# Thread: Joint probability function

1. ## Joint probability function

OK i have a question that is on a lab, i missed class the other day and didnt buy the book (it was like 250 bucks) and all the sudden im lost. We were working on p.f.'s and p.d.f.'s and then all the sudden i got 2 variables and stuff looks weird. Here is the question.

Let f(x,y) = {(x+y)/32 for x=1,2 and y=1,2,3,4

Find P(X>Y)

are you just supposed to integrate with respect to x and then y and use 1-2 for the dx integral and 1-4 for the dy and then how would the probability X>Y be found? There are other questions like P(Y=2X) and P(X+Y=3) and P(X<3-Y) but im just trying to figure out the method to solving these

2. Originally Posted by ChrisBickle
Let f(x,y) = {(x+y)/32 for x=1,2 and y=1,2,3,4
Find P(X>Y)
The Boolean function [x>y] is 1 if true and 0 if false.
$P(X > Y) = \sum\limits_{x = 1}^2 {\sum\limits_{y = 1}^4 {\left[ {x > y}\right]\left( {\frac{{x + y}}{{32}}} \right)} }$