10件产品中有2件次品，每次从中取3件，用x 表示取出的3件中所含次品件数。（1）求x的分布列和数学期望... 从一批由45件正品，5件次品组成的产品中任取3件求恰好有一件...
\u5df2\u77e5\u67d0\u79cd\u4ea7\u54c1\u517110\u4ef6\uff0c\u5176\u4e2d2\u4ef6\u662f\u6b21\u54c1\uff0c\u4ece\u4e2d\u53d63\u6b21\uff0c\u6bcf\u6b21\u53d6\u4e00\u4e2a\uff0c\u7528y\u8868\u793a\u53d6\u51fa\u4ea7\u54c1\u4e2d\u6b21\u54c1\u7684\u6570\u76ee\u8fd9\u6837
\u81f3\u5c11\u67092\u4ef6\u6b21\u54c1\u5c31\u662f\u8bf4\u67092\u4ef6\u6b21\u54c1\u548c3\u4ef6\u6b21\u54c1\u4e24\u79cd\u53ef\u80fd
2\u4ef6\u6b21\u54c1\u7684\u60c5\u51b5\uff1a
P1={C(2,5)*C(1,45)}/C(3,50)=9/392
3\u4ef6\u6b21\u54c1\u7684\u60c5\u51b5\uff1a
P2=C(3,5)/C(3,50)=1/1960
\u6240\u4ee5P\uff08\u81f3\u5c11\u67092\u4ef6\u6b21\u54c1\uff09=P1+P2=9/392+1/1960=23/980
\u62d3\u5c55\u8d44\u6599\uff1a
\u5df2\u77e5\u51fd\u6570f(x)=ax-(2a+1)lnx-2/x g(x)=-2alnx-2/x
\u5f53a=2\u65f6\u6c42y=f\uff08x\uff09\u5728\uff081\uff0cf\uff081\uff09\uff09\u7684\u5207\u7ebf\u65b9\u7a0b
\u5f53a\uff1e0\u65f6\uff0c\u6c42f\uff08x\uff09\u5355\u8c03\u533a\u95f4
\u82e5x\u5c5e\u4e8e[1\uff0fe\uff0ce^2]\uff0c\u4f7ff\uff08x\uff09\u5927\u4e8e\u7b49\u4e8eg\uff08x\uff09\u6210\u7acb\uff0c\u6c42a\u7684\u8303\u56f4\u3002
\u89e3\u7b54\uff1a
\uff081.\uff09\u5f53a=2\u65f6\uff0cf(x)=2x-5lnx-(2/x)\uff0c\u5b9a\u4e49\u57dfx\uff1e0
f(1)=2-0-2=0
\u4e14f'(x)=2-(5/x)+(2/x²)
\u6240\u4ee5\uff0cf'(1)=2-5+2=-1
\u5219\u5728\u70b9(1,f(1))\u7684\u5207\u7ebf\u65b9\u7a0b\u4e3a\uff1ay-0=-1(x-1)
\u5373\uff1ax+y-1=0
\uff082.\uff09\u5f53a>0\u65f6\uff0c\u56e0\u4e3af'(x)=a+2/(x^2)-(2a+1)/x =[ax^2-(2a+1)x+2]/(x^2)=(ax-1)(x-2)/(x^2)
\u4e8e\u662f\uff1a
\u5f531/a=2\u65f6\uff0c\u5373a=1/2\u65f6\uff0cf'(x)\u22650\uff0c\u5373y=f(x)\u5728R\u4e0a\u589e\u51fd\u6570\uff1b
\u5f531/a>2\u65f6\uff0c\u53730 \u3002
X~ 0 1 2
7/15 7/15 1/15
数学期望为
E(X)=0+7/15+2/15=3/5
（2）Y的分布列为
Y~-4 -2 0
1/15 7/15 7/15
所以EY=-4/15-14/15+0=-6/5
又E(Y^2)=16/15+4*7/15+0=44/15
故DY=44/15-(-6/5)^2=112/75
绛旓細锛1锛夌敤缁勫悎璁＄畻寰楀垎甯冨垪涓 X~ 0 1 2 7/15 7/15 1/15 鏁板鏈熸湜涓 E(X)=0+7/15+2/15=3/5 锛2锛塝鐨勫垎甯冨垪涓 Y~-4 -2 0 1/15 7/15 7/15 鎵浠Y=-4/15-14/15+0=-6/5 鍙圗(Y^2)=16/15+4*7/15+0=44/15 鏁匘Y=44/15-(-6/5)^2=112...
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绛旓細瑙ｏ細绗竴棰橈紝鍥犱负10浠朵骇鍝佷腑锛屾湁8浠舵鍝侊紝2浠舵鍝侊紝鏁呮娊鍒颁簡涓浠舵鍝佺殑缁勫悎鏄紝C(2)涓婃爣(8)涓嬫爣*C(1)涓婃爣(2)涓嬫爣=28锛浠庡崄浠朵骇鍝佷腑閫変笁浠剁殑缁勫悎鏄紝C(3)涓婃爣(10)涓嬫爣=120,鎵浠ワ紝鎭版湁涓浠舵鍝佺殑姒傜巼鏄紝28/120=23.333 绗簩棰橈紝鍚岀悊锛屽洜涓鏈変袱浠舵鍝侊紝鏁呮湁鎶藉埌涓浠舵鍝佺殑鎯呭喌锛屼篃鏈...
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