设随机变量X服从二项式分布B(2,p),随机变量Y服从二项分布B(3,P)，若P{X>=1}=5/9,试求P{Y>=1} 设随机变量X服从二项分布b(2,p),随机变量Y服从二项分布...
\u8bbe\u968f\u673a\u53d8\u91cfX\u670d\u4ece\u4e8c\u9879\u5206\u5e03b(2,p),\u968f\u673a\u53d8\u91cfY\u670d\u4ece\u4e8c\u9879\u5206\u5e03b(4,p)\uff0c\u82e5P(X\u22651)=5/9,\u5219P(Y\u22651)=?\u8bbe\u968f\u673a\u53d8\u91cfX\u670d\u4ece\u4e8c\u9879\u5206\u5e03b\uff082\uff0cp\uff09\uff0c\u968f\u673a\u53d8\u91cfY\u670d\u4ece\u4e8c\u9879\u5206\u5e03b\uff084\uff0cp\uff09\uff0c\u82e5P\uff08X\u22651\uff09\uff1d5\uff0f9\uff0c\u5219P\uff08Y\u22651\uff09\uff1d65\uff0f81\u3002
\u56e0\u4e3aP\uff08X\u22651\uff09\uff1d5\uff0f9\uff0c\u6240\u4ee5\u6ee1\u8db3\u7b49\u5f0fP\uff08X\u22651\uff09\uff1d1\uff0dP\uff08X\uff1d0\uff09\uff1d1\uff0d\uff081\uff0dp\uff09\uff3e2\uff1d5\uff0f9\u3002\u5316\u7b80\u53ef\u5f97P\uff1d1\uff0f3\u3002\u6240\u4ee5\uff0cP\uff08Y\u22651\uff09\uff1d1\uff0dP\uff08Y\uff1d0\uff09\uff1d1\uff0d\uff081\uff0dp\uff09\uff3e4\uff1d1\uff0d\uff081\uff0d1\uff0f3\uff09\uff3e4\uff1d1\uff0d\uff082\uff0f3\uff09\uff3e4\uff1d65\uff0f81\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u5bf9\u4e8e\u56fa\u5b9a\u7684n\u4ee5\u53cap\uff0c\u5f53k\u589e\u52a0\u65f6\uff0c\u6982\u7387P\uff5bX\uff1dk\uff5d\u5148\u662f\u968f\u4e4b\u589e\u52a0\u76f4\u81f3\u8fbe\u5230\u6700\u5927\u503c\uff0c\u968f\u540e\u5355\u8c03\u51cf\u5c11\u3002\u5f53\uff08n\uff0b1\uff09p\u4e0d\u4e3a\u6574\u6570\u65f6\uff0c\u4e8c\u9879\u6982\u7387P\uff5bX\uff1dk\uff5d\u5728k\uff1d\uff3b\uff08n\uff0b1\uff09p\uff3d\u65f6\u8fbe\u5230\u6700\u5927\u503c\uff1b\u5f53\uff08n\uff0b1\uff09p\u4e3a\u6574\u6570\u65f6\uff0c\u4e8c\u9879\u6982\u7387P\uff5bX\uff1dk\uff5d\u5728k\uff1d\uff08n\uff0b1\uff09p\u548ck\uff1d\uff08n\uff0b1\uff09p\uff0d1\u65f6\u8fbe\u5230\u6700\u5927\u503c\u3002
\u5f53\u8bd5\u9a8c\u7684\u6b21\u6570\u8d8b\u4e8e\u65e0\u7a77\u5927\uff0c\u800c\u4e58\u79efnp\u56fa\u5b9a\u65f6\uff0c\u4e8c\u9879\u5206\u5e03\u6536\u655b\u4e8e\u6cca\u677e\u5206\u5e03\u3002\u56e0\u6b64\u53c2\u6570\u4e3a\u03bb\uff1dnp\u7684\u6cca\u677e\u5206\u5e03\u53ef\u4ee5\u4f5c\u4e3a\u4e8c\u9879\u5206\u5e03B\uff08n\uff0cp\uff09\u7684\u8fd1\u4f3c\uff0c\u8fd1\u4f3c\u6210\u7acb\u7684\u524d\u63d0\u8981\u6c42n\u8db3\u591f\u5927\uff0c\u800cp\u8db3\u591f\u5c0f\uff0cnp\u4e0d\u662f\u5f88\u5c0f\u3002
(1)\u7531P(X\u22651)=5/9\uff0c\u53ef\u5f97P(X=0)=4/9=(1-p)^2,\u6545p=1/3,\u4ece\u800c
P(Y\u22651)=1-(1-p)^3=26/27
(2)np \u4e58(1-p)^{n-1}= n(n-1)/2 \u4e58p^2 \u4e58(1-p)^{n-2}=2\u4e58n(n-1)(n-2)/6\u4e58p^3\u4e58(1-p)^{n-3}
\u89e3\u5f97n=5, p=1/3
P(X>=1)=1-P(X<1)=1-P(X=0)
这样应该可以做了，从上述式子中得到P，再求P(Y>=1)就不难了。。。。。
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