设随机变量(X,Y)的概率密度为 f(x,y)={k(6-x-y),0
人气:100 ℃ 时间:2019-08-18 21:31:56
解答
∫∫f(x,y)dxdy=1,可得k=1/8
P{X+Y≤4﹜=∫∫f(x,y)dxdy=∫dx∫k(6-x-y)dy=2/3,(前面的积分下上限为0和2,后面的积分下上限为2和x-4)
积分限的确定要画图 0积分出来的P{X+Y≤4﹜=∫∫f(x,y)dxdy=∫dx∫k(6-x-y)dy=k∫(x^2/2-4x+6)dx=1/8*(x^3/6-2x^3+6x)|(0,2)=1/8*16/3=2/3
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