Answer to Question #265464 in Statistics and Probability for Susan

Question

A company advertised for a position of managing director the number of applications for position were 50 ,out which 10 were shortlisted for interview,. Determine the probability that an applicant selected at random will be :

A, shortlisted for interviews

B, shortlisted for interviews and employed as the managing director

C , employed as managing director, given that he has been shortlisted for interviews

D, shortlisted and not employed as managing director

E, not shortlisted for interviews.

Answer 1

Let $S$ be the event that an application is shortlisted for interview and $M$ be the event that an applicant is employed as the managing director.

Number of applications is 50 and number of shortlisted applications is 10. This shows that the probability of an application being shortlisted is $p(S)=10/50=1/5$.

The conditional probability that given that an applicant is shortlisted, the probability that they are employed as the managing director is given as,

$p(M/S)=1/10$ since there are 10 applications and only 1 position.

a.

As stated above, the probability that an applicant is shortlisted is given as,

$p(S)=10/50=1/5$

b.

The probability that an applicant is shortlisted for interviews and employed as the managing director is defined by $p(M\cap S)$ and it is given by,

$p(M\cap S)=P(S)*p(M|S)=1/5*1/10=1/50$ ,

Therefore, the probability that an applicant is shortlisted for interviews and employed as the managing director is $1/50$

c.

 The conditional probability that an applicant is employed as managing director, given that he has been shortlisted for interviews is written as $p(M|S)$.

By definition of conditional probability, $p(M|S)=p(M\cap S)/p(S)=(1/50)/(1/5)=1/10.$

Thus, given that an applicant is shortlisted the probability that the applicant is employed as the managing director is $1/10$

d.

The probability that an applicant is  shortlisted and not employed as managing director is given as $p(S\cap M' )=P(M'|S)*p(S)$

Now, $p(M'|S)=1-p(M|S)=1-1/10=9/10$

Therefore,

$p(S\cap M')=9/10*1/5=9/50$

Hence $9/50$ is the probability that an applicant is  shortlisted and not employed as managing director.

e.

Probability that an applicant is  not shortlisted for interviews is given as $p(S')$. Following the rules of probability, $p(S')=1-p(S)=1-1/5=4/5$ .

Therefore, the probability that an applicant is not shortlisted for interviews is 4/5.