\( 31 \% \) 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) \( P(2)=0.222 \) (Round to the nearest thousandth as needed.) (b) \( P(x>2)=\square \) (Round to the nearest thousandth as needed.)

To solve this problem, we can use the binomial probability formula, where the number of trials \( n \) is 10 (the number of college students), the probability of success \( p \) is 0.31 (the probability that a student uses credit cards for rewards), and the probability of failure \( q \) is \( 1 - p = 0.69 \). (a) We already have the probability that exactly 2 out of the 10 students say they use credit cards because of the rewards program: \[ P(X = 2) = \binom{n}{k} p^k q^{n-k} \] Where \( \binom{n}{k} \) is the binomial coefficient, \( k \) is the number of successes (2 in this case), and \( n \) is the total number of trials (10): \[ P(2) = \binom{10}{2} \cdot (0.31)^2 \cdot (0.69)^{10-2} \approx 0.222 \] (b) To find \( P(X > 2) \), we use the complementary probability: \[ P(X > 2) = 1 - P(X \leq 2) \] \( P(X \leq 2) \) can be calculated as \( P(X = 0) + P(X = 1) + P(X = 2) \). - For \( P(X = 0) \): \[ P(0) = \binom{10}{0} \cdot (0.31)^0 \cdot (0.69)^{10} \approx 0.014 \] - For \( P(X = 1) \): \[ P(1) = \binom{10}{1} \cdot (0.31)^1 \cdot (0.69)^{9} \approx 0.107 \] Now, we can calculate \( P(X \leq 2) \): \[ P(X \leq 2) = P(0) + P(1) + P(2) \approx 0.014 + 0.107 + 0.222 \approx 0.343 \] Finally, we find \( P(X > 2) \): \[ P(X > 2) = 1 - P(X \leq 2) = 1 - 0.343 = 0.657 \] So, rounding to the nearest thousandth as required: (b) \( P(X > 2) = 0.657 \)