Can social media mistakes hurt your chances of finding a job? According to a survey of 1,000 hiring managers across many different industries, 74% claim that they use social media sites to research prospective candidates for any job. Calculate the probabilities of the following events.

(a) Out of 30 job listings, at least 21 will conduct social media screening.

(b) Out of 30 job listings, fewer than 19 will conduct social media screening.

(c) Out of 30 job listings, exactly between 21 and 25 (including 21 and 25) will conduct social media screening.

Answer :

Given :

Probability of success $(p)=0.74$

Sample size $(n)=30$

Solution :

A) The Probability of At Least 21 Job Listings Conducting Social Media Screening :

$$\text{P}(x\ge 21)=\sum _{21}^{30}(\genfrac{}{}{0ex}{}{n}{x})\cdot p^x\cdot (1-p{)}^{n-x}$$ $$=\sum _{21}^{30}(\genfrac{}{}{0ex}{}{30}{x})\cdot {0.74}^x\cdot (1–0.74{)}^{30–x}$$ $$=0.765$$

B) The Probability of Fewer Than 19 Job Listings Conducting Social Media Screening :

$$\text{P}(x\le 19)=\sum _0^{18}(\genfrac{}{}{0ex}{}{n}{x})\cdot p^x\cdot (1-p{)}^{n-x}$$ $$=\sum _0^{18}(\genfrac{}{}{0ex}{}{30}{x})\cdot {0.74}^x\cdot (1–0.74{)}^{30–x}$$ $$=0.066$$

C) The Probability of Exactly Between 21 and 25 Job Listings Conducting Social Media Screening :

$$\text{P}(21\le x\le 25)=\sum _{21}^{25}(\genfrac{}{}{0ex}{}{n}{x})\cdot p^x\cdot (1-p{)}^{n-x}$$ $$=\sum _{21}^{25}(\genfrac{}{}{0ex}{}{30}{x})\cdot {0.74}^x\cdot (1–0.74{)}^{30–x}$$ $$=0.687$$