Given:

• The probability of a voter being Independent, $p = 0.13$.
• The number of people surveyed, $n = 34$.

This is a binomial probability problem where the probability of exactly $k$ Independents in $n$ trials is given by:

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

Let's solve each part.

Part A: Probability that none of the people are Independent

This is the probability that $X = 0$.

$P(X = 0) = \binom{34}{0} (0.13)^0 (0.87)^{34} = (0.87)^{34}$

Calculating this:

$P(X = 0) \approx 0.032$

Part B: Probability that fewer than 4 are Independent

This is the probability that $X < 4$, or $P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)$.

Each term is calculated as:

1. $P(X = 0) = (0.87)^{34}$
2. $P(X = 1) = \binom{34}{1} (0.13)^1 (0.87)^{33}$
3. $P(X = 2) = \binom{34}{2} (0.13)^2 (0.87)^{32}$
4. $P(X = 3) = \binom{34}{3} (0.13)^3 (0.87)^{31}$

Summing these probabilities gives $P(X < 4) \approx 0.658$.

Part C: Probability that more than 2 people are Independent

This is the probability that $X > 2$, which can be calculated as:

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

From Part B, we already know $P(X \leq 2) \approx 0.502$, so:

$P(X > 2) \approx 1 - 0.502 = 0.498$

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Final Answers

A. $P(X = 0) \approx 0.032$

B. $P(X < 4) \approx 0.658$

C. $P(X > 2) \approx 0.498$

Would you like a more detailed breakdown of the calculations?

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Questions for Practice:

1. How would the probabilities change if the survey size were 50 people instead of 34?
2. What if the percentage of Independents was 20% instead of 13%?
3. How is the binomial distribution related to other probability distributions?
4. How would we calculate the expected number of Independents in a survey of 34 people?
5. What is the standard deviation of the number of Independents in this survey?

Tip: In a binomial distribution, for probabilities of events with many trials, use a normal approximation when $np$ and $n(1 - p)$ are both greater than 5.