Questions: 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 six and (b) exactly two.

Solution Steps

We are tasked with finding the probabilities of U.S. adults having very little confidence in newspapers, given that \( p = 0.39 \) (the probability of success) and \( n = 9 \) (the number of trials). We need to calculate the probabilities for two specific cases: (a) exactly 6 adults and (b) exactly 2 adults.

To find the probability that exactly 6 out of 9 adults have very little confidence in newspapers, we use the binomial probability formula:

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

where:

• \( n = 9 \)
• \( x = 6 \)
• \( p = 0.39 \)
• \( q = 1 - p = 0.61 \)

Substituting the values, we calculate:

\[ P(X = 6) = \binom{9}{6} \cdot (0.39)^6 \cdot (0.61)^{3} = 0.067 \]

Thus, the probability that the number who have very little confidence in newspapers is exactly six is \( 0.067 \).

Next, we calculate the probability that exactly 2 out of 9 adults have very little confidence in newspapers using the same binomial probability formula:

\[ P(X = 2) = \binom{9}{2} \cdot (0.39)^2 \cdot (0.61)^{7} = 0.172 \]

Thus, the probability that the number who have very little confidence in newspapers is exactly two is \( 0.172 \).

Final Answer