(1)盒子里装有3只黑球、2只红球、2只白球,在其中任取4只球.以X-|||-表示取到黑球的只数,以Y表示取到红球的只数.求X和Y的联合分布律.-|||-(2)在(1)中求 Xgt Y , Y=2X , X+Y=3 Xlt 3-Y .

步骤 1：确定X和Y的取值范围
X的取值范围为0,1,2,3，Y的取值范围为0,1,2。由于总共取4只球，所以X和Y的取值必须满足X+Y≤4。

步骤 2：计算联合分布律
根据组合数计算每种情况的概率，即从3只黑球中取X只，从2只红球中取Y只，从2只白球中取4-X-Y只，除以总的取法数C(7,4)。
- P{X=0,Y=2} = C(3,0)C(2,2)C(2,0)/C(7,4) = 1/35
- P{X=1,Y=1} = C(3,1)C(2,1)C(2,2)/C(7,4) = 6/35
- P{X=1,Y=2} = C(3,1)C(2,2)C(2,1)/C(7,4) = 6/35
- P{X=2,Y=0} = C(3,2)C(2,0)C(2,2)/C(7,4) = 3/35
- P{X=2,Y=1} = C(3,2)C(2,1)C(2,1)/C(7,4) = 12/35
- P{X=2,Y=2} = C(3,2)C(2,2)C(2,0)/C(7,4) = 3/35
- P{X=3,Y=0} = C(3,3)C(2,0)C(2,1)/C(7,4) = 2/35
- P{X=3,Y=1} = C(3,3)C(2,1)C(2,0)/C(7,4) = 2/35
其他情况的概率为0。

步骤 3：计算指定概率
- P{X>Y} = P{X=2,Y=0} + P{X=2,Y=1} + P{X=3,Y=0} + P{X=3,Y=1} = 3/35 + 12/35 + 2/35 + 2/35 = 19/35
- P{Y=2X} = P{X=1,Y=2} = 6/35
- P{X+Y=3} = P{X=1,Y=2} + P{X=2,Y=1} = 6/35 + 12/35 = 18/35
- P{X<3-Y} = P{X=0,Y=2} + P{X=1,Y=1} + P{X=2,Y=0} = 1/35 + 6/35 + 3/35 = 10/35