# Math Help - probability 1

1. ## 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.

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

Does that have a solution?

3. ## 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.