A personnel director has two lists of applicants for jobs

[Guest]

Here's a question I found doing my stats homework (don't worry, I solved it myself, I'm not asking for homework help) that I thought was pretty interesting. I wonder if you all agree:

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 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 [from list 2] is that of a man, what is the probability that a woman's name was originally selected from list 1?

[Guest]

Here's a question I found doing my stats homework (don't worry, I solved it myself, I'm not asking for homework help) that I thought was pretty interesting. I wonder if you all agree:

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 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 [from list 2] is that of a man, what is the probability that a woman's name was originally selected from list 1?

! = not

Event A = A woman's name was selected from list 1.

P[A] = 5/7

Event B = A man's name is selected from list 2.

P = P[b|A]*P[A] + P[b|!A]*P[!A]

P[b|A] = 6/9 = 2/3

P[b|!A] = 7/9

P = (2/3)*(5/7) + (7/9)*(2/7) = 10/21 + 2/9 = (30 + 14)/63 = 44/63

What is P[A|B]?

P[A|B]*P = P[A and B] = P[b|A]*P[A] = 10/21

P[A|B]*(44/63) = (10/21)

P[A|B] = 630/924 = 315/462

This is the answer unless I made some arithmetic mistake (which I am prone to do sometimes).

[Molly Mae]

15/22

[Guest]

Because although you have been given a very eloquent answer to this, and it appears correct, in my mind event 1 occured first and completely seperate so my answer is simply 5/7

[Guest]

Id have to agree with peanut.

A name is first picked from list 1. The probability of it being a woman is 5/7.

Whatever happens after that doesnt matter because your question still asks what the probability of a womans name being picked at first.

[EventHorizon]

Question (and a hint) for slyde87 and peanut.

If list 1 had 100 women and 1 man, and list 2 had 100 women and no men, and the name selected from the augmented list 2 was a man, would you still think probability that the man's name was selected from list 1 initially was 1/101? If a man's name wasn't selected, then the observed final result would be impossible... so the probability that a man's name was selected initially must be 1.

The extra information does in fact change the probability of the unobserved initial selection. The type of reasoning that this problem requires is called Bayesian logic.

It's actually a very interesting thing to learn. Try looking up the monty hall problem for another example.

[Guest]

Question (and a hint) for slyde87 and peanut.

If list 1 had 100 women and 1 man, and list 2 had 100 women and no men, and the name selected from the augmented list 2 was a man, would you still think probability that the man's name was selected from list 1 initially was 1/101? If a man's name wasn't selected, then the observed final result would be impossible... so the probability that a man's name was selected initially must be 1.

The extra information does in fact change the probability of the unobserved initial selection. The type of reasoning that this problem requires is called Bayesian logic.

It's actually a very interesting thing to learn. Try looking up the monty hall problem for another example.

You are correct, I know. Even though I understand the logic...my thinking can be a bit stuborn(my college professors can attest to that). You would have enjoyed listening to me debate the impossibility of an infinite straight line.

[Guest]

15/22

Yep! Explanation is below:

Let's use lowercase letters to represent the first group and uppercase to represent the second. We have groups

wwwwwmm

WWMMMMMM

We know (and will use):

P(w)=5/7

P(M|w)=6/9 (if you add one W to the second list you get 6/9 chance of picking M)

P(M|m)=7/9 (same, but add one M)

We need to find P(w|M) (probability of having picked w given we picked M).

Conditional probability:

P(w|M) = P(w and M)/P(M)

P(w and M) = P(M and w) = P(w)*P(M|w) = 5/7*6/9

P(M) = P(M and w)+P(M and m) = P(w)*P(M|w)+P(m)*P(M|m) = 5/7*6/9 + 2/7*7/9

P(w|M) = P(w and M)/P(M) = (5/7*6/9)/(5/7*6/9 + 2/7*7/9)

=15/22

[Guest]

Yep! Explanation is below:

Let's use lowercase letters to represent the first group and uppercase to represent the second. We have groups

wwwwwmm

WWMMMMMM

We know (and will use):

P(w)=5/7

P(M|w)=6/9 (if you add one W to the second list you get 6/9 chance of picking M)

P(M|m)=7/9 (same, but add one M)

We need to find P(w|M) (probability of having picked w given we picked M).

Conditional probability:

P(w|M) = P(w and M)/P(M)

P(w and M) = P(M and w) = P(w)*P(M|w) = 5/7*6/9

P(M) = P(M and w)+P(M and m) = P(w)*P(M|w)+P(m)*P(M|m) = 5/7*6/9 + 2/7*7/9

P(w|M) = P(w and M)/P(M) = (5/7*6/9)/(5/7*6/9 + 2/7*7/9)

=15/22

Cool, I forgot to completely reduce the fraction to lowest terms.

Nice puzzle.

[Guest]

Uhh, I hope this "solved" poll works. First time!

[Guest]

I think the answer should be 5/7 as event 1 (selection from list A) happens first and hence it is independent of the second event.

[Guest]

I think the answer should be 5/7 as event 1 (selection from list A) happens first and hence it is independent of the second event.

I'm going to steal EventHorizon's example to respond to this.

If list 1 had 100 women and 1 man, and list 2 had 100 women and zero men,

and you know that a man was selected from list 2, then what is the probability a man was selected from list 1?

Since there are zero men originally in list 2, then the man selected from list 2 HAD to have come from list 1. This means that a man HAD to have been selected from list 1.

Therefore the probability a man was selected from list 1 given that a man was selected from list 2 is not 1/101, but 1.

Edited September 27, 2010 by mmiguel1