\( 31 \% \) 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 indusive. 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, which is given by: \[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k} \] where: - \(n\) = total number of trials (in this case, students surveyed) = 10 - \(k\) = number of successful outcomes (students that use credit cards for rewards) - \(p\) = probability of success on a single trial (in this case, the proportion of students who use credit cards for rewards) = 0.31 - \(\binom{n}{k}\) = combination of \(n\) taken \(k\) at a time ### (a) Probability that exactly 2 students use credit cards for rewards Using \(n = 10\), \(k = 2\), and \(p = 0.31\): \[ P(X = 2) = \binom{10}{2} (0.31)^2 (1 - 0.31)^{10 - 2} \] Calculating the binomial coefficient: \(\binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45\) Now substituting values into the formula: \[ P(X = 2) = 45 \times (0.31)^2 \times (0.69)^8 \] Calculating \((0.31)^2 \approx 0.0961\) and \((0.69)^8 \approx 0.0453\): \[ P(X = 2) \approx 45 \times 0.0961 \times 0.0453 \approx 0.1959 \] Thus, rounding to the nearest thousandth yields: \[ P(2) \approx \boxed{0.196} \] ### (b) Probability that more than 2 students use credit cards for rewards To find \(P(X > 2)\): \[ P(X > 2) = 1 - P(X \leq 2) = 1 - (P(0) + P(1) + P(2)) \] Calculating \(P(0)\) and \(P(1)\): **For \(P(0)\):** \[ P(X = 0) = \binom{10}{0} (0.31)^0 (0.69)^{10} = 1 \times 1 \times (0.69)^{10} \] \((0.69)^{10} \approx 0.0233\) So, \(P(0) \approx 0.023\). **For \(P(1)\):** \[ P(X = 1) = \binom{10}{1} (0.31)^1 (0.69)^{9} = 10 \times 0.31 \times (0.69)^{9} \] \((0.69)^9 \approx 0.0337\): \[ P(1) \approx 10 \times 0.31 \times 0.0337 \approx 0.1047 \] Now summarizing: \[ P(X \leq 2) = P(0) + P(1) + P(2) \approx 0.023 + 0.105 + 0.196 \approx 0.324 \] Thus, \[ P(X > 2) \approx 1 - 0.324 = 0.676 \] ### (c) Probability that between 2 and 5 students use credit cards for rewards To find \(P(2 \leq X \leq 5)\): \[ P(2 \leq X \leq 5) = P(2) + P(3) + P(4) + P(5) \] Using similar calculations as above: **For \(P(3)\):** \[ P(X = 3) = \binom{10}{3} (0.31)^3 (0.69)^{7} \approx 120 \cdot 0.0299 \cdot 0.0726 \approx 0.2616 \] **For \(P(4)\):** \[ P(X = 4) = \binom{10}{4} (0.31)^4 (0.69)^{6} \approx 210 \cdot 0.0093 \cdot 0.1060 \approx 0.1992 \] **For \(P(5)\