(1)设A表示事件“从第三箱中有放回地抽取3次(每次一件),恰有两次取到二等品”,
依题意知,每次抽到二等品的概率为
2
5 ,
故 P(A)=
C 23 (
2
5 ) 2 ×
3
5 =
36
125 .
(2)ξ可能的取值为0,1,2,3.
P(ξ=0)=
C 24
C 23
C 25
C 25 =
18
100 =
9
50 ,P(ξ=1)=
C 14
C 23
C 25
C 25 +
C 24
C 13
C 12
C 25
C 25 =
12
25 ,
P(ξ=2)=
C 14
C 13 •
C 12
C 25
C 25 +
C 24
C 22
C 25
C 25 =
15
50 =
3
10 ,P(ξ=3)=
C 14
C 22
C 25
C 25 =
1
25 .
ξ的分布列为
ξ 0 1 2 3
P
9
50
12
25
15
50
1
25 数学期望为Eξ=1×
12
25 +2×
15
50 +3×
1
25 =1.2.