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Math Help - More expected value

  1. #1
    Senior Member Twig's Avatar
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    More expected value

    Hi!

    A family has children until it has a boy or until it has three children, whichever comes first. Assume the child is a boy with probability \frac{1}{2}. Find the expected number of boys in this family, and the expected number of girls.


    Let X denote the number of boys in the family.
    Let Y denote the number of girls in the family.

    We are seeking E(X) and  E(Y).

    There are the following possible "child outcomes": G = girl B = boy

    \Omega = \left(GGG,GGB,GB,B\right)

    It is not possible to have for example BBG, since the family stops having children when they have gotten a boy.
    Let F be any of the events in \Omega.
    Hence,

    E(X)=\displaystyle \sum_{\Omega} E(X|F)P(F) = \frac{3}{4}

    E(Y)=\displaystyle \sum_{\Omega} E(Y|F)P(F) = \frac{3}{2}

    Did I go wrong somewhere, if so, what is the error?

    Thanks!
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  2. #2
    Junior Member
    Joined
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    Quote Originally Posted by Twig View Post
    Hi!

    A family has children until it has a boy or until it has three children, whichever comes first. Assume the child is a boy with probability \frac{1}{2}. Find the expected number of boys in this family, and the expected number of girls.


    Let X denote the number of boys in the family.
    Let Y denote the number of girls in the family.

    We are seeking E(X) and  E(Y).

    There are the following possible "child outcomes": G = girl B = boy

    \Omega = \left(GGG,GGB,GB,B\right)

    It is not possible to have for example BBG, since the family stops having children when they have gotten a boy.
    Let F be any of the events in \Omega.
    Hence,

    E(X)=\displaystyle \sum_{\Omega} E(X|F)P(F) = \frac{3}{4}

    E(Y)=\displaystyle \sum_{\Omega} E(Y|F)P(F) = \frac{3}{2}

    Did I go wrong somewhere, if so, what is the error?

    Thanks!
    i think, you better construct the probability distribution first before find the expected value....
    probability distibution from
    \Omega = \left(GGG,GGB,GB,B\right)
    and Let X denote the number of boys in the family.

    the probability distribution
    X P(X=x)
    0 1/8
    1 7/8

    you can find expected value easily from the probability distribution table...
    E(X)=\sum(Xi Pi)


    ......hope it help..........
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  3. #3
    Senior Member Twig's Avatar
    Joined
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    Hi

    Thanks you´re right.
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