# Expected value of random variable

I have this question:

What's the expected value of a random variable $X$ if $P(X=1)=1/3$, $P(X=2)=1/3$, and $P(X=6)=1/3$?

I am very confused as to how I can work this problem out. I was thinking it would be something like:

$$E[X] = P(X=1) \cdot (1/3) + P(X=2) \cdot 1/3 + P(X=6) \cdot 1/3.$$

I am not sure this is correct because then I do not have values for $P(X=1)$, $P(X=2)$, and $P(X=6)$. Should I just do the calculation like this:

$$E[x] = (1/3)+(1/3)+(1/3)$$

I am not sure exactly how Expected value for random variables should be calculated. Should $E[x]$ always add up to $1$?

Thank you.

• Your first expression is not right. You want $1\cdot\Pr(X=1)+2\cdot\Pr(X=2)+6\cdot\Pr(X=6)$. – André Nicolas Mar 3 '14 at 21:51

$$1\cdot 1/3 + 2\cdot 1/3 + 6\cdot 1/3 = 3$$
$$E[X] = \sum_i P(X_i)X_i$$