Based on a​ survey, 

3131​% 

of likely voters would be willing to vote by internet instead of the​ in-person traditional method of voting. For each of the​ following, assume that 

1414 

likely voters are randomly selected. Complete parts​ (a) through​ (c) below.

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Part 1

**a.** What is the probability that exactly 

1111 

of those selected would do internet​ voting?

enter your response here

​(Round to five decimal places as​ needed.)

To solve this problem, we can use the **binomial probability formula**. The binomial distribution is appropriate because we have a fixed number of trials (14 likely voters), and each voter has only two possible outcomes: either they would vote via internet (success) or they would not (failure).

### The binomial probability formula is:

\[
P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n-k}
\]

where:
- \( n \) is the number of trials (in this case, 14 voters),
- \( k \) is the number of successes (in this case, 11 voters),
- \( p \) is the probability of success (31% or 0.31),
- \( \binom{n}{k} \) is the number of combinations of \( n \) trials taken \( k \) at a time, which is computed as:
  
  \[
  \binom{n}{k} = \frac{n!}{k!(n-k)!}
  \]

### Step-by-step solution:

1. **Given values:**
   \[
   n = 14, \quad k = 11, \quad p = 0.3131
   \]

2. **Calculate the binomial coefficient** \(\binom{n}{k}\):
   \[
   \binom{14}{11} = \frac{14!}{11!(14-11)!} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 364
   \]

3. **Calculate the probability**:
   \[
   P(X = 11) = 364 \cdot (0.3131)^{11} \cdot (1 - 0.3131)^{3}
   \]

   First, calculate the powers:
   \[
   (0.3131)^{11} \approx 5.1825 \times 10^{-6}, \quad (0.6869)^3 \approx 0.3240
   \]

   Then, multiply these values:
   \[
   P(X = 11) = 364 \cdot 5.1825 \times 10^{-6} \cdot 0.3240
   \]
   \[
   P(X = 11) \approx 364 \times 1.6792 \times 10^{-6} = 6.111 \times 10^{-4}
   \]

Thus, the probability that exactly 11 out of 14 likely voters would choose internet voting is approximately:

\[
P(X = 11) \approx 0.00061
\]

### Final answer:
\[
P(X = 11) \approx 0.00061
\]

Would you like further details or have any questions? Here are five related questions to expand on this topic:

1. What is the probability that fewer than 11 voters would choose internet voting?
2. What is the expected number of voters who would choose internet voting?
3. How does the variance of this binomial distribution help in understanding voting patterns?
4. What is the probability that more than 11 voters would choose internet voting?
5. How does changing the sample size from 14 to 20 affect the probability?

**Tip:** When using binomial probability, double-check the calculation of powers and factorials for accuracy.

Based on a survey, 31% of likely voters would be willing to vote by internet instead of the in-person traditional method of voting. For each of the following, assume that 14 likely voters are randomly selected. What is the probability that exactly 11 of those selected would do internet voting?

This solution explains how to calculate the probability that exactly 11 out of 14 likely voters would choose internet voting using the binomial probability formula. It covers key probability concepts and detailed step-by-step calculations, ideal for college-level students.

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