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Math Help - Hypergeometrical Dist

  1. #1
    Member
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    Mar 2006
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    Hypergeometrical Dist

    Hi,

    I need some help with this question regarding Hypergeometric distribution.

    We mix in an urn 15 coins. 5 coins of 1 pound, 5 coins of 50 pens and 5 coins of 20 pens.
    we choose randomly and without replacement 4 coins. X is defined as the number of coins of 1 pound.

    a. what is the probability function of X, calculate E(X) and Var(X)
    b. what is the probability to have maximum 2 coins of 20 pens ?
    c. what is the probability to have at least 1 coin of 50 pens ?

    I started with:
    P(X=k)=((5 over k)*(10over 4-k)) / (15 over 4)
    E(X)=4*5=20
    Var(X)=[4*5*(15-4)*(15-5)] / [225*(15-1)]

    is it correct ? how to solve b and c ? a hint would be appreciated !
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  2. #2
    Senior Member
    Joined
    Feb 2008
    Posts
    410
    Quote Originally Posted by WeeG View Post
    Hi,

    I need some help with this question regarding Hypergeometric distribution.

    We mix in an urn 15 coins. 5 coins of 1 pound, 5 coins of 50 pens and 5 coins of 20 pens.
    we choose randomly and without replacement 4 coins. X is defined as the number of coins of 1 pound.

    a. what is the probability function of X, calculate E(X) and Var(X)
    b. what is the probability to have maximum 2 coins of 20 pens ?
    c. what is the probability to have at least 1 coin of 50 pens ?

    I started with:
    P(X=k)=((5 over k)*(10over 4-k)) / (15 over 4)
    E(X)=4*5=20
    Var(X)=[4*5*(15-4)*(15-5)] / [225*(15-1)]

    is it correct ? how to solve b and c ? a hint would be appreciated !
    Your pmf and variance are correct, but you forgot to divide 20 by 15 to get E(X)=4/3.

    Let U denote the number of 20-pen coins drawn and V denote the number of 50-pen coins drawn. It is obvious that U\sim V\sim X. So

    P(U\leq 2) = P(U=0)+P(U=1)+P(U=2), and

    P(V\geq 1)=1-P(V<1)=1-P(V=0).

    I'll let you do the actual computations.
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