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.

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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.

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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.