由题设知ξ=0,1,2,3
∵P(ξ=0)=(1−
1
2)×(1−
1
2)×(1−
2
3)=[1/12],
P(ξ=1)=[1/2×(1−
1
2)×(1−
2
3)+(1−
1
2)×
1
2×(1−
2
3)+(1−
1
2)×(1−
1
2)×
2
3]=[1/3],
P(ξ=3)=[1/2×
1
2×
2
3]=[1/6],
∴P(ξ=2)=1-[1/12]-[1/3]=[5/12],
∴Eξ=0×
1
12+1×
1
3+2×
5
12+3×
1
6 =
5
3.
故答案为:[5/3].