Question

Joint density functiin from random variables xyz is f(xyz) is cxyz if 0<= x<=2 0<=y<=1 and 0<=z<=2 and f(xyz) =0 utherwise then find p(x+y+z<=1

Answer 1

Since fXYZ(x,y,z)����(�,�,�) is a joint probability density function:

∭X,Y,Zf(x,y,x) dV=1.∭�,�,��(�,�,�) ��=1.

Now,

∭X,Y,Zf(x,y,x) dV=∭0≤x,y,x≤2Cxyz dx dy dz=C(∫20x dx)(∫20y dy)(∫20z dz)=C(x2/2 ]20(y2/2 ]20(z2/2 ]20=C(22/2−0)(22/2−0)(22/2−0)=C(2)(2)(2)=8C.∭�,�,��(�,�,�) ��=∭0≤�,�,�≤2���� �� �� ��=�(∫02� ��)(∫02� ��)(∫02� ��)=�(�2/2 ]02(�2/2 ]02(�2/2 ]02=�(22/2−0)(22/2−0)(22/2−0)=�(2)(2)(2)=8�.

Thus,

8C=1C=1/8,8�=1�=1/8,

and

fX,Y,Z(x,y,z)={xyz/8,0≤x,y,