\( 26 \% \) 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) \( \mathrm{P}(2)=\square \) (Round to the nearest thousandth as needed.)

To solve this problem, we can use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] where: - \( n \) = number of trials (students, in this case) = 10 - \( k \) = number of successes (students who say they use credit cards for rewards) - \( p \) = probability of success = 0.26 - \( \binom{n}{k} \) = "n choose k" = the number of combinations of n items taken k at a time. Given: - \( n = 10 \) - \( p = 0.26 \) - \( 1 - p = 0.74 \) (a) Calculate \( P(2) \): \[ P(2) = \binom{10}{2} (0.26)^2 (0.74)^{10 - 2} \] First, calculate \( \binom{10}{2} \): \[ \binom{10}{2} = \frac{10!}{2!(10 - 2)!} = \frac{10 \times 9}{2 \times 1} = 45 \] Now plug the values into the equation: \[ P(2) = 45 \times (0.26)^2 \times (0.74)^8 \] \[ = 45 \times 0.0676 \times 0.1002 \approx 0.30397 \] So, rounding to the nearest thousandth: \[ \mathrm{P}(2) \approx 0.304 \] (b) To find \( P(X > 2) \), we can calculate it as: \[ P(X > 2) = 1 - P(X \leq 2) = 1 - (P(0) + P(1) + P(2)) \] Calculating \( P(0) \) and \( P(1) \): 1. **Calculate \( P(0) \)**: \[ P(0) = \binom{10}{0} (0.26)^0 (0.74)^{10} = 1 \times 1 \times (0.74)^{10} \approx 0.0199 \] 2. **Calculate \( P(1) \)**: \[ P(1) = \binom{10}{1} (0.26)^1 (0.74)^{9} = 10 \times 0.26 \times (0.74)^{9} \approx 10 \times 0.26 \times 0.0264 \approx 0.0682 \] Now summing these probabilities: \[ P(X \leq 2) = P(0) + P(1) + P(2) \approx 0.0199 + 0.0682 + 0.304 = 0.3921 \] Thus, \[ P(X > 2) = 1 - 0.3921 \approx 0.6079 \] (c) For \( P(2 \leq X \leq 5) \): Calculate \( P(3), P(4), P(5) \) similarly and add these probabilities. 1. **Calculate \( P(3) \)**: \[ P(3) = \binom{10}{3} (0.26)^3 (0.74)^{7} \approx 120 \times 0.017576 \times 0.1606 \approx 0.3376 \] 2. **Calculate \( P(4) \)**: \[ P(4) = \binom{10}{4} (0.26)^4 (0.74)^{6} \approx 210 \times 0.004569 \times 0.2170 \approx 0.2094 \] 3. **Calculate \( P(5) \)**: \[ P(5) = \binom{10}{5} (0.26)^5 (0.74)^{5} \approx 252 \times 0.001186 \times 0.3370 \approx 0.0962 \] Now sum these probabilities: \[ P(2 \leq X \leq 5) \approx P(2) + P(3) + P(4) + P(5) = 0.304 + 0.3376 + 0.2094 + 0.0962 \approx 0