I will assume my comment about you meaning f(0,1) instead of f(0.1) was correct.
Think about it this way: if X, Y can each be only 0 or 1 there are only 4 possibilities: (0,0), (1,0), (0,1), (1,1). You have the probabilities assigned to three of the four, so you can make a table with two columns (labeled 0 and 1) and two rows (also labeled 0 and 1). In each cell put the appropriate probability:
In cell (0,1) put .1, in cell (1,0) put .1, in cell (1,1) put 0.344. (Since the four probabilities have to sum to 1 you can find the probability for cell (0,0) yourself.)
Now you have two ways to go.
Method A: To calculate the expectation work out the sum
[tex]
E(X+Y) = \sum f(m,n) (m+n)[/tex]
- it will have as many terms as there are cells in the table.
Method B:
The rows of your table have the values for X, the columns the values for Y, so the edges of the table give the marginal distributions of X and Y. Then
[tex]
E(X+Y) = E(X) + E(Y)[/tex]
can be calculated using the appropriate marginal distributions.
Methods A and B give the same answer (when applied correctly) so you can use whichever you prefer.
Set things up and try the work before posting more questions.