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I'm having trouble reasoning this problem:

There's a business that has 3 different offices, "A", "B" and "C". Each one has 50, 75 and 100 employees respectively and 50%, 60% and 70% of the employees of each office are women. The probability that a employee quits his job is the same for women and men.

If you know that a employee (a women) has quit her job, what is the probability that this women has worked in the "C" office?

I know i have to use Bayes somehow, but i don't know how!

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Let $O$ be office, $S$ be gender. Then $\Pr(O=A) = 50/225$ $\Pr(O=B)=75/225$ and $\Pr(O=C)= 100/225$

$\Pr(S=W|O=A)=50\%$ $\Pr(S=W|O=B)=60\%$ and $\Pr(S=W|O=C)=75\%$

Then by Bayes $\Pr(O=C|S=W) = \frac{\Pr(O=C)\Pr(S=W|O=C)}{\sum_{x=A,B,C}\Pr(O=x)\Pr(S=W|O=x)} =\frac{ 100/225*75\%} {(100*0.75 +75*0.60 + 50*0.50)/225} $

Another way: Total # of women in business is 100∗0.75+75∗0.60+50∗0.50, among them 100*0.75 are from C. So the probability of a woman from business is from C is 100∗0.75/(100∗0.75+75∗0.60+50∗0.50), given this woman is randomly drawn from all of the women in that business.

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  • $\begingroup$ When users submit homework questions, it's preferred that you help guide them to the answers on their own -- not simply provide them with the answers. $\endgroup$ Jun 3 '17 at 20:25
  • $\begingroup$ But i don't understand why you didn't use the fact of quiting their job. Does that doesn't play a role in the solution of the problem? Why? $\endgroup$
    – S. Cow
    Jun 4 '17 at 16:41
  • $\begingroup$ I think teacher wanted to say sampling a woman from all of women randomly with equal probability. Suddenly "quit" come to his mind, and was used in question. But quitting job is not so random in a company. So I prefer "woman is randomly drawn from all of the women". $\endgroup$
    – user158565
    Jun 6 '17 at 1:31

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