In the United States, voters who are neither Democrat nor Republican are called Independent. It is believed that 9% of voters are Independent. A survey asked 15 people to identify themselves as Democrat, Republican, or Independent.

A. What is the probability that none of the people are Independent?

Probability = 

B. What is the probability that fewer than 6 are Independent?

Probability = 

C. What is the probability that more than 2 people are Independent?

Probability = 


This problem involves a binomial distribution since we are dealing with a fixed number of trials (15 people), each of which results in one of two outcomes: a person being Independent or not. The probability of an individual being Independent is 9%, or \( p = 0.09 \), and the number of trials is \( n = 15 \).

### Part A: Probability that none of the people are Independent

The probability that none of the 15 people are Independent is the probability that all 15 people are either Democrat or Republican. This is a binomial probability where \( k = 0 \) (none are Independent):

\[
P(X = 0) = \binom{15}{0} \cdot (0.09)^0 \cdot (0.91)^{15} = (0.91)^{15}
\]

Calculating:

\[
P(X = 0) = (0.91)^{15} \approx 0.239
\]

Thus, the probability that none of the people are Independent is approximately \( 0.239 \).

### Part B: Probability that fewer than 6 are Independent

To find the probability that fewer than 6 people are Independent, we need to calculate the cumulative probability for \( X = 0, 1, 2, 3, 4, 5 \):

\[
P(X < 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)
\]

We use the binomial probability formula:

\[
P(X = k) = \binom{15}{k} \cdot (0.09)^k \cdot (0.91)^{15-k}
\]

Calculating for each value of \( k \) from 0 to 5 (using a binomial calculator or software for exact values):

\[
P(X < 6) \approx 0.239 + 0.354 + 0.261 + 0.106 + 0.031 + 0.007 = 0.998
\]

Thus, the probability that fewer than 6 people are Independent is approximately \( 0.998 \).

### Part C: Probability that more than 2 people are Independent

The probability that more than 2 people are Independent is:

\[
P(X > 2) = 1 - P(X \leq 2)
\]

We already know \( P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) \), which is:

\[
P(X \leq 2) = 0.239 + 0.354 + 0.261 = 0.854
\]

Therefore:

\[
P(X > 2) = 1 - 0.854 = 0.146
\]

Thus, the probability that more than 2 people are Independent is approximately \( 0.146 \).

---

Do you want more details or have any questions about this?  
Here are 5 related questions for deeper understanding:

1. What is the expected number of Independent voters in the sample of 15 people?
2. How would the probability change if 20 people were surveyed instead of 15?
3. What is the probability that exactly 3 people are Independent?
4. How does the binomial distribution apply to this kind of survey problem?
5. What happens to the probabilities if the percentage of Independent voters increases to 15%?

**Tip:** When solving binomial probability problems, using cumulative probability tables or software tools like a calculator can save a lot of time!

In the United States, voters who are neither Democrat nor Republican are called Independent. It is believed that 9% of voters are Independent. A survey asked 15 people to identify themselves as Democrat, Republican, or Independent.

A. What is the probability that none of the people are Independent?

B. What is the probability that fewer than 6 are Independent?

C. What is the probability that more than 2 people are Independent?

Solve a binomial probability problem where 9% of voters are Independent. Learn how to calculate the probabilities that none, fewer than 6, or more than 2 people in a 15-person survey are Independent. This problem uses the binomial probability formula and cumulative probability, suitable for Grades 9-12.

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