Question

跪求此数理统计题的解答过程，万分感谢

设二维随机变量（X，Y）的概率密度是f（x）=1/8（x+y），0<=x<=2，0<=y<=2 求①P{X<=1,Y<=1} ②P｛X+Y<=3} 麻烦高手帮帮忙写出详细解答过程，谢谢 满意加分

Answer 1

(1)P(X<=1,Y<=1)=∫[0,1]dx∫[0,1]1/8(x+Y)dy
=∫[0,1](1/8*x+/16)dx
=1/8
(2)P(X+Y<=3)==1-P(X+Y>3)
=1-∫[1,2]dx∫[3-x,2]1/8(x+Y)dy
=1-∫[1,2][1/8(2x+2)-1/8x(3-x)-1/16(3-x)^2]dx
=1/2