高二数学问题概率
\u9ad8\u4e8c\u6570\u5b66\u6982\u7387\u95ee\u9898\u6700\u76f4\u63a5\u7684\u65b9\u6cd5\uff0c\u8bbe\u7537\u751fX\u4eba\uff0c\u5973\u751fY\u4eba\uff0c\u6839\u636e\u9898\u610f\u53ef\u77e5\uff1a
1\u3001Y/(X+Y)=3/5\uff0c(Y/(X+Y))*((Y-I)/(X+Y-1))=1/3,\u89e3\u65b9\u7a0b\u7ec4\u5f97X=4\uff0cY=6
2\u3001C103=120\uff0c\u4e09\u4eba\u90fd\u662f\u7537\u751f\u7684\u65b9\u6cd5\uff1aC43=4\uff0c\u90fd\u662f\u5973\u751fC63=20\uff0c\u5219\u7537\u5973\u751f\u90fd\u6709\u7684\u6982\u7387\uff1a
\uff081-\uff084+20\uff09/120\uff09*100%=80%
3\u3001\u4e00\u4eba\u90fd\u4e0d\u901a\u8fc7\u7684\u6982\u7387\u4e3aX=(2/5)*(2/5)*(1/5),\u53ea\u6709\u4e00\u4e2a\u7537\u751f\u901a\u8fc7\u7684\u6982\u7387\u4e3a\uff1aY=(2*3/5)*(2/5)*(1/5)\uff0c\u53ea\u6709\u4e19\u901a\u8fc7\u7684\u6982\u7387\u4e3aZ=(2/5)*(2/5)*(4/5)\uff0c
\u5219\u81f3\u5c11\u4e24\u4e2a\u4eba\u901a\u8fc7\u7684\u6982\u7387\u4e3a1-\uff08X+Y+Z\uff09=93/125=74.4%
\u76f4\u63a5\u8ba1\u7b97\u66f4\u597d\uff0c\u5148\u7b97\u51fa\u4e09\u4e2a\u4eba\u90fd\u901a\u8fc7\u7684\u6982\u7387\uff0c\u518d\u7b97\u51fa\u53ea\u6709\u4e24\u4e2a\u7537\u751f\u901a\u8fc7\u7684\u6982\u7387\uff0c\u518d\u7b97\u51fa\u4e00\u4e2a\u7537\u751f\u4e00\u4e2a\u5973\u751f\u901a\u8fc7\u7684\u6982\u7387\uff0c\u4e09\u8005\u76f8\u52a0\u5f9793/125=74.4%
\uff081\uff09\u2235x+y+z=3 \uff0cx\uff0cy\uff0cz\uff1c(\u5c5e\u4e8e)N \uff0cx=y= z=1\u6216x=0\uff0cy=1\uff0cz=2\u6216x=2\uff0c
2y=x+z
y=1\uff0cz=0
\u2460\u5f53x=y= z=1\u65f6\uff0c3\u6b21\u6295\u63b7\u4e2d\u6709\u4e00\u6b21\u51fa\u73b01\u70b9\uff0c\u53c8\u6709\u4e00\u6b21\u51fa\u73b02\u70b9\u62163\u70b9\uff0c\u8fd8\u6709\u4e00\u6b21\u51fa\u73b04\u70b9\u30015\u70b9\u62166\u70b9\u3002\u2234\u6b64\u65f6\u7684\u6982\u7387P\uff08A1\uff09=C3^1\uff081/6\uff09C2^1(2/6)C1^1(3/6)=1/6
\u2461\u5f53x=0\uff0cy=1\uff0cz=2\u65f6\uff0c3\u6b21\u6295\u63b7\u4e2d\u6709\u4e00\u6b21\u51fa\u73b02\u70b9\u62163\u70b9\uff0c\u6709\u4e24\u6b21\u51fa\u73b04\u70b9\u30015\u70b9\u62166\u70b9\u3002\u2234\u6b64\u65f6\u7684\u6982\u7387P\uff08A2)=C3^1(2/6)C2^2(3/6)=1/4
\u2462\u5f53x=2\uff0cy=1\uff0cz=0\u65f6\uff0c P\uff08A3\uff09
\u2234P=P\uff08A1+A2+A3\uff09= P\uff08A1\uff09+ P\uff08A2\uff09+ P\uff08A3)=4/9
\uff082\uff09\u7531\u6761\u4ef6\u5f97x+y+z=6 \uff0cx\uff0cy\uff0cz \uff0c\u2234x=y=z=2
y^2=xz
\u2234\u57286\u6b21\u6295\u63b7\u4e2d\uff0c\u67092\u6b21\u90fd\u63b7\u51fa1 \u70b9\uff0c\u4e14\u53c8\u67092\u6b21\u90fd\u63b7\u51fa2\u70b9\u62163\u70b9\uff0c\u53e6\u67092\u6b21\u90fd\u63b7\u51fa4\u70b9\u30015\u70b9\u62166\u70b9\u3002
\u2234\u6240\u6c42\u6982\u7387\u4e3aC6^2(1/6)^2C4^2(2/6)^2C2^2(3/6)^2=5/72
2、1-0.4的五次方=0.98976
1.取得白球3次的概率:
C5(3)*(3/5)³*(2/5)²=216/625=0.3456
2.至少有一次取得白球的概率:
1-(2/5)^5=1-0.01024=0.98976
1 首先，五次选三次，有C5 3中选发，即10种。袋子中取到白球的概率为6/10=3/5，故独立的三次取到白球的概率为27/125；而且其他两次要去到黑球，概率为4/10*4/10=4/25
故取到白球三次的概率为 10*27/125*4/25=0.3456
2考虑反面，全取到黑球的概率为 （2/5）^5=0.01024
故，至少取到一次白球的概率为1-0.01024=0.98976
不好表述，仔细看一下
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