49% 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.

Step 1 ：This problem is a binomial probability problem. The binomial distribution model is appropriate for a statistical experiment if the following conditions are met: The experiment consists of n repeated trials. Each trial can result in just two possible outcomes. We call one of these outcomes a success and the other, a failure. The probability of success, denoted by P, is the same on every trial. The trials are independent; that is, the outcome on one trial does not affect the outcome on other trials.

Step 2 ：In this case, we have n=10 trials (10 adults are selected), and each trial can result in just two possible outcomes (an adult either has very little confidence in newspapers or not). The probability of success is 0.49 (49% adults have very little confidence in newspapers), and the trials are independent (the adults are selected randomly).

Step 3 ：We can use the binomial probability formula to calculate the probabilities for (a), (b), and (c). The binomial probability formula is: \(P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\), where: \(P(X=k)\) is the probability of k successes in n trials, \(C(n, k)\) is the number of combinations of n items taken k at a time, p is the probability of success on any given trial, n is the number of trials, k is the number of successes.

Step 4 ：For (a), we need to find the probability that exactly 5 out of 10 adults have very little confidence in newspapers. So, we need to calculate \(P(X=5)\).

Step 5 ：For (b), we need to find the probability that at least 6 out of 10 adults have very little confidence in newspapers. This is equivalent to 1 minus the probability that 5 or fewer adults have very little confidence in newspapers. So, we need to calculate \(1 - P(X<=5)\).

Step 6 ：For (c), we need to find the probability that less than 4 out of 10 adults have very little confidence in newspapers. This is equivalent to the probability that 3 or fewer adults have very little confidence in newspapers. So, we need to calculate \(P(X<=3)\).

Step 7 ：Using the binomial probability formula, we find that the probability that the number of U.S. adults who have very little confidence in newspapers is (a) exactly five is approximately \(\boxed{0.246}\), (b) at least six is approximately \(\boxed{0.353}\), and (c) less than four is approximately \(\boxed{0.189}\).