This problem involves binomial probability because each person surveyed either identifies as Independent (success) or not (failure), with a fixed probability of success (12%).

Given:

• The probability that a person is Independent: $p = 0.12$.
• The number of people surveyed: $n = 29$.
• The number of successes (people identifying as Independent) is denoted by $X$.

The binomial distribution formula is:

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

where $\binom{n}{k}$ is the binomial coefficient, $k$ is the number of successes, $p$ is the probability of success, and $n$ is the total number of trials.

Let's solve each part:

A. Probability that none of the people are Independent

This corresponds to $P(X = 0)$.

$P(X = 0) = \binom{29}{0} (0.12)^0 (0.88)^{29} = (1) \cdot (1) \cdot (0.88)^{29}$

Calculating $(0.88)^{29}$:

$P(X = 0) \approx 0.072$

B. Probability that fewer than 6 are Independent

This corresponds to $P(X < 6)$, which is the sum of probabilities for $X = 0, 1, 2, 3, 4, 5$.

$P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

Each of these terms can be computed using the binomial formula. For example:

$P(X = 1) = \binom{29}{1} (0.12)^1 (0.88)^{28}$ $P(X = 2) = \binom{29}{2} (0.12)^2 (0.88)^{27}$

Summing up these probabilities gives the total:

$P(X < 6) \approx 0.998$

C. Probability that more than 2 people are Independent

This corresponds to $P(X > 2)$, which is the complement of $P(X \leq 2)$.

$P(X > 2) = 1 - P(X \leq 2)$

First, we calculate $P(X \leq 2)$:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

Using the binomial probabilities for each term:

$P(X = 0) \approx 0.072, \quad P(X = 1) \approx 0.283, \quad P(X = 2) \approx 0.396$

Summing these:

$P(X \leq 2) \approx 0.072 + 0.283 + 0.396 = 0.751$

Thus:

$P(X > 2) = 1 - 0.751 = 0.249$

Summary of Answers:

A. Probability that none of the people are Independent: $P(X = 0) \approx 0.072$

B. Probability that fewer than 6 are Independent: $P(X < 6) \approx 0.998$

C. Probability that more than 2 people are Independent: $P(X > 2) \approx 0.249$

Would you like more details on any part of the solution?

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Here are 5 questions related to this problem:

1. How does the binomial distribution change if the probability of success increases?
2. What is the cumulative distribution function in binomial probability?
3. How would the results change if the survey size was increased to 50 people?
4. How can the normal approximation be applied to binomial distributions?
5. What is the probability of exactly 6 Independents in the same scenario?

Tip: In binomial distributions, probabilities of "at least" or "more than" are often easier to compute using complementary probabilities.