Multiple-choice questions have five possible answers (a,b,c,d,e), one of which is correct. Assume that you guess the answers to three such questions. Beginning with CWC, make a complete list of the different possible arrangements of two correct answers and one wrong answer, then find the probability for each entry in the list.

Let's begin by solving the problem step by step:

### Step 1: **Understanding the problem**
- We have three multiple-choice questions.
- Each question has five possible answers (a, b, c, d, e), and only one of them is correct.
- We are supposed to make a list of all possible arrangements where there are **two correct answers (C)** and **one wrong answer (W)**.
- Then, we need to calculate the probability for each arrangement.

### Step 2: **Listing all possible arrangements of two correct answers and one wrong answer**
The arrangement will have two C's (correct answers) and one W (wrong answer). We can arrange these in different ways by placing the "W" in different positions.

Here are all possible arrangements (denoted by C for correct and W for wrong):
1. CWC
2. CCW
3. WCC

These are the only three possible arrangements of two correct answers and one wrong answer.

### Step 3: **Finding the probability for each arrangement**
To calculate the probability of each arrangement, we need to understand the probabilities for correct and wrong answers:
- The probability of getting a correct answer (C) on any question is \( \frac{1}{5} \) (since only one out of five options is correct).
- The probability of getting a wrong answer (W) is \( \frac{4}{5} \) (since four out of five options are wrong).

For each arrangement:
1. **CWC**:
   - The probability is:
     \[
     P(CWC) = \left( \frac{1}{5} \right) \times \left( \frac{4}{5} \right) \times \left( \frac{1}{5} \right) = \frac{4}{125}
     \]

2. **CCW**:
   - The probability is:
     \[
     P(CCW) = \left( \frac{1}{5} \right) \times \left( \frac{1}{5} \right) \times \left( \frac{4}{5} \right) = \frac{4}{125}
     \]

3. **WCC**:
   - The probability is:
     \[
     P(WCC) = \left( \frac{4}{5} \right) \times \left( \frac{1}{5} \right) \times \left( \frac{1}{5} \right) = \frac{4}{125}
     \]

### Step 4: **Conclusion**
- There are 3 different possible arrangements of two correct answers and one wrong answer: **CWC, CCW, WCC**.
- The probability for each arrangement is \( \frac{4}{125} \).

Would you like more details or have any questions about this solution?

Here are five related questions to consider:
1. What is the probability of getting all three questions correct?
2. How would the probability change if there were 4 multiple-choice answers instead of 5?
3. What is the probability of getting exactly one correct answer out of three?
4. How does the probability change if you know one correct answer in advance?
5. How does the total probability sum up across all different arrangements?

**Tip:** When working with probability, ensure that all possible outcomes are considered to validate that probabilities sum up to 1.

Explore how to calculate the probability of getting two correct answers and one wrong answer in a set of three multiple-choice questions. This problem highlights key concepts in probability and combinatorics, including the probability of correct and wrong guesses. Suitable for students in Grades 9-12.

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