Thirty​-nine percent of U.S. adults have very little confidence in newspapers. You randomly select eleven U.S. adults. Find the probability that the number who have very little confidence in newspapers is​ (a) exactly seven and​ (b) exactly three.
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Part 1
​(a) The probability that the number who have very little confidence in newspapers is exactly seven is

enter your response here.
​(Round to three decimal places as​ needed.) 

This problem involves a **binomial distribution** because we are dealing with a fixed number of trials (11 adults), and each trial (adult) results in either "success" (having very little confidence) or "failure" (not having very little confidence). 

The formula for the **binomial probability** is:  
\[
P(X = k) = \binom{n}{k} \cdot p^k \cdot (1 - p)^{n - k}
\]
Where:
- \(n\) = number of trials = 11  
- \(k\) = number of successes (the specific number we are interested in)  
- \(p\) = probability of success (39% = 0.39)  
- \(\binom{n}{k} = \frac{n!}{k! \cdot (n - k)!}\) is the binomial coefficient.  

---

### **(a) Exactly 7 adults have very little confidence**

\[
P(X = 7) = \binom{11}{7} \cdot (0.39)^7 \cdot (1 - 0.39)^4
\]
First, calculate the binomial coefficient:
\[
\binom{11}{7} = \frac{11!}{7! \cdot (11 - 7)!} = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330
\]

Now calculate the powers of probabilities:
- \(0.39^7 \approx 0.002764\)  
- \(0.61^4 \approx 0.138458\)  

Now multiply everything together:
\[
P(X = 7) = 330 \cdot 0.002764 \cdot 0.138458 \approx 0.126
\]

So, the probability that exactly 7 adults have very little confidence is:
\[
\boxed{0.126}
\]

---

### **(b) Exactly 3 adults have very little confidence**

\[
P(X = 3) = \binom{11}{3} \cdot (0.39)^3 \cdot (1 - 0.39)^8
\]
First, calculate the binomial coefficient:
\[
\binom{11}{3} = \frac{11!}{3! \cdot (11 - 3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165
\]

Now calculate the powers of probabilities:
- \(0.39^3 \approx 0.059319\)  
- \(0.61^8 \approx 0.052922\)  

Now multiply everything together:
\[
P(X = 3) = 165 \cdot 0.059319 \cdot 0.052922 \approx 0.517
\]

So, the probability that exactly 3 adults have very little confidence is:
\[
\boxed{0.517}
\]

---

Let me know if you need more steps or details!  

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**Follow-up Questions:**  
1. What is the mean and variance of the binomial distribution in this problem?  
2. How would the probabilities change if the sample size increased to 20 adults?  
3. What is the probability that fewer than 5 adults have very little confidence?  
4. Can you use the normal approximation to the binomial distribution here? Why or why not?  
5. What is the cumulative probability of having between 2 and 6 adults with little confidence?

**Tip:** When working with binomial probabilities, using a calculator or software helps avoid rounding errors.

Thirty-nine percent of U.S. adults have very little confidence in newspapers. You randomly select eleven U.S. adults. Find the probability that the number who have very little confidence in newspapers is (a) exactly seven and (b) exactly three.

Find the probability that exactly 7 or 3 out of 11 U.S. adults have very little confidence in newspapers using the binomial distribution. This problem covers probability concepts, including the binomial probability formula and binomial coefficients. Suitable for students in Grades 10-12.

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