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**Robb** A personnel director has two lists of applicants for jobs. List 1 contains the names of five women and two men, whereas list 2 contains the names of two women and 6 men. A name is randomly selected from list 1 and added to list 2. A name is then randomly selected from the augmented list 2. Given that the name selected is that of a man, what is the probabliity that a woman's name was originally selected from list 1?

This question is confusing me a little bit, but this is how I have done it;

$\displaystyle P(W|M)= \frac{P(W) \cdot P(M|W)}{P(M)}$

with $\displaystyle P(M|W) = \frac{6}{9}$ (That is, a man is selected from the second list, given a women select from first)

So the total probability of a man being selected from the second list is: $\displaystyle P(M)=P(W)\cdot P(M|W)+P(\bar{W}) \cdot P(M| \bar{W})=\frac{5}{7} \cdot \frac{6}{9} + \frac{2}{7} \cdot \frac{7}{9}=0.698413$

So

$\displaystyle P(W|M)=\frac{\frac{5}{7}\cdot \frac{6}{9}}{0.698413}=.69843$

I am a little bit confused as to if I have calculated the probablities of the events happening, the 2 lists thing is confusing me a bit.. Thanks!