$21 \%$ of college students say they use credit cards because of the rewards program. You randomly select 10 college students and ask each to name the reason he or she uses credit cards. Find the probability that the number of college students who say they use credit cards because of the rewards program is (a) exactly two, (b) more than two, and (c) between two and five inclusive. If convenient, use technology to find the probabilities. (a) $P(2)=\square($ Round to the nearest thousandth as needed)

Step 1 ：This is a binomial probability problem. We have a fixed number of trials (10 students), each trial is independent (one student's response doesn't affect another's), there are only two possible outcomes (the student uses credit cards for rewards or not), and the probability of success (the student uses credit cards for rewards) is constant (21%).

Step 2 ：We can use the binomial probability formula to solve this problem: \(P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\) where: \(P(X=k)\) is the probability we want to find, \(C(n, k)\) is the combination of n items taken k at a time, p is the probability of success, n is the number of trials, and k is the number of successes we want.

Step 3 ：For part (a), we want to find the probability that exactly 2 students use credit cards for rewards, so k=2.

Step 4 ：Given that n = 10, k = 2, p = 0.21, and the combination of n items taken k at a time is 45, we can substitute these values into the binomial probability formula.

Step 5 ：The probability is calculated as \(P(2) = C(10, 2) * (0.21^2) * ((1-0.21)^(10-2)) = 0.3010702433259571\).

Step 6 ：Rounding to the nearest thousandth, the final answer is \(P(2)=\boxed{0.301}\).