A.126/125 B.6/5 C.168/125 D.7/5

　　【解析】X的可能取值为0,1,2,3.

　　①大正方体8个顶点处的8个小正方体涂有3个面,所以P(X=3)=8/125;

　　②大正方体每条棱上对应的小正方体除了两个顶点处的还有3个,一共3×12=36个小正方体涂有2个面,所以P(X=2)=36/125;

　　③大正方体每个面上对应的小正方体除去棱上的还有9个,一共9×6=54个小正方体涂有1个面,所以P(X=1)=54/125;

　　④还有125-(8+36+54)=27个没有涂漆的小正方体,所以P(X=0)=27/125.

　　故E(X)=0×27/125+1×54/125+2×36/125+3×8/125=6/5.

　　【答案】B

9.体育课的排球发球项目考试的规则是:每位学生最多可发3次球,一旦发球成功,则停止发球,否则一直发到3次为止.设某学生一次发球成功的概率为p(p≠0),发球次数为X,若X的数学期望E(X)>1.75,则p的取值范围是(　　).

　　A.(0"," 7/12)　　　 B.(7/12 "," 1) C.(0"," 1/2) D.(1/2 "," 1)

　　【解析】由已知可得P(X=1)=p,P(X=2)=(1-p)p,P(X=3)=(1-p)2,

　　则E(X)=1×p+2×(1-p)p+3×(1-p)2=p2-3p+3>1.75,解得p>5/2或p<1/2.

　　又p∈(0,1),所以p∈(0"," 1/2).

　　【答案】C

10.已知离散型随机变量X满足P(X=x1)=2/3,P(X=x2)=1/3,且x1

　　【解析】由题意得

　　{■(x_1×2/3+x_2×1/3=4/3 "," @(x_1 "-" 4/3)^2×2/3+(x_2 "-" 4/3)^2×1/3=2/9 "," )┤

　　即{■(2x_1+x_2=4"," @2(x_1 "-" 4/3)^2+(x_2 "-" 4/3)^2=2/3 "," )┤

　　解得{■(x_1=5/3 "," @x_2=2/3)┤或{■(x_1=1"," @x_2=2"." )┤

　　∵x1

　　【答案】3

11.从一批产品中抽取4件做检验,这4件产品中优质品的件数记为n,如果n=3,再从这批产品中任取4件做检验,若都为优质品,则这批产品通过检验;如果n=4,再从这批产品中任取1件做检验,若为优质品,则这批