Question

A personnel director has two lists of applicants for jobs. List 1 contains the names of five womenandtwomen,whereaslist2containsthenamesoftwowomenandsixmen.Anameis randomlyselected from list 1 and added to list 2. A name is then randomlyselected from the augmented list 2. Given that the name selected is that of a man, what is the probabilitythat a woman’s name was originallyselected from list 1?

Answer 1

http://akk.li/pics/anne.jpg check it out

Answer 2

^Don't click that unless you want nightmares for life :o

Answer 3

List 1: 5 women, 2 men
List 2: 2 women, 6 men

Let \(A\) = a woman's name is picked from list 1
\(B\) = a man's name is picked from list 2
(These events are identified this way to make the problem easier, as this is information required to find your conditional probability).

What you want to know is \(P(A|B)\)

So...
\[P(A|B)=\frac{P(A \cap B)}{P(B)}\]
Notice that \[ P(A \cap B)=P(B|A)P(A)=\frac{6}{9}\cdot \frac{5}{7}=\frac{10}{21} \]Notice that \(P(B|A)=6/9\) since if you know a woman was picked first, then there are 3 women and 6 men in list 2, so the probability of picking a man then is just 6/9.

In the next step, notice how \(\overline{A}\)=picking a man from list 1

And \(\begin{align}P(B)&=P(B \cap A)+P(B \cap \overline{A})\\&=P(B|A)P(A)+P(B|\overline{A})P(\overline{A})\\&=\frac{10}{21}+ \frac{7}{9}\cdot \frac{2}{7}\end{align}\)

And back to the original equation now:
\[P(A|B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{10}{21}}{\frac{44}{63}}=\frac{15}{22}\]

Answer 4

What I really did is just Bayes' formula in multiple steps, using more fundamental formulas. But, if you wanted, you could have done it all with Bayes' formula in one step:\[P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|\overline{A})P(\overline{A})}\]

Answer 5

lol how did you know @kirbykirby

Answer 6

do you mean the answer, or the link?