Question 1198689
Certainly, let's calculate the probabilities.

**Assumptions:**

* The number of Independents follows a binomial distribution.
* The probability of a voter being Independent is constant (p = 0.10).
* The voters are independent of each other.

**A. Probability that none of the people are Independent**

* In this case, the number of successes (Independent voters) is 0.

* Using the binomial probability formula:
   P(X = k) = (nCk) * p^k * (1-p)^(n-k)

   where:
      * n = sample size = 23
      * k = number of successes = 0
      * p = probability of success (Independent) = 0.10
      * q = probability of failure (Not Independent) = 1 - p = 0.90

* P(X = 0) = (23C0) * (0.10)^0 * (0.90)^(23-0) 
             = 1 * 1 * (0.90)^23 
             ≈ 0.0948

**Therefore, the probability that none of the people are Independent is approximately 0.0948.**

**B. Probability that fewer than 6 are Independent**

* We need to find the probability of 0, 1, 2, 3, 4, or 5 Independents.

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

* Calculate each probability using the binomial probability formula and sum them up.

* Using a calculator or statistical software, we can find:
   P(X < 6) ≈ 0.9917

**Therefore, the probability that fewer than 6 people are Independent is approximately 0.9917.**

**C. Probability that more than 2 people are Independent**

* We need to find the probability of 3, 4, 5, ..., 23 Independents.

* P(X > 2) = 1 - P(X ≤ 2) 
             = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]

* Calculate each probability using the binomial probability formula and sum them up.

* Using a calculator or statistical software, we can find:
   P(X > 2) ≈ 0.0593

**Therefore, the probability that more than 2 people are Independent is approximately 0.0593.**

I hope this helps! Let me know if you have any other questions.