\( 1 \leftarrow \) Assume that when human resource managers are randomly selected, \( 45 \% \) say job applicants should follow up within two weeks. If 9 human resource managers are randomly selected, find the probability that exactly 6 of them say job applicants should follow up within two weeks. The probability is \( \square \) (Round to four decimal places as needed.)

Given that 45% of human resource managers say job applicants should follow up within two weeks, we can calculate the probability of exactly 6 out of 9 randomly selected human resource managers saying the same. Let's denote the probability of a human resource manager saying job applicants should follow up within two weeks as \( p = 0.45 \). The probability of exactly 6 out of 9 human resource managers saying job applicants should follow up within two weeks can be calculated using the binomial probability formula: \[ P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \] where: - \( n = 9 \) (total number of human resource managers) - \( k = 6 \) (number of human resource managers saying job applicants should follow up within two weeks) - \( p = 0.45 \) (probability of a human resource manager saying job applicants should follow up within two weeks) Substitute the values into the formula to find the probability. Calculate the value by following steps: - step0: Calculate: \( { }_{9}C_{6}\times 0.45^{6}\left(1-0.45\right)^{9-6}\) - step1: Subtract the numbers: \( { }_{9}C_{6}\times 0.45^{6}\times 0.55^{9-6}\) - step2: Subtract the numbers: \( { }_{9}C_{6}\times 0.45^{6}\times 0.55^{3}\) - step3: Expand the expression: \(\frac{9!}{6!\times \left(9-6\right)!}\times 0.45^{6}\times 0.55^{3}\) - step4: Subtract the numbers: \(\frac{9!}{6!\times 3!}\times 0.45^{6}\times 0.55^{3}\) - step5: Reduce the fraction: \(\frac{9\times 8\times 7}{3!}\times 0.45^{6}\times 0.55^{3}\) - step6: Multiply the terms: \(\frac{504}{3!}\times 0.45^{6}\times 0.55^{3}\) - step7: Divide the terms: \(84\times 0.45^{6}\times 0.55^{3}\) - step8: Convert the expressions: \(84\left(\frac{9}{20}\right)^{6}\times 0.55^{3}\) - step9: Convert the expressions: \(84\left(\frac{9}{20}\right)^{6}\left(\frac{11}{20}\right)^{3}\) - step10: Simplify: \(84\times \frac{9^{6}}{20^{6}}\times \left(\frac{11}{20}\right)^{3}\) - step11: Multiply the terms: \(\frac{21\times 9^{6}}{16000000}\left(\frac{11}{20}\right)^{3}\) - step12: Evaluate the power: \(\frac{21\times 9^{6}}{16000000}\times \frac{11^{3}}{20^{3}}\) - step13: Multiply the fractions: \(\frac{21\times 9^{6}\times 11^{3}}{16000000\times 20^{3}}\) - step14: Multiply: \(\frac{27951\times 9^{6}}{128000000000}\) The probability that exactly 6 out of 9 randomly selected human resource managers say job applicants should follow up within two weeks is approximately 0.1160 (rounded to four decimal places).