Question

assume that when human resource managers are randomly selected, 49% say job applicants should follow up within two weeks. if 5 human resource managers are randomly selected, find the probability that exactly 2 of them say job applicants should follow up within two weeks. the probability is (round to four decimal places as needed.)

Explanation:

Step1: Identify the binomial probability formula

The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(n\) is the number of trials, \(k\) is the number of successes, \(p\) is the probability of success in a single trial, and \(C(n,k)=\frac{n!}{k!(n - k)!}\)

Here, \(n = 5\) (5 human - resource managers selected), \(k = 2\) (exactly 2 say job applicants should follow up), \(p=0.49\) (probability that a manager says job applicants should follow up), and \(1 - p = 0.51\)

Step2: Calculate the combination \(C(n,k)\)

$$
LATEXBLOCK0
$$

Step3: Calculate \(p^{k}\) and \((1 - p)^{n - k}\)

\(p^{k}=(0.49)^{2}=0.2401\)
\((1 - p)^{n - k}=(0.51)^{3}=0.51\times0.51\times0.51 = 0.132651\)

Step4: Calculate the probability \(P(X = 2)\)

$$
LATEXBLOCK1
$$

Answer:

\(0.3185\)