Question 5, 4.2.19
HW Score: $50.97 \%, 6.12$ of 12 points
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$56 \%$ of U.S. adults have very little confidence in newspapers. You randomly select 10 U.S. adults. Find the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five, (b) at least six, and (c) less than four.
(a) $P(5)=0.229$ (Round to three decimal places as needed.)
(b) $P(x \geq 6)=\square$ (Round to three decimal places as needed)

Step 1 ：Given that 56% of U.S. adults have very little confidence in newspapers, we are asked to find the probability that the number of U.S. adults who have very little confidence in newspapers is at least six when randomly selecting 10 U.S. adults.

Step 2 ：We can solve this problem using the binomial probability formula, which is used when there are exactly two mutually exclusive outcomes of a trial. In this case, the two outcomes are 'having very little confidence in newspapers' and 'not having very little confidence in newspapers'.

Step 3 ：The binomial probability formula is given by: \(P(x) = C(n, x) \cdot p^x \cdot (1-p)^{n-x}\), where \(P(x)\) is the probability of \(x\) successes in \(n\) trials, \(C(n, x)\) is the number of combinations of \(n\) items taken \(x\) at a time, \(p\) is the probability of success on a single trial, and \(1-p\) is the probability of failure on a single trial.

Step 4 ：In this case, \(p = 0.56\), \(n = 10\), and we want to find \(P(x \geq 6)\), which is the probability that the number of U.S. adults who have very little confidence in newspapers is at least six.

Step 5 ：To find \(P(x \geq 6)\), we need to calculate the binomial probability for each case from 6 to 10 and then sum them up.

Step 6 ：Using the binomial probability formula, we find that \(P(x \geq 6) = 0.530\).

Step 7 ：\(\boxed{0.530}\) is the probability that the number of U.S. adults who have very little confidence in newspapers is at least six when randomly selecting 10 U.S. adults.