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Of all the job applicants at the Apex Company who come for a job interview ,$30$% are given the interview right away. The other $70$% are asked to wait in a waiting room. About $40$% of the time, people who are asked to wait will leave the waiting room before being given the job interview, and will not return.

(i) What percent of the job applicants who come for a job interview will actually have the interview?

I am not sure about this question. Would we have to use conditional probability or Baye's Formula for this. If so, can some someone please show me?

(ii) Given that an applicant is given an interview, what is the probability that the applicant had to wait in the waiting room?

Here is my attempt:

$P$ (given an interview right away)$*$ $P$ (had to wait in waiting room) $+$ $P$ (not given an interview right away) $*$ $P$ (did not have to wait in a waiting room)

$(0.3)*(0.7)+(0.7)*(0.3)=.42$

I suppose we would have to divide this answer by the one we got in (i).

I was reading ahead about Baye's Formula and saw this question. I am not sure about part (i) and I tried on part (ii). I was thinking that we would need part (i) to use in part (ii) to get the answer. Is there a certain way I should do this? Is there a better way that I can solve this so I can see it clearer. Can someone please show me how to do this?

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  • $\begingroup$ (1) The probability is $0.3 + 0.7*0.6$ $\endgroup$
    – user130512
    Feb 25 '14 at 18:08
  • $\begingroup$ @user130512 Can we show this by writing out a formula similar to what I did in part (ii). $\endgroup$
    – 9599
    Feb 25 '14 at 18:19
  • $\begingroup$ P(given an interview right away) + P(waiting room)*P(stays for waiting room given waiting room) $\endgroup$
    – user130512
    Feb 25 '14 at 18:21
  • $\begingroup$ i think that answer for part i is 72% $\endgroup$
    – User8976
    Feb 25 '14 at 18:22
  • $\begingroup$ and for the second part it is 7/12 $\endgroup$
    – User8976
    Feb 25 '14 at 18:29
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Part i)

$30$% get an interview straight away, so $0.3$

$70$% have to wait, of which only $60$% stay, $0.7 * 0.6 = 0.42$ Adding these together gives $0.72$

ie. $72$% of applicants have an interview.

Part ii)

Use Bayes $P(A|B)= \frac{P(B|A)P(A)}{P(B)}$ where | denotes 'given'. So;

$P(Had.to.wait|Had.interview)=\frac{P(Had.interview|Had.to.wait)P(Had.to.wait)}{P(Had.interview)}=\frac{0.6*0.7}{0.72}=\frac{0.42}{0.72}=\frac{7}{12}=0.58333$

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