To solve the first part of your question, we can use the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$

where:

• $n = 12$ (the number of trials),
• $k = 2$ (the number of successes),
• $p = 0.26$ (the probability of success),
• $\binom{n}{k}$ is the binomial coefficient.

Calculating this step-by-step:

1. Calculate the binomial coefficient $\binom{12}{2}$:

$\binom{12}{2} = \frac{12!}{2!(12-2)!} = \frac{12 \times 11}{2 \times 1} = 66$

2. Calculate $p^k$ and $(1-p)^{n-k}$:

$p^k = 0.26^2 = 0.0676$ $1 - p = 0.74 \quad \Rightarrow \quad (1-p)^{n-k} = 0.74^{10} \approx 0.0563$

3. Combine these values into the binomial formula:

$P(X = 2) = 66 \times 0.0676 \times 0.0563$

Calculating this gives:

$P(X = 2) \approx 66 \times 0.0676 \times 0.0563 \approx 0.2522$

Rounding to three decimal places, we find:

$P(X = 2) \approx 0.252$

Answer for the First Part:

The probability that exactly 2 of them do not require vision correction is approximately 0.252.

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Part 2:

Now, for the second part of your question regarding the probability $P(CWW)$:

Assuming you are guessing the answers to three questions, with:

• Probability of a correct answer $P(C) = \frac{1}{4}$ (since there is one correct answer out of four),
• Probability of a wrong answer $P(W) = \frac{3}{4}$.

Using the multiplication rule, the probability $P(CWW)$ is calculated as follows:

$P(CWW) = P(C) \times P(W) \times P(W) = \left(\frac{1}{4}\right) \times \left(\frac{3}{4}\right) \times \left(\frac{3}{4}\right)$

Calculating this gives:

$P(CWW) = \frac{1}{4} \times \frac{3}{4} \times \frac{3}{4} = \frac{1 \times 3 \times 3}{4 \times 4 \times 4} = \frac{9}{64}$

Answer for the Second Part:

The exact answer for $P(CWW)$ is $\frac{9}{64}$.

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Would you like any further details or explanations on these calculations? Here are some follow-up questions you might find interesting:

1. How does changing the probability of requiring vision correction affect the outcome?
2. What would be the probability of no adults requiring vision correction?
3. How can the binomial distribution be applied in other real-world scenarios?
4. What is the cumulative probability of having 2 or fewer adults not requiring vision correction?
5. How does the multiplication rule apply to more than three questions with different probabilities?

Tip: Always double-check your calculations when dealing with probabilities, especially when rounding.