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

This is a binomial probability problem because we're dealing with a fixed number of independent trials (selecting 9 adults), each with two possible outcomes: "has very little confidence in newspapers" or "does not." The binomial probability formula is:

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

where:
- \( n \) is the number of trials (9 adults),
- \( k \) is the number of successes (e.g., the number of adults with very little confidence in newspapers),
- \( p \) is the probability of success on each trial (39% or 0.39),
- \( \binom{n}{k} \) is the binomial coefficient, which calculates the number of ways to choose \( k \) successes out of \( n \) trials.

Let's calculate the probabilities for both parts of the question.

### Part (a): Probability that exactly 4 adults have very little confidence
We need to find \( P(X = 4) \), where \( n = 9 \), \( k = 4 \), and \( p = 0.39 \):

\[
P(X = 4) = \binom{9}{4} (0.39)^4 (1-0.39)^{9-4}
\]

### Part (b): Probability that exactly 1 adult has very little confidence
Now, we find \( P(X = 1) \), where \( n = 9 \), \( k = 1 \), and \( p = 0.39 \):

\[
P(X = 1) = \binom{9}{1} (0.39)^1 (1-0.39)^{9-1}
\]

I'll calculate both values for you.The probabilities are as follows:

- (a) The probability that exactly 4 adults have very little confidence in newspapers is approximately **0.246** (or 24.6%).
- (b) The probability that exactly 1 adult has very little confidence in newspapers is approximately **0.067** (or 6.7%).

Would you like further details on how these probabilities are computed or have any other questions?

Here are 5 related questions to deepen your understanding:
1. What is the probability that exactly 0 adults have very little confidence in newspapers?
2. How would the probability change if we increased the number of adults sampled?
3. What is the cumulative probability for 4 or fewer adults having very little confidence in newspapers?
4. Can you find the probability that at least 2 adults have very little confidence in newspapers?
5. How does changing the confidence level (e.g., to 50%) affect the binomial probability?

**Tip:** The sum of all possible probabilities for a binomial distribution equals 1, as it accounts for every possible outcome.

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

Find the probability that exactly 4 out of 9 U.S. adults have very little confidence in newspapers using the binomial probability formula. This detailed explanation covers core concepts like the binomial distribution, probability calculation, and key formulas, making it suitable for Grades 9-12.

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