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Math Help - probability 1

  1. #1
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    probability 1

    Can anyone help me with this problem? Thanks in advance.

    A random variable X has {1,3,6} as only possible values. E(x) = 4 and Var(x) = 2. Find Pr (4=1), Pr (4=3), Pr (4=6) or explain why no such variable exists.
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  2. #2
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    Quote Originally Posted by mhitch03 View Post
    A random variable X has {1,3,6} as only possible values. E(x) = 4 and Var(x) = 2. Find Pr (4=1), Pr (4=3), Pr (4=6) or explain why no such variable exists.
    You know that the given implies:
    \begin{gathered}<br />
  P(X = 1) + P(X = 3) + P(X = 6) = 1 \hfill \\<br />
  E(X) = 1 \cdot P(X = 1) + 3 \cdot P(X = 3) + 6 \cdot P(X = 6) = 4 \hfill \\<br />
  V(X) = E(X^2 ) - E^2 (X) = 2 \hfill \\ <br />
\end{gathered} .

    Does that have a solution?
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  3. #3
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    algebra

    Write a=P(1), b=P(3), c=P(6).
    Then, as Plato suggests, we get:

    a+b+c = 1

    a+3b+6c = 4

    a*(1-4)^2 + b*(3-4)^2 + c*(6-4)^2 = 2, which is equivalent to:
    9a+b+4c=2.

    Solving gives a=0, b=2/3 and c = 1/3.
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