Assume that when human resource managers are randomly selected, $46 \%$ say job applicants should follow up within two weeks. If 10 human resource managers are randomly selected, find the probability that fewer than 3 of them say job applicants should follow up within two weeks. The probability is (Round to four decimal places as needed.)

Step 1 ：We are given a problem where 46% of human resource managers say job applicants should follow up within two weeks. We are asked to find the probability that fewer than 3 out of 10 human resource managers say this.

Step 2 ：This is a binomial probability problem. The probability of success (a human resource manager saying job applicants should follow up within two weeks) is 0.46. We need to find the probability of 0, 1, or 2 successes out of 10 trials.

Step 3 ：The formula for binomial probability is: \(P(X=k) = C(n, k) * (p^k) * ((1-p)^(n-k))\), where \(P(X=k)\) is the probability of \(k\) successes in \(n\) trials, \(C(n, k)\) is the combination of \(n\) items taken \(k\) at a time, \(p\) is the probability of success, and \(n\) is the number of trials.

Step 4 ：We can calculate this by summing the probabilities for 0, 1, and 2 successes.

Step 5 ：The calculated probability is approximately 0.08891405997641595.

Step 6 ：Final Answer: The probability that fewer than 3 out of 10 human resource managers say job applicants should follow up within two weeks is approximately \(\boxed{0.0889}\).