Based on a​ survey, 

3838​% 

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?

In a binomial probability​ distribution, probabilities can be calculated by using​ technology, a table of binomial​ probabilities, or the binomial probability​ formula, shown below where n is the number of​ trials, x is the number of successes among n​ trials, p is the probability of success in any one​ trial, and q is the probability of failure in any one trial 

​(qequals=1minus−​p).

​P(x)equals=StartFraction n exclamation mark Over left parenthesis n minus x right parenthesis exclamation mark x exclamation mark EndFraction times p Superscript x Baseline times left parenthesis 1 minus p right parenthesis Superscript n minus xn!(n−x)!x!•px•(1−p)n−x​, 

for 

xequals=​0, 

​1, 2,​ ..., n

While either technology or the binomial probability formula can be used to find the simple binomial​ probabilities, for this​ exercise, use technology. 

To solve this binomial probability problem, let's break it down into the key elements needed for the binomial probability formula or using technology to calculate the probability:

### Given Information:
- The sample size (number of trials): \( n = 14 \)
- The number of successes (people who would vote by internet): \( x = 11 \)
- The probability of success (people who would vote by internet, given by the survey): \( p = 38\% = 0.38 \)
- The probability of failure (people who would not vote by internet): \( q = 1 - p = 1 - 0.38 = 0.62 \)

### Binomial Probability Formula:

\[
P(x = 11) = \binom{14}{11} \cdot (0.38)^{11} \cdot (0.62)^{3}
\]

Where:
\[
\binom{14}{11} = \frac{14!}{11!(14-11)!} = \frac{14!}{11!3!}
\]
is the binomial coefficient (number of ways to choose 11 successes from 14 trials).

However, since the problem asks to use **technology** for the calculation, we can use a calculator or statistical software to compute the probability.

#### Using a Calculator or Statistical Software (like Excel, R, or Python):

The binomial probability formula can be computed using functions like `BINOM.DIST` in Excel or `binom.pmf` in Python. Using technology, you would input the following:

- Number of trials (\(n = 14\))
- Number of successes (\(x = 11\))
- Probability of success (\(p = 0.38\))

For example, in Excel, you would use:
```
=BINOM.DIST(11, 14, 0.38, FALSE)
```

When computed, the result is approximately:
\[
P(x = 11) \approx 0.0043
\]

Thus, the probability that exactly 11 of the 14 voters would prefer to vote online is approximately **0.0043** or **0.43%**.

---

Would you like further details on this calculation or help with parts (b) and (c)?

### 5 Related Questions:
1. How does changing the sample size affect the probability in a binomial distribution?
2. What is the probability that fewer than 11 voters would prefer online voting?
3. How can you calculate cumulative binomial probabilities using technology?
4. How does the binomial distribution compare to the normal distribution for large \(n\)?
5. What assumptions must hold for the binomial distribution to be valid?

**Tip**: When calculating binomial probabilities, make sure to correctly identify the probability of success \(p\) and the number of trials \(n\) for accurate results.

Based on a​ survey, 38% 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. Complete parts​ (a) through​ (c) below. **a.** What is the probability that exactly 11 of those selected would do internet voting?

Calculate the binomial probability that exactly 11 out of 14 voters would choose internet voting, given a 38% chance of success. Learn how to apply the binomial distribution formula and use technology to find the solution. Suitable for high school students in Grades 10-12.

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