid C) $$. $$ \mathrm{P}(D \mid C)=1 $$

Question

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 $$ D) 
eq 0 $$. C $$ C $$ and $$ D $$ are not independent because $$ \mathrm{P}(C \mid D) 
eq \mathrm{P}(C) $$ are independent because $$ \mathrm{P}(C \mid D) 
eq \mathrm{P}(C)+\mathrm{P}(D) $$. $$ \mathrm{D} ~ $$ B and $$ D $$ are not independent because $$ \mathrm{P}(C) 
eq \mathrm{P}(D) $$. (d) Find $$ \mathrm{P}(C $$ OR $$ D) $$. $$ \mathrm{P}(C $$ OR $$ D)=C $$ (e) Find $$ \mathrm{P}(D \mid C) $$. $$ \mathrm{P}(D \mid C)=1 $$

UpStudy AI · Accepted Answer

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!
**Part (c):** - **Independent Events:** - **Definition:** Two events $$ C $$ and $$ D $$ are independent if $$ \mathrm{P}(C \text{ AND } D) = \mathrm{P}(C) \times \mathrm{P}(D) $$. - **Check:** - If $$ \mathrm{P}(C \text{ AND } D) = \mathrm{P}(C) \times \mathrm{P}(D) $$, then $$ C $$ and $$ D $$ are independent. - If $$ \mathrm{P}(C \mid D) \neq \mathrm{P}(C) $$, then $$ C $$ and $$ D $$ are not independent. **Part (d):** $$ \mathrm{P}(C \text{ OR } D) = \mathrm{P}(C) + \mathrm{P}(D) - \mathrm{P}(C \text{ AND } D) $$ **Part (e):** $$ \mathrm{P}(D \mid C) = \frac{\mathrm{P}(C \text{ AND } D)}{\mathrm{P}(C)} $$

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