Question

assume that when human - resource managers are randomly selected, 52% say job applicants should follow up within two weeks. if 15 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 (square) (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 $C(n,k)=\frac{n!}{k!(n - k)!}$. We want to find $P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)$. Here, $n = 15$, $p=0.52$, and $1 - p = 0.48$.

Step2: Calculate $P(X = 0)$

$C(15,0)=\frac{15!}{0!(15 - 0)!}=1$. Then $P(X = 0)=C(15,0)\times(0.52)^{0}\times(0.48)^{15}=1\times1\times(0.48)^{15}\approx0.000005$.

Step3: Calculate $P(X = 1)$

$C(15,1)=\frac{15!}{1!(15 - 1)!}=\frac{15!}{1!14!}=15$. Then $P(X = 1)=C(15,1)\times(0.52)^{1}\times(0.48)^{14}=15\times0.52\times(0.48)^{14}\approx0.00009$.

Step4: Calculate $P(X = 2)$

$C(15,2)=\frac{15!}{2!(15 - 2)!}=\frac{15\times14}{2\times1}=105$. Then $P(X = 2)=C(15,2)\times(0.52)^{2}\times(0.48)^{13}=105\times0.2704\times(0.48)^{13}\approx0.0007$.

Step5: Calculate $P(X\lt3)$

$P(X\lt3)=P(X = 0)+P(X = 1)+P(X = 2)\approx0.000005 + 0.00009+0.0007=0.000795\approx0.0008$.

Answer:

$0.0008$