(1)由题意,ξ的取值为0,1,2,则 P(ξ=0)=
1
2 ×
1
2 =
1
4 ,P(ξ=1)=
1
2 ×
1
2 +
1
2 ×
1
3 =
5
12 ,P(ξ=2)=
1
2 ×
2
3 =
1
3
∴ξ的分布列为
ξ 0 1 2
P
1
4
5
12
1
3 ∴ Eξ=0×
1
4 +1×
5
12 +2×
1
3 =
13
12
(2)由已知可得 a n = a n-1 •
1
2 +(1- a n-1 )•
1
3 (n∈ N * ,n≥2)
∴ a n =
1
6 a n-1 +
1
3 ,
∴
lim
n→∞ a n =
lim
n→∞ (
1
6 a n-1 +
1
3 ) =
1
6
lim
n→∞ a n +
1
3
∴
lim
n→∞ a n =
2
5