Points: 0 of 1 Save Multiple-choice questions each have four possible answers ( \( a, b, c, d \) ), one of which is correct. Assume that you guess the answers to three such questions. a. Use the multiplication rule to find \( \mathrm{P}(\mathrm{CWC}) \), where C denotes a correct answer and W denotes a wrong answer. \[ P(C W C)=\frac{3}{64} \text { (Type an exact answer.) } \] b. 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. \[ \begin{array}{l} \mathrm{P}(\mathrm{CWC})-\text { see above } \\ \mathrm{P}(\mathrm{WCC})=0.046875 \\ \mathrm{P}(\mathrm{CCW})=0.046875 \end{array} \] (Type exact answers.) c. Based on the preceding results, what is the probability of getting exactly two correct answers when three guesses are made? \( \square \) (Type an exact answer.)

To solve the problem, we will follow the steps outlined in the question. ### Part a: Calculate \( P(CWC) \) 1. **Understanding the Problem**: We have three questions, each with four possible answers (a, b, c, d). The probability of guessing a correct answer (C) is \( \frac{1}{4} \), and the probability of guessing a wrong answer (W) is \( \frac{3}{4} \). 2. **Using the Multiplication Rule**: The probability of a specific sequence of answers can be calculated by multiplying the probabilities of each individual answer. For \( P(CWC) \): \[ P(CWC) = P(C) \times P(W) \times P(C) = \left(\frac{1}{4}\right) \times \left(\frac{3}{4}\right) \times \left(\frac{1}{4}\right) \] 3. **Calculating**: \[ P(CWC) = \frac{1}{4} \times \frac{3}{4} \times \frac{1}{4} = \frac{3}{64} \] Thus, the answer for part a is: \[ P(CWC) = \frac{3}{64} \] ### Part b: List Arrangements and Calculate Probabilities 1. **Arrangements of Two Correct and One Wrong**: The different arrangements of two correct answers (C) and one wrong answer (W) are: - CWC - WCC - CCW 2. **Calculating Probabilities**: - We already calculated \( P(CWC) = \frac{3}{64} \). - For \( P(WCC) \): \[ P(WCC) = P(W) \times P(C) \times P(C) = \left(\frac{3}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) = \frac{3}{64} \] - For \( P(CCW) \): \[ P(CCW) = P(C) \times P(C) \times P(W) = \left(\frac{1}{4}\right) \times \left(\frac{1}{4}\right) \times \left(\frac{3}{4}\right) = \frac{3}{64} \] Thus, the probabilities for part b are: \[ \begin{array}{l} P(CWC) = \frac{3}{64} \\ P(WCC) = \frac{3}{64} \\ P(CCW) = \frac{3}{64} \end{array} \] ### Part c: Probability of Getting Exactly Two Correct Answers 1. **Summing the Probabilities**: The total probability of getting exactly two correct answers is the sum of the probabilities of the arrangements: \[ P(\text{exactly 2 correct}) = P(CWC) + P(WCC) + P(CCW) \] \[ P(\text{exactly 2 correct}) = \frac{3}{64} + \frac{3}{64} + \frac{3}{64} = \frac{9}{64} \] Thus, the answer for part c is: \[ P(\text{exactly 2 correct}) = \frac{9}{64} \] ### Final Answers - a. \( P(CWC) = \frac{3}{64} \) - b. \[ \begin{array}{l} P(CWC) = \frac{3}{64} \\ P(WCC) = \frac{3}{64} \\ P(CCW) = \frac{3}{64} \end{array} \] - c. \( P(\text{exactly 2 correct}) = \frac{9}{64} \)