(1)由题意得:2×
1
2×(1−
1
2)=a2
∴a=
2
2
(2)ξ=0,1,2,3,4
P(ξ=0)=C20(1−
1
2)2C20(1−a)2=
1
4(1−a)2,
P(ξ=1)=C21
1
2(1−
1
2) C20(1−a)2+C20(1−
1
2)2C21a(1−a)=
1
2(1−a)
P(ξ=2)=C22
1
22C20(1−a)2+C21
1
2(1−
1
2) C21a(1−a)+C20(1−
1
2)2C22a2=
1
4(1+2a−2a2)
P(ξ=3)=C22
1
22C21a(1−a) +C21
1
2(1−
1
2) C22a2=
a
2
P(ξ=4)=C22(
1
2