Discussion Problem_(30)

[duck]

Suppose the joint pdf of X and Y is given by:
f(x,y) = 6(1-x-y) for x+y<1, x>0 , y>0
         = 0 elsewhere.
Find P(2X+3Y<1)

[rohan]

5/36

[Sumit]

Hi duck,
I'm getting 11/72...plz tell the correct answer....

[Mauli]

13/36 . that is what i am getting
my method :
fixing x first . so integral for x will be over 0 to 1/2.
then for y it is 0 to (1-2x)/3
which implies
integral from 0 to 1/2 .{ integral from 0 to (1-2x)/3 [6(1-x-y) dy]} dx
solving then gives 13/36.
PLEASE cnfrm .

[Ayushya Kaul]

I'm muddled up while solving this.
@Duck- Could you please solve this?

[Sinistral]

plotting x+y<1 [the domain of f(x,y) ] and 2x+3y<1 (the region whose probability needs to be calculated).
seeing the overlapping area.

by observing the overlapping area it is seen that the area of interest is  2x+3y<1 in the first quadrant (since it is given that x>0 , y>0)

so P(2X+3Y<1) = Integral [x= 0 to 1/2] Integral [ y=0 to (1-2x)/3] f(x,y) dy dx.
                     = 13/36

[Sumit]

Hey mauli,
I also did it in a same way as you did.... but I'm getting 11/72...

[Sumit]

Hey I got it...It was a calculation mistake...

[duck]

Great.. :)

Correct Answer:
13/36

[Ayushya Kaul]

Ah got it.
Had made a huge calculation mistake. Thanks @Sinistral.