# Probability questions

• Sep 16th 2009, 04:55 PM
Danneedshelp
Probability questions
Q: A personal 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 six 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 probability that a woman's name was originally selected from list 1?

We just covered total probability and Baye's theorem. Even so, I am completely lost. I can't find any examples similar to this question. Some guidence would be appreciated.

Thank you.
• Sep 17th 2009, 03:42 AM
The Second Solution
Quote:

Originally Posted by Danneedshelp
Q: A personal 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 six 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 probability that a woman's name was originally selected from list 1?

We just covered total probability and Baye's theorem. Even so, I am completely lost. I can't find any examples similar to this question. Some guidence would be appreciated.

Thank you.

After the choice there are two possible scenarios for list 2:

List A: 3 W, 6M
List B: 2W, 7M

Pr(List A) = 1/2 and Pr(List B) = 1/2.

Pr(M | List A) = 6/9.

Pr(M | List B) = 7/9.

Pr(M) = Pr(M | List A) x Pr(List A) + Pr(M | List B) x Pr(List B) = 13/18.

Pr(List A | M) = Pr(List A and M)/Pr(M) = (6/18)/(13/18) = 6/13.
• Sep 18th 2009, 03:42 AM
kin
Quote:

Originally Posted by Danneedshelp
Q: A personal 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 six 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 probability that a woman's name was originally selected from list 1?

We just covered total probability and Baye's theorem. Even so, I am completely lost. I can't find any examples similar to this question. Some guidence would be appreciated.

Thank you.

let (W) be the event of selecting a woman from list 1 to list 2 ;
so selecting a man from list1 to list2 is (W');
let (AM) be the event of selecting a man from the augmented list2.

thus, the required probability
=P(W | AM)
=P( W and AM)/P(AM)

where
P( W and AM)= P(AM|W) P(W)= (6/9)(5/7);

P(AM) =P(AM and W) + P(AM and W')
=P(AM | W)P(W) +P(AM | W') P(W') -->total probability theorem
=(6/9)(5/7) +(7/9)(2/7)

so =P(W | AM)= (6/9)(5/7) / (6/9)(5/7) +(7/9)(2/7) =15/22
• Sep 18th 2009, 03:54 AM
kin
Quote:

Originally Posted by The Second Solution
After the choice there are two possible scenarios for list 2:

List A: 3 W, 6M
List B: 2W, 7M

Pr(List A) = 1/2 and Pr(List B) = 1/2.

Pr(M | List A) = 6/9.

Pr(M | List B) = 7/9.

Pr(M) = Pr(M | List A) x Pr(List A) + Pr(M | List B) x Pr(List B) = 13/18.

Pr(List A | M) = Pr(List A and M)/Pr(M) = (6/18)/(13/18) = 6/13.

Pr(List A) not = Pr(List B) not = 1/2.
• Sep 18th 2009, 06:30 AM
mr fantastic
Quote:

Originally Posted by The Second Solution
After the choice there are two possible scenarios for list 2:

List A: 3 W, 6M
List B: 2W, 7M

Pr(List A) = 1/2 and Pr(List B) = 1/2.
[snip]

Correction: Pr(List A) = 5/7 and Pr(List B) = 2/7.

Then:

Pr(M) = Pr(M | List A) x Pr(List A) + Pr(M | List B) x Pr(List B) = 44/63.

Pr(List A and M) = Pr(M | List A) x Pr(List A) = 30/63.

Pr(List A | M) = Pr(List A and M)/Pr(M) = (30/63)/(44/63) = 30/44.