Question content area top
Part 1
​Multiple-choice questions each have fourfour possible answers left parenthesis a comma b comma c comma d right parenthesis(a, b, c, d)​, one of which is correct. Assume that you guess the answers to three such questions.
Question content area bottom
Part 1
a. Use the multiplication rule to find ​P(CWW​), where C denotes a correct answer and W denotes a wrong answer.
​P(CWW​)=StartFraction 9 Over 64 EndFraction
9
64 ​(Type an exact​ answer.)
Part 2
b. Beginning with CWW​, make a complete list of the different possible arrangements of oneone correct answer and two wrong answers​, then find the probability for each entry in the list.
​P(CWWCWW​)minus−see above
​P(WWC​)=enter your response here
​P(WCW​)=enter your response here
 ​(Type exact​ answers.)
Assume that when an adult is randomly​ selected, the probability that they do not require vision correction is 24​%. If 8 adults are randomly​ selected, find the probability that exactly 2 of them do not require a vision correction.
Question content area bottom
Part 1
If 8 adults are randomly​ selected, the probability that exactly 2 of them do not require a vision correction is enter your response here.
​(Round to three decimal places as​ needed.)

Let's go through each part of the question step-by-step.

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### Part 1a
#### Problem:
We have three multiple-choice questions, each with four possible answers: \( a \), \( b \), \( c \), and \( d \). One of these answers is correct. We are guessing the answers, so the probability of getting a correct answer (\( C \)) on any given question is \( \frac{1}{4} \), and the probability of getting a wrong answer (\( W \)) is \( \frac{3}{4} \).

We want to find the probability \( P(CWW) \), where we get one correct answer followed by two wrong answers.

#### Solution:
Using the multiplication rule, we calculate:
\[
P(CWW) = P(C) \times P(W) \times P(W) = \frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{9}{64}
\]

So,
\[
P(CWW) = \frac{9}{64}
\]

---

### Part 1b
#### Problem:
We need to list all possible arrangements of one correct answer and two wrong answers. Then, we need to find the probability for each arrangement.

#### Solution:
The different possible arrangements with one correct answer and two wrong answers are:
1. \( CWW \)
2. \( WCW \)
3. \( WWC \)

Now, let's calculate the probability for each arrangement. Since each question is independent, each arrangement will have the same probability as \( P(CWW) \) because we are just changing the order of outcomes:

\[
P(CWW) = \frac{9}{64}
\]
\[
P(WCW) = \frac{9}{64}
\]
\[
P(WWC) = \frac{9}{64}
\]

Each arrangement has the same probability, so:
\[
P(CWW) = P(WCW) = P(WWC) = \frac{9}{64}
\]

---

### Part 2
#### Problem:
Now, we are given that the probability an adult does not require vision correction is 24%, or \( 0.24 \). We need to find the probability that exactly 2 out of 8 randomly selected adults do not require vision correction.

#### Solution:
This is a binomial probability problem where:
- \( n = 8 \) (number of trials),
- \( k = 2 \) (number of successful outcomes),
- \( p = 0.24 \) (probability of success for each trial).

The binomial probability formula is:
\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]

Substitute the values:
\[
P(X = 2) = \binom{8}{2} (0.24)^2 (1 - 0.24)^{8 - 2}
\]

1. **Calculate \( \binom{8}{2} \):**
   \[
   \binom{8}{2} = \frac{8!}{2!(8 - 2)!} = \frac{8 \times 7}{2 \times 1} = 28
   \]

2. **Calculate \( (0.24)^2 \):**
   \[
   (0.24)^2 = 0.0576
   \]

3. **Calculate \( (1 - 0.24)^6 \):**
   \[
   (1 - 0.24) = 0.76
   \]
   \[
   (0.76)^6 \approx 0.2205
   \]

4. **Combine everything:**
   \[
   P(X = 2) = 28 \times 0.0576 \times 0.2205 \approx 0.354
   \]

So, the probability that exactly 2 out of 8 adults do not require vision correction is approximately:
\[
P(X = 2) \approx 0.354
\]

---

### Summary of Answers:
1. **Part 1a**: \( P(CWW) = \frac{9}{64} \)
2. **Part 1b**:
   - \( P(CWW) = \frac{9}{64} \)
   - \( P(WCW) = \frac{9}{64} \)
   - \( P(WWC) = \frac{9}{64} \)
3. **Part 2**: \( P(\text{exactly 2 out of 8}) \approx 0.354 \)

---

Would you like further details on any of the steps? 

Here are some related questions you might find useful:
1. What is the general multiplication rule in probability?
2. How do you calculate combinations for binomial probability problems?
3. Can the binomial formula be used when probabilities change between trials?
4. How would you approach the problem if the probability of getting a correct answer was different for each question?
5. What other types of probability distributions are there besides binomial?

**Tip**: When dealing with multiple events, always check if they are independent; this simplifies calculations significantly.

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. Use the multiplication rule to find P(CWW), where C denotes a correct answer and W denotes a wrong answer. Also, list the different possible arrangements of one correct answer and two wrong answers, then find the probability for each arrangement. Finally, find the probability that exactly 2 out of 8 randomly selected adults do not require vision correction.

Explore the calculation of probabilities for multiple-choice questions and the binomial distribution. Learn how to find the probability of specific outcomes when guessing answers and the likelihood of adults needing vision correction. This guide is suitable for Grades 9-12 students.

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