Y12OX0.20.10.220.10.30.1解(1)Z,=X+Y可能的取值为1,2,3,4共四个值,相应的概率为P[Z, = 1) = P[X = 1,Y = 0) = 0.2P[Z, = 2) = P[X = 1,Y =1) + P[X = 2,Y = 0)= 0.1+0.1= 0.2P[Z, = 3) = P[X = 2,Y =1)+P(X =1,Y = 2)= 0.3+0.2= 0.5P[Z, = 4) = P[X = 2,Y = 2) = 0.1洗阳师范大学ShenYang Normal Un
解 (1) Z X Y 1 = + 可能的取值为 1,2,3,4 共四个值,相应的概率为 2 P Z P X Y 1 = = = = = 1 1, 0 0.2 1 2 1, 1 2, 0 0.1 0.1 0.2 P Z P X Y P X Y = = = = + = = = + = 1 3 2, 1 1, 2 0.3 0.2 0.5 P Z P X Y P X Y = = = = + = = = + = P Z P X Y 1 = = = = = 4 2, 2 0.1 X Y 0 1 2 10.2 0.1 0.2 0.1 0.3 0.1
Y102X0.10.20.220.10.30.1(2)Z,=XY可能的取值为0,1,2,4,相应的概率为P[Z, = 0)= P(X = 1,Y = 0)+ P(X = 2,Y = 0) = 0.2 +0.1= 0.3PZ, = 1) = P(X =1,Y = 1) = 0.1P[Z, = 2) = P(X = 2,Y =1)+ P[X = 1,Y = 2) = 0.3+0.2 = 0.5P[Z, = 4) = P[X = 2,Y = 2) = 0. 1洗阳师范大学ShenYangNormalUniv
(2) Z XY 2 = 可能的取值为 0,1,2,4,相应的概率为 2 X Y 0 1 2 10.2 0.1 0.2 0.1 0.3 0.1 P Z P X Y P X Y 2 = = = = + = = = + = 0 1, 0 2, 0 0.2 0.1 0.3 P Z P X Y 2 = = = = = 1 1, 1 0.1 P Z P X Y P X Y 2 = = = = + = = = + = 2 2, 1 1, 2 0.3 0.2 0.5 P Z P X Y 2 = = = = = 4 2, 2 0.1
Y102X0.20.10.220.10.30.1(3)Z,=max(X,Y)可能的取值为1,2,相应的概率为P(Z, =1) = PX =1,Y = 0}+P[X = 1,Y =1)= 0.2+0.1= 0.3P[Z, = 2) = 1 - P(Z, =1) =1 -0.3 = 0.7洗阳师范大学ShenYangNoemal Unsivenit
2 X Y 0 1 2 10.2 0.1 0.2 0.1 0.3 0.1 (3) Z X Y 3 = max{ , }可能的取值为 1,2,相应的概率为 P Z P X Y P X Y 3 = = = = + = = = + = 1 1, 0 1, 1 0.2 0.1 0.3 P Z P Z 3 3 = = − = = − = 2 1 1 1 0.3 0.7
结论若二维离散型随机变量的联合分布律为P(X =x,Y = y;} = Pj, i,j= 1,2,.",则随机变量函数 Z = g(X,Y)的分布律为P(Z = zk) = P(g(X,Y) = zh)py,k = 1,2,....-zk=g(xiyj)沈阳师范大学ShenYangNoemal Univenit
结论 若二维离散型随机变量的联合分布律为 P{X = x ,Y = y } = p , i, j = 1,2, , i j ij 则随机变量函数 Z = g(X,Y )的分布律为 { } { ( , ) } k k P Z = z = P g X Y = z , 1,2, . ( ) = = = p k k i j z g x y ij 7
单选题O3设置1分设二维离散型随机变量(X,Y)的概率分布如下,Y231X00.20.10.210.10.30.1).则 P[min(X,Y = O}C)0. 2B) 0. 30. 4D) 0. 5A)公沈阳师范大学ShentangNiomal Univesth
A B C D 提交 设二维离散型随机变量( , ) X Y 的概率分布如下, Y X 1 2 3 0 0.2 0.1 0.2 1 0.1 0.3 0.1 则 P X Y {min{ , } 0} = ( ). A) 0.2 B) 0.3 C) 0.4 D)0.5 单选题 1分