设X，Y相互独立，且X，Y的分布律分别为
X012
P1/42/41/4
X-1123
P1/45/121/41/12
求：(1)(X，Y)的分布律；(2)Z=XY的分布律．
【正确答案】：P{X=0，Y=-1}=P{X=0}•P{Y=-1}=(1/4)×(1/4)=1/16 P{X=0，Y=1}=P{X=0}•P{Y=1}=(1/4)×(5/12)=5/48 P{X=0，Y=2}=P{X=0}•P{Y=2}=(1/4)×(1/4)=1/16 P{X=0，Y=3}=P{X=0}•P{Y=3}=(1/4)×(1/12)=1/48 P{X=1，Y=-1}=P{X=1}•P{Y=-1}=(2/4)×(1/4)=1/8 P{X=1，Y=1}=P{X=1}•P{Y=1}=(2/4)×(5/12)=5/24 P{X=1，Y=2}=P{X=1}•P{Y=2}=(2/4)×(1/4)=1/8 P{X=1，Y=3}=P{X=1}•P{Y=3}=(2/4)×(1/12)=1/24 P{X=2，Y=-1}=P{X=2}•P{Y=-1}=(1/4)×(1/4)=1/16 P{X=2，Y=1}=P{X=2}•P{Y=1}=(1/4)×(5/12)=5/48 P{X=2，Y=2}=P{X=2}•P{Y=2}=(1/4)×(1/4)=1/16 P{X=2，Y=3}=P{X=2}•P{Y=3}=(1/4)×(1/12)=1/48 所以 (X，Y)的分布律为 X\Y -1 1 2 3 0 1/16 5/48 1/16 1/48 1 1/8 5/24 1/8 1/24 2 1/16 5/48 1/16 1/48 (2)Z可能的取值为-2，-1，0，1，2，3，4，6 P{Z=-2}=P{X=2，Y=-1}=P{X=2}•P{Y=-1}=(1/4)×(1/4)=1/16 P{Z=-1}=P{X=1，Y=-1}=P{X=1}•P{Y=-1}=(2/4)×(1/4)=1/8 P{Z=0}=P{Z=0，Y=-1}+P{Z=0，Y=1}+P{Z=0，Y=2}+P{Z=0，Y=3} =(1/4)×(1/4)+(1/4)×(5/12)+(1/4)×(1/4)+(1/4)×(1/12)=1/4 P{Z=1}=P{X=1，Y=1}=P{X=1}•P{Y=1}=(2/4)×(5/12)=5/24 P{Z=2}=P{X=1，Y=2}+P{X=2}•P{Y=1}=(2/4)×(1/4)+(1/4)×(5/12)=11/48 P{Z=3}=P{X=1，Y=3}=P{X=1}•P{Y=3}=(2/4)×(1/12)=1/24 P{Z=4}=P{X=2，Y=2}=P{X=2}•P{Y=2}=(1/4)×(1/4)=1/16 P{Z=6}=P{X=2，Y=3}=P{X=2}•P{Y=3}=(1/4)×(1/12)=1/48 所以 Z的分布律为 Z -2 -1 0 1 2 3 4 6 P 1/16 1/8 1/4 5/24 11/48 1/24 1/16 1/48
