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Math Help - Probability - Finding P(X = 0)

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    Probability - Finding P(X = 0)

    Suppose that P(X=0)=1-P(X=1). If E[X]=3Var(X), nd P(X=0).
    Well the answer is 1/3 which i cannot seem to get.

    I calculate P(X = 1) = p
    thus P(X = 0) = 1 - p

    Var(x) = (1 - p)/p^2
    E(x) = 1 / p

    Solving this u get

    p = 3(1 - p) and thus p = 3/4...so P(x = 0) = 1/4 and not 1/3 which is the answer.

    Any help would be appreciated.

    Thank-you
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    Quote Originally Posted by AshleyT View Post
    Well the answer is 1/3 which i cannot seem to get.
    I calculate P(X = 1) = p
    thus P(X = 0) = 1 - p
    Var(x) = (1 - p)/p^2
    E(x) = 1 / p
    Those expressions in red are wrong.
    E(X)=1(p)+0(1-p)=p
    and Var(X)=E(X^2)-E^2(X).
    If you solve that you will get p=\frac{2}{3}
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    Quote Originally Posted by Plato View Post
    Those expressions in red are wrong.
    E(X)=1(p)+0(1-p)=p
    and Var(X)=E(X^2)-E^2(X).
    If you solve that you will get p=\frac{2}{3}
    How comes they are wrong?
    I was assuming from the above it would be a Geometric distribution and thus the equations in the red.
    If it's not a geometric distribution, what would it be?
    Thankyou
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    Quote Originally Posted by AshleyT View Post
    How comes they are wrong?
    I was assuming from the above it would be a Geometric distribution and thus the equations in the red. If it's not a geometric distribution, what would it be?
    You have no right to assume anything about the distribution.
    It is just a distribution on two points X=0~\&~X=1.

    This question is designed to test your knowledge of the definitions of E(X)~\&~Var(X).
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    Quote Originally Posted by Plato View Post
    You have no right to assume anything about the distribution.
    It is just a distribution on two points X=0~\&~X=1.

    This question is designed to test your knowledge of the definitions of E(X)~\&~Var(X).
    Sorry. previous questions was on geometric which is why i assumed so... Sorry.
    Thankyou for the reply
    Last edited by AshleyT; May 11th 2011 at 01:23 PM.
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