Question

Assume that when human resource managers are randomly solected, 63% say job applicants should follow up within two weeks. it 6 human resource managers are randomly selected, find the probability that exactly 4 of them say job applicants should folow up within two woeks: The probability is (Round to four decimal places as needed.)

Answer 1

To find the probability that exactly 4 out of 6 randomly selected human resource managers say job applicants should follow up within two weeks, you can use the binomial probability formula. In this case:

n = 6 (number of trials)

p = 0.63 (probability of success - HR managers saying applicants should follow up within two weeks)

k = 4 (number of successful outcomes)

The formula for the probability of k successes in n trials is:

P(X = k) = (n choose k) * p^k * (1 - p)^(n - k)

Here, "n choose k" represents the binomial coefficient, which can be calculated as:

(n choose k) = n! / (k! * (n - k)!)

Let's calculate it step by step:

1. Calculate (n choose k):

(6 choose 4) = 6! / (4!(6 - 4)!) = (6! / (4! * 2!)) = (6 * 5 / (2 * 1)) = 15

2. Now, apply the binomial probability formula:

P(X = 4) = (15) * (0.63^4) * (0.37^2)

P(X = 4) ≈ 15 * 0.63^4 * 0.37^2

P(X = 4) ≈ 15 * 0.17918577 * 0.1369

P(X = 4) ≈ 0.36870965 (rounded to four decimal places)

So, the probability that exactly 4 out of 6 randomly selected human resource managers say job applicants should follow up within two weeks is approximately 0.3687 when rounded to four decimal places.