- #1
tronter
- 185
- 1
If [tex] f(x,y) = e^{-x-2y} [/tex] find [tex] P(X<Y) [/tex].
So is this equaled to [tex] 1 - P(X>Y) = 1 - \int\limits_{0}^{\infty} \int\limits_{0}^{x} e^{-x-2y} \ dy \ dx = \frac{1}{3}?[/tex]
So is this equaled to [tex] 1 - P(X>Y) = 1 - \int\limits_{0}^{\infty} \int\limits_{0}^{x} e^{-x-2y} \ dy \ dx = \frac{1}{3}?[/tex]