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I have the following question:

$10$% of applications for a job possess the right skills. A company has 3 positions to fill and they interview applicants one at a time until they fill all $3$ positions. The company takes $3$ hours to interview an unqualified applicant and $5$ hours to interview a qualified applicant. Calculate the mean of the time to conduct all interviews.

I got a bit lost considering that they wanted to mean of all interviews however I just did $0.10*3+0.90*5=3.2$ which seemed too simplistic, any help in correction/verification would be appreciated.

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    $\begingroup$ Maybe a good idea to start with a (slightly) simpler problem. Suppose the firm only needs to hire one (qualified) applicant. How many hours is that expected to take? $\endgroup$ – lulu Nov 30 '20 at 15:05
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First, we can note that, no matter how many applicants are interviewed, $3$ qualified applicants will be interviewed, taking a total of $3*5=15$ hours.

Next, we need to find how many unqualified applicants are interviewed. Since each applicant can be assumed to be independently equally likely to be qualified, and you stop interviewing after a fixed number of qualified applicants, the number of unqualified applicants interviewed is given by a Negative Binomial random variable, with parameters $p=0.9$ (the probability of an unqualified candidate) and $r=3$ (the number of qualified candidates). The mean of this is given by $\dfrac{pr}{1-p}=\dfrac{0.9*3}{1-0.9}=27$.

Therefore, the mean time taken for all interviews is $3$ hours for each unqualified candidate plus $5$ hours for each qualified candidate, giving a total of $$27*3 + 3*5 = 96$$

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  • $\begingroup$ thank you, alll clear! $\endgroup$ – Sergio Nov 30 '20 at 15:08
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I think you were assuming that only one interview was done.

What happens if the company conducts interviews until 3 qualified applicants are found ?

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  • $\begingroup$ No idea how to do this since the total number isn't mentioned, maybe 0.10*3*5+0.9*3? $\endgroup$ – Sergio Nov 30 '20 at 14:56

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