$23 \%$ of U.S. adults say they are more likely to make purchases during a sales tax holiday. You randomly select 10 adults. Find the probability that the number of adults who say they are more likely to make purchases during a sales tax holiday is (a) exactly two, (b) more than two, and (c) between two and five, inclusive. (a) $P(2)=\square(R o u n d$ to the nearest thousandth as needed.)

Step 1 ：This is a binomial probability problem. We have a fixed number of trials (10 adults), each trial is independent (one adult's preference doesn't affect another's), there are two outcomes (either an adult is more likely to make purchases during a sales tax holiday or they're not), and the probability of success (an adult is more likely to make purchases during a sales tax holiday) is constant (23%).

Step 2 ：We can use the binomial probability formula to solve this problem: \(P(x) = C(n, x) * (p^x) * ((1-p)^(n-x))\) where: \(P(x)\) is the probability we're trying to find, \(C(n, x)\) is the number of combinations of \(n\) items taken \(x\) at a time, \(p\) is the probability of success, \(n\) is the number of trials, and \(x\) is the number of successes we're interested in.

Step 3 ：For part (a), we're trying to find the probability that exactly 2 adults are more likely to make purchases during a sales tax holiday. So, \(n=10\), \(x=2\), and \(p=0.23\).

Step 4 ：Using the binomial probability formula, we find that the number of combinations is 45 and the probability is approximately 0.294.

Step 5 ：Final Answer: \(P(2)=\boxed{0.294}\)