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

This problem involves binomial probability because we are dealing with a fixed number of trials (8 U.S. adults) and two possible outcomes for each trial (either an adult has very little confidence in newspapers or they do not). 

The binomial probability formula is given by:

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

Where:
- \( P(X = k) \) is the probability of getting exactly \( k \) successes (in this case, adults with little confidence),
- \( n \) is the number of trials (8 adults),
- \( p \) is the probability of success (0.39),
- \( k \) is the number of successes we're interested in (6 for part (a) and 1 for part (b)),
- \( \binom{n}{k} \) is the binomial coefficient, calculated as \( \binom{n}{k} = \frac{n!}{k!(n - k)!} \).

### Part (a): Finding \( P(X = 6) \)

Substitute into the binomial formula:

\[
P(X = 6) = \binom{8}{6} (0.39)^6 (1 - 0.39)^{8 - 6}
\]

Let's break this down:
- \( \binom{8}{6} = \frac{8!}{6!2!} = 28 \),
- \( p = 0.39 \),
- \( 1 - p = 0.61 \).

So the probability becomes:

\[
P(X = 6) = 28 \cdot (0.39)^6 \cdot (0.61)^2
\]

I'll calculate this next.The probability that exactly six out of eight U.S. adults have very little confidence in newspapers is approximately **0.0367** (or 3.67%).

### Part (b): Finding \( P(X = 1) \)

Now, we use the same binomial probability formula to find the probability that exactly one adult out of eight has very little confidence in newspapers:

\[
P(X = 1) = \binom{8}{1} (0.39)^1 (1 - 0.39)^{8 - 1}
\]

I'll calculate this next.The probability that exactly one out of eight U.S. adults has very little confidence in newspapers is approximately **0.0981** (or 9.81%).

### Summary:
- **(a)** The probability that exactly six adults have very little confidence is **0.0367**.
- **(b)** The probability that exactly one adult has very little confidence is **0.0981**.

Would you like further details on these calculations or have any questions?

### Related questions:
1. What is the probability that exactly three adults have very little confidence in newspapers?
2. What is the probability that at least five adults have very little confidence in newspapers?
3. How would the probability change if you selected ten adults instead of eight?
4. What is the probability that none of the adults have very little confidence in newspapers?
5. What is the expected number of adults who have very little confidence, based on this selection?

### Tip:
When working with binomial distributions, always check if the conditions for the distribution apply: fixed number of trials, two possible outcomes, and constant probability of success across trials.

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

Find the probability that six or one out of eight U.S. adults have very little confidence in newspapers, using binomial distribution. This problem involves key probability concepts, including the binomial probability formula and binomial coefficient calculation. Suitable for students in Grades 10-12.

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