1. joint pdf probability

suppose X and Y have the joint pdf
$
f_{X, Y} = x + y, 0 < x, y < 1
$

Find P(X < 2Y)

2. Originally Posted by wik_chick88
suppose X and Y have the joint pdf
$
f_{X, Y} = x + y, 0 < x, y < 1
$

Find P(X < 2Y)
$\displaystyle \Pr(X < 2Y) = \int_{y=0}^{y = 1} \int_{x = 2y}^{x = 2} f(x, y) \, dx \, dy$.

3. Right, but X cannot exceed 1, with any probability, so the imposter must draw the line in the unit square.

I'd switch the order of integration...

$1-\int_0^1\int_0^{x/2} (x+y)dydx$

4. Originally Posted by matheagle
Right, but X cannot exceed 1, with any probability, so the imposter must draw the line in the unit square.

I'd switch the order of integration...

$1-\int_0^1\int_0^{x/2} (x+y)dydx$
Right you are. I misread the support.

5. thats why I'm the math beagle

6. the final answer i got is $\frac{19}{24}$ is this correct? thanks so much!