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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- Nov 20th 2009, 08:27 AMmhitch03probability 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. - Nov 20th 2009, 09:25 AMPlato
You know that the given implies:

$\displaystyle \begin{gathered}

P(X = 1) + P(X = 3) + P(X = 6) = 1 \hfill \\

E(X) = 1 \cdot P(X = 1) + 3 \cdot P(X = 3) + 6 \cdot P(X = 6) = 4 \hfill \\

V(X) = E(X^2 ) - E^2 (X) = 2 \hfill \\

\end{gathered} $.

Does that have a solution? - Nov 20th 2009, 02:03 PMqmechalgebra
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.