To solve this problem, we'll use the binomial distribution where:

• The number of trials $n = 33$ (the number of surveyed voters).
• The probability of a voter being Independent $p = 0.11$.
• The probability of not being Independent $q = 1 - p = 0.89$.

A. Probability that none of the people are Independent

This is $P(X = 0)$:

$P(X = 0) = \binom{33}{0} (0.11)^0 (0.89)^{33} = 1 \cdot 1 \cdot (0.89)^{33}$

Calculating $(0.89)^{33}$:

$(0.89)^{33} \approx 0.0312$

So,

$P(X = 0) \approx 0.0312$

B. Probability that fewer than 7 are Independent

This is $P(X < 7)$, which can be calculated as:

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

Calculating each term:

1. For $k = 1$: $P(X = 1) = \binom{33}{1} (0.11)^1 (0.89)^{32} = 33 \cdot 0.11 \cdot (0.89)^{32} \approx 33 \cdot 0.11 \cdot 0.0351 \approx 0.1277$

2. For $k = 2$: $P(X = 2) = \binom{33}{2} (0.11)^2 (0.89)^{31} = \frac{33 \cdot 32}{2} \cdot 0.0121 \cdot (0.89)^{31} \approx 528 \cdot 0.0121 \cdot 0.0394 \approx 0.2472$

3. For $k = 3$: $P(X = 3) = \binom{33}{3} (0.11)^3 (0.89)^{30} = \frac{33 \cdot 32 \cdot 31}{6} \cdot 0.001331 \cdot (0.89)^{30} \approx 5456 \cdot 0.001331 \cdot 0.0442 \approx 0.2422$

4. For $k = 4$: $P(X = 4) = \binom{33}{4} (0.11)^4 (0.89)^{29} \approx 35960 \cdot 0.000146 \cdot (0.89)^{29} \approx 35960 \cdot 0.000146 \cdot 0.0495 \approx 0.0326$

5. For $k = 5$: $P(X = 5) = \binom{33}{5} (0.11)^5 (0.89)^{28} \approx 237336 \cdot 0.0000161 \cdot (0.89)^{28} \approx 0.0022$

6. For $k = 6$: $P(X = 6) = \binom{33}{6} (0.11)^6 (0.89)^{27} \approx 1345968 \cdot 0.000001771 \cdot (0.89)^{27} \approx 0.0003$

Now summing these probabilities:

$P(X < 7) \approx 0.0312 + 0.1277 + 0.2472 + 0.2422 + 0.0326 + 0.0022 + 0.0003 \approx 0.6832$

C. Probability that more than 2 people are Independent

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

Calculating $P(X \leq 2)$:

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \approx 0.0312 + 0.1277 + 0.2472 \approx 0.4061$

Thus,

$P(X > 2) = 1 - 0.4061 \approx 0.5939$

Summary of Results

A. Probability that none are Independent: $P(X = 0) \approx 0.031$

B. Probability that fewer than 7 are Independent: $P(X < 7) \approx 0.683$

C. Probability that more than 2 are Independent: $P(X > 2) \approx 0.594$

Would you like more details or have any questions? Here are some related questions you might consider:

1. How would the probabilities change if the percentage of Independents were to rise to 20%?
2. What assumptions are made when using the binomial distribution in this context?
3. How does increasing the sample size affect the probabilities in a binomial distribution?
4. Can you explain the significance of the complement rule in probability?
5. What are some practical implications of these probability findings for political campaigns?