(Ⅰ)ξ的可能取值为0,1,2
∴P(ξ=0)= (1-
1
2 )×(1-
2
3 ) =
1
6 ,P(ξ=1)= (1-
1
2 )×
2
3 +
1
2 ×(1-
2
3 ) =
1
2 ,P(ξ=2)=
1
2 ×
2
3 =
1
3
∴ξ的分布列为
ξ 0 1 2
P
1
6
1
2
1
3 ∴Eξ=0×
1
6 +1×
1
2 +2×
1
3 =
7
6 ;
(Ⅱ)记事件A为四次投球中至少一次命中,则
∵P(
.
A )=
1
2 ×
1
2 ×
1
3 ×
1
3 =
1
36 ,
∴P(A)=1-P(
.
A )=
35
36 .