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幼苗
共回答了16个问题采纳率:87.5% 举报
(I)X的取值为0,1,2,
P(X=0)=
C02
C313
C315=
22
35,
P(X=1)=
C12
C213
C315=
12
35,
P(X=2)=
C22
C113
C315=
1
35
故X的分布列为:
X 0 1 2
P
22
35
12
35
1
35(II)E(X)=1×
12
35+2×
1
35=
2
5;
D(X)=
22
35×(0-
2
5)2+
12
35×(1-
2
5)2+
1
35×(2-
2
5)2=
52
175.
1年前
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