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Part 1
Assume that when an adult is randomly​ selected, the probability that they do not require vision correction is 26​%. If 12 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 12 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.).          what is the next questions Assume that random guesses are made for eleveneleven multiple choice questions on a medical admissions​ test, so that there are n=11 ​trials, each with a probability of success​ (correct) given by p =0.20. Find the probability that the number x of correct answers is fewer than 3.
Question content area bottom
Part 1
The probability that the number x of correct answers is fewer than 3 is enter your response here.
​(Round to three decimal places as​ needed.)

### Problem 1: Probability that exactly 2 adults do not require vision correction
Given:
- Probability that an adult does **not** require vision correction, \( p = 0.26 \).
- Number of adults selected, \( n = 12 \).
- We want the probability that exactly 2 of them do not require vision correction.

This problem follows a **binomial distribution** with parameters \( n = 12 \) and \( p = 0.26 \). The probability mass function for a binomial distribution is given by:
\[
P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
\]
where \( X \) is the number of successes (adults who do not require vision correction), \( k = 2 \), \( n = 12 \), and \( p = 0.26 \).

Substituting the values:
\[
P(X = 2) = \binom{12}{2} (0.26)^2 (0.74)^{10}
\]

1. **Calculate** \( \binom{12}{2} \):
   \[
   \binom{12}{2} = \frac{12!}{2!(12 - 2)!} = 66
   \]

2. **Calculate** \( (0.26)^2 \):
   \[
   (0.26)^2 = 0.0676
   \]

3. **Calculate** \( (0.74)^{10} \):
   \[
   (0.74)^{10} \approx 0.0527
   \]

4. **Combine**:
   \[
   P(X = 2) \approx 66 \times 0.0676 \times 0.0527 \approx 0.234
   \]

So, the probability that exactly 2 of the 12 adults do not require vision correction is approximately **0.234**.

---

### Problem 2: Probability that fewer than 3 answers are correct
Given:
- Number of questions, \( n = 11 \).
- Probability of a correct answer by guessing, \( p = 0.20 \).
- We want the probability that fewer than 3 answers are correct, i.e., \( P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) \).

This also follows a **binomial distribution**. We will calculate \( P(X = 0) \), \( P(X = 1) \), and \( P(X = 2) \) separately and then add them together.

1. **Calculate \( P(X = 0) \):**
   \[
   P(X = 0) = \binom{11}{0} (0.20)^0 (0.80)^{11} = 1 \times 1 \times (0.80)^{11} \approx 0.0859
   \]

2. **Calculate \( P(X = 1) \):**
   \[
   P(X = 1) = \binom{11}{1} (0.20)^1 (0.80)^{10} = 11 \times 0.20 \times 0.1074 \approx 0.2363
   \]

3. **Calculate \( P(X = 2) \):**
   \[
   P(X = 2) = \binom{11}{2} (0.20)^2 (0.80)^9 = 55 \times 0.04 \times 0.1342 \approx 0.295
   \]

4. **Combine**:
   \[
   P(X < 3) = 0.0859 + 0.2363 + 0.295 \approx 0.617
   \]

So, the probability that fewer than 3 answers are correct is approximately **0.617**.

---

Let me know if you need further clarification!

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**Additional Related Questions:**
1. How would the probability change if the probability of success was different?
2. What is the probability of getting exactly 3 correct answers?
3. How can we determine the expected number of successes in a binomial distribution?
4. How do we calculate the variance of a binomial distribution?
5. What is the probability of getting at least 5 correct answers in 11 trials?

**Tip:** In binomial distributions, as the probability \( p \) approaches 0.5, the distribution tends to become more symmetric.

Assume that when an adult is randomly selected, the probability that they do not require vision correction is 26%. If 12 adults are randomly selected, find the probability that exactly 2 of them do not require a vision correction. Also, assume that random guesses are made for eleven multiple choice questions on a medical admissions test, with n=11 trials, each with a probability of success (correct) given by p = 0.20. Find the probability that the number x of correct answers is fewer than 3.

Solve binomial probability problems involving real-world scenarios: finding the probability that exactly 2 out of 12 adults do not need vision correction, and calculating the probability of fewer than 3 correct answers in 11 multiple-choice questions with random guessing (p=0.20). Step-by-step solutions with calculations.

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