f(x,y) = x2 + xy3 for 0 < x < 1, 0 < y < 2
and 0 otherwise.
Calculate P(X+Y < 1)
The Attempt at a Solution
P(X+Y < 1) = P(X < 1-Y) which means y is now bounded by 0:1 instead of 0:2 and x is bounded by 0:y.
So we get ∫[0-1]∫[0-y] x2 + xy3 dx dy
Integrating twice I get the answer P(X < 1-Y) = 1/8, which is incorrect.
What am I doing wrong?