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Math Help - Probability questions

  1. #1
    Senior Member Danneedshelp's Avatar
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    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.
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  2. #2
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    Quote Originally Posted by Danneedshelp View Post
    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.
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  3. #3
    kin
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    Quote Originally Posted by Danneedshelp View Post
    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
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  4. #4
    kin
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    Quote Originally Posted by The Second Solution View Post
    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.
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  5. #5
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    Quote Originally Posted by The Second Solution View Post
    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.
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