Assume that when human resource managers are randomly selected, 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.)

Question

Assume that when human resource managers are randomly selected,

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.)

UpStudy AI · Accepted Answer

Given that $$ 42\% $$ of human resource managers say job applicants should follow up within two weeks, we can calculate the probability that fewer than 3 of 10 randomly selected human resource managers say the same.

Let's denote the probability that a human resource manager says job applicants should follow up within two weeks as $$ p = 0.42 $$.

The probability that fewer than 3 of 10 human resource managers say job applicants should follow up within two weeks can be calculated using the binomial distribution formula:

$$ P(X < 3) = \sum_{k=0}^{2} \binom{10}{k} p^k (1-p)^{10-k} $$

where:
- $$ \binom{10}{k} $$ is the binomial coefficient, which represents the number of ways to choose $$ k $$ items from a set of 10 items.
- $$ p^k $$ is the probability that $$ k $$ human resource managers say job applicants should follow up within two weeks.
- $$ (1-p)^{10-k} $$ is the probability that the remaining $$ 10-k $$ human resource managers do not say job applicants should follow up within two weeks.

Let's calculate this probability.
Find the sum by following steps:
- step0: Solution:
  $$\sum _{k=0}^{2} { }_{10}C_{k}\times 0.42^{k}\left(1-0.42\right)^{10-k}$$
- step1: Subtract the numbers:
  $$\sum _{k=0}^{2} { }_{10}C_{k}\times 0.42^{k}\times 0.58^{10-k}$$
- step2: Expand the expression:
  $$\sum _{k=0}^{2}\frac{10!}{k!\times \left(10-k\right)!}\times 0.42^{k}\times 0.58^{10-k}$$
- step3: Calculate the value:
  $$\sum _{k=0}^{2}\frac{3628800}{k!\times \left(10-k\right)!}\times 0.42^{k}\times 0.58^{10-k}$$
- step4: Convert the expressions:
  $$\sum _{k=0}^{2}\frac{3628800}{k!\times \left(10-k\right)!}\times \left(\frac{21}{50}\right)^{k}\times 0.58^{10-k}$$
- step5: Convert the expressions:
  $$\sum _{k=0}^{2}\frac{3628800}{k!\times \left(10-k\right)!}\times \left(\frac{21}{50}\right)^{k}\left(\frac{29}{50}\right)^{10-k}$$
- step6: Multiply the terms:
  $$\sum _{k=0}^{2}\frac{3628800\left(\frac{21}{50}\right)^{k}\left(\frac{29}{50}\right)^{10-k}}{k!\times \left(10-k\right)!}$$
- step7: Rewrite the expression:
  $$\sum _{k=0}^{2}\frac{3628800\left(\frac{21}{50}\right)^{k}\left(\frac{29}{50}\right)^{10}\left(\frac{29}{50}\right)^{-k}}{k!\times \left(10-k\right)!}$$
- step8: Rewrite the expression:
  $$\sum _{k=0}^{2}567\times 29^{10}\times 3906250^{-2}\left(k!\right)^{-1}\left(\left(10-k\right)!\right)^{-1}\left(\left(\frac{29}{50}\right)^{k}\right)^{-1}\left(\frac{21}{50}\right)^{k}$$
- step9: Convert the expressions:
  $$567\times 29^{10}\times 3906250^{-2}\left(0!\right)^{-1}\left(\left(10-0\right)!\right)^{-1}\left(\left(\frac{29}{50}\right)^{0}\right)^{-1}\left(\frac{21}{50}\right)^{0}+567\times 29^{10}\times 3906250^{-2}\left(1!\right)^{-1}\left(\left(10-1\right)!\right)^{-1}\left(\left(\frac{29}{50}\right)^{1}\right)^{-1}\left(\frac{21}{50}\right)^{1}+567\times 29^{10}\times 3906250^{-2}\left(2!\right)^{-1}\left(\left(10-2\right)!\right)^{-1}\left(\left(\frac{29}{50}\right)^{2}\right)^{-1}\left(\frac{21}{50}\right)^{2}$$
- step10: Calculate:
  $$\frac{29^{10}}{3906250^{2}\times 6400}+\frac{21\times 29^{9}}{3906250^{2}\times 640}+\frac{3969\times 29^{8}}{1280\times 3906250^{2}}$$
- step11: Calculate:
  $$\frac{3347\times 29^{8}}{800\times 3906250^{2}}$$
The probability that fewer than 3 of 10 human resource managers say job applicants should follow up within two weeks is approximately 0.1372 (rounded to four decimal places).
The probability is approximately 0.1372.

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