解析:(Ⅰ)在抽检的6件产品中恰有一件二等品的概率为:
P(ξ=1)=
C 14
C 25 ?
C 23
C 25 +
C 24
C 25 ?
C 13
C 12
C 25 =
12
25 (3分)
(Ⅱ)ξ可能的取值为0,1,2,3.(4分)
P(ξ=0)=
C 24
C 25 ?
C 23
C 25 =
18
100 =
9
50 ;
P(ξ=1)=
C 14
C 25 ?
C 23
C 25 +
C 24
C 25 ?
C 13
C 12
C 25 =
12
25 ;
P(ξ=2)=
C 14
C 25 ?
C 13
C 12
C 25 +
C 24
C 25 ?
C 22
C 25 =
15
50 =
3
10 ;
P(ξ=3)=
C 14
C 25 ?
C 22
C 25 =
1
25 .(7分)
ξ的分布列为
ξ 0 1 2 3
P
9
50
12
25
3
10
1
25 (8分)
E(ξ)=0×
9
50 +1×
12
25 +2×
3
10 +3×
1
25 =
6
5 (9分)
(Ⅲ)所求的概率为 P(ξ≥2)=P(ξ=2)+P(ξ=3)=
15
50 +
1
25 =
17
50 .(12分)