$23 \%$ 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) $P(2)=0.294$ (Round to the nearest thousandth as needed.)
(b) $P(x> 2)=\square($ Round to the nearest thousandth as needed.)

Step 1 ：This is a binomial probability problem. We are given the probability of success (p), the number of trials (n), and we are asked to find the probability of a certain number of successes (k).

Step 2 ：For part (b), we need to find the probability that the number of successes is more than 2. This is equivalent to finding the probability of 3, 4, ..., up to 10 successes and summing them up.

Step 3 ：We can use the formula for binomial probability to calculate this: \(P(k) = C(n, k) * p^k * (1-p)^(n-k)\) where C(n, k) is the binomial coefficient, which gives the number of ways to choose k successes out of n trials.

Step 4 ：Given that p = 0.23 and n = 10, we calculate the probability of having more than 2 successes.

Step 5 ：The calculated probability is approximately 0.4137172738381184.

Step 6 ：Rounding to the nearest thousandth, the final answer is \(\boxed{0.414}\).