# probability 1

• Nov 20th 2009, 08:27 AM
mhitch03
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.
• Nov 20th 2009, 09:25 AM
Plato
Quote:

Originally Posted by mhitch03
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:
$\displaystyle \begin{gathered} P(X = 1) + P(X = 3) + P(X = 6) = 1 \hfill \\ E(X) = 1 \cdot P(X = 1) + 3 \cdot P(X = 3) + 6 \cdot P(X = 6) = 4 \hfill \\ V(X) = E(X^2 ) - E^2 (X) = 2 \hfill \\ \end{gathered}$.

Does that have a solution?
• Nov 20th 2009, 02:03 PM
qmech
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.