Assume that when human resource managers are randomly selected, $59 \%$ say job applicants should follow up within two weeks. If 9 human resource managers are randomly selected, find the probability that exactly 5 of them say job applicants should follow up within two weeks.
The probability is

Step 1 ：We are given a problem that involves the binomial distribution. The binomial distribution model is suitable if the following conditions are met: (1) The trials are independent. (2) The number of trials, n, is fixed. (3) Each trial outcome can be classified as a success or failure. (4) The probability of a success, p, is the same for each trial.

Step 2 ：In this case, we have n=9 trials (the 9 human resource managers), and each trial is a success if the manager says job applicants should follow up within two weeks. The probability of success is p=0.59. We want to find the probability that exactly 5 of the trials are a success, i.e., k=5.

Step 3 ：The formula for the binomial probability is: \(P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\), where \(C(n, k)\) is the number of combinations of n items taken k at a time, also known as 'n choose k'.

Step 4 ：Let's calculate this. We have n=9, k=5, and p=0.59. The number of combinations, \(C(n, k)\), is 126. The probability of k successes, \(p^k\), is approximately 0.0715. The probability of n-k failures, \((1-p)^(n-k)\), is approximately 0.0283.

Step 5 ：Multiplying these values together, we find that the probability that exactly 5 out of 9 randomly selected human resource managers say job applicants should follow up within two weeks is approximately 0.255.

Step 6 ：Final Answer: The probability is approximately \(\boxed{0.255}\).