Expectation of joint probability mass function Let the joint probabilty mass function of discrete random variables X and Y be given by
$f(x,y)=\frac{x^2+y^2}{25}$, for $(x,y) = (1,1), (1,3), (2,3)$
The value of E(Y) is  ?
Attempt
$E(Y) = \sum_{x,y} y\cdot\frac{x^2 + y^2}{25}$
$E(Y) = \sum_{x,y}\frac{x^2y + y^3}{25}$
Substituting for $(x,y) = (1,1), (1,3), (2,3)$
$E(Y) = \frac1{25} + \frac{30}{25} + \frac{39}{25}$
$E(Y) = 2.80$
Is this right?

In relation to the (1,1) point you seem to be claiming that $1\cdot \frac{1^2+1^2}{25}=\frac{1}{25}$.
It's more usually thought to be the case that $1^2+1^2>1$*. This seems to be the cause of your problem here. 
* (Some people - reckless people perhaps - even claim that $1^2+1^2$ could be as much as $2$.)
