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.
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Part 1
If 88 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.)

This question involves a binomial probability scenario, as we're given a fixed number of trials (8 adults) and a probability of success (not requiring vision correction).

### Step-by-Step Solution

1. **Identify parameters:**
   - Let \( p = 0.24 \) be the probability that an adult does not require vision correction.
   - Let \( n = 8 \) be the number of adults selected.
   - We are asked to find the probability that **exactly** 2 out of these 8 adults do not require vision correction.

2. **Set up the Binomial Probability Formula:**
   The probability of exactly \( k \) successes in \( n \) trials, where each trial has a success probability \( p \), is given by:
   \[
   P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
   \]
   where \( \binom{n}{k} \) is the binomial coefficient, calculated as:
   \[
   \binom{n}{k} = \frac{n!}{k!(n - k)!}
   \]

3. **Substitute values:**
   For \( k = 2 \), \( n = 8 \), and \( p = 0.24 \):
   \[
   P(X = 2) = \binom{8}{2} (0.24)^2 (1 - 0.24)^{8 - 2}
   \]
   Simplifying further:
   \[
   \binom{8}{2} = \frac{8!}{2!(8 - 2)!} = \frac{8 \times 7}{2 \times 1} = 28
   \]
   Now, calculate each part:
   - \( (0.24)^2 = 0.0576 \)
   - \( (1 - 0.24) = 0.76 \)
   - \( (0.76)^6 \approx 0.2088 \)

   Putting it all together:
   \[
   P(X = 2) \approx 28 \times 0.0576 \times 0.2088
   \]
   \[
   P(X = 2) \approx 0.336
   \]

Thus, the probability that exactly 2 out of the 8 adults do not require vision correction is approximately **0.336** (rounded to three decimal places).

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### Related Questions

1. What is the probability that exactly 3 adults out of 8 do not require vision correction?
2. How would the probability change if we selected 10 adults instead of 8?
3. What is the probability that none of the 8 adults require vision correction?
4. What is the probability that at least one adult out of 8 does not require vision correction?
5. How would you calculate the expected number of adults not requiring vision correction in a group of 8?

#### Tip:
In binomial probability problems, always ensure that the probability \( p \) and \( 1 - p \) sum to 1.

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 vision correction.

Learn how to calculate the probability that exactly 2 out of 8 randomly selected adults do not require vision correction, given a 24% probability per adult. This example explores binomial probability concepts and provides a step-by-step solution, ideal for high school students studying probability.

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