Question

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 4 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 on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Determine the values of $n$, $k$, and $p$

Here, $n = 9$ (the number of human - resource managers selected), $k = 4$ (the number of managers who say job applicants should follow up within two weeks), and $p=0.45$ (the probability that a single manager says job applicants should follow up within two weeks), and $1 - p = 0.55$.

Step3: Calculate the combination $C(n,k)$

$C(9,4)=\frac{9!}{4!(9 - 4)!}=\frac{9!}{4!5!}=\frac{9\times8\times7\times6\times5!}{4\times3\times2\times1\times5!}=126$.

Step4: Calculate the binomial probability

$P(X = 4)=C(9,4)\times(0.45)^{4}\times(0.55)^{9 - 4}=126\times(0.45)^{4}\times(0.55)^{5}$.
$(0.45)^{4}=0.45\times0.45\times0.45\times0.45 = 0.04100625$.
$(0.55)^{5}=0.55\times0.55\times0.55\times0.55\times0.55=0.0503284375$.
$P(X = 4)=126\times0.04100625\times0.0503284375\approx0.2600$.

Answer:

$0.2600$