设随机变量x服从[0,5]均匀分布,用Y表示X的3次独立重复观察中事件(x>2)出现的次数,则Y服从什么分布 设随机变量X的概率密度函数为f(x)=2x,0<x<1,以Y...

\u8bbe\u968f\u673a\u53d8\u91cfX\u5728\u533a\u95f4\u30102\uff0c5\u3011\u4e0a\u670d\u4ece\u5747\u5300\u5206\u5e03\uff0c\u6c42\u5bf9X\u76843\u6b21\u72ec\u7acb\u89c2\u6d4b\u4e2d\uff0c\u81f3\u5c11\u4e24\u6b21\u7684\u89c2\u6d4b\u503c\u5927\u4e8e3\u7684\u6982\u7387\u3002

x>3\u7684\u6982\u7387\u662f\u3010\uff085-3\uff09/\uff085-2\uff09\u3011=2/3
\u90a3\u4e48\u81f3\u5c11\u6709\u4e24\u6b21\u5927\u4e8e3\u7684\u6982\u7387\u662f\uff1a
\uff083*2*1/2/1\uff09\uff082/3\uff09*\uff082/3\uff09*\uff081/3\uff09+\uff082/3\uff09*\uff082/3\uff09*\uff082/3\uff09=12/27+8/27=20/27
\u5373\u5927\u4e8e3\u7684\u6982\u7387\u662f20/27
\u5747\u5300\u5206\u5e03\u7684\u968f\u673a\u53d8\u91cf\u843d\u5728\u56fa\u5b9a\u957f\u5ea6\u7684\u4efb\u4f55\u95f4\u9694\u5185\u7684\u6982\u7387\u4e0e\u533a\u95f4\u672c\u8eab\u7684\u4f4d\u7f6e\u65e0\u5173\uff08\u4f46\u53d6\u51b3\u4e8e\u95f4\u9694\u5927\u5c0f\uff09\uff0c\u53ea\u8981\u95f4\u9694\u5305\u542b\u5728\u5206\u5e03\u7684\u652f\u6301\u4e2d\u5373\u53ef\u3002
\u6269\u5c55\u8d44\u6599\uff1a
\u968f\u673a\u53d8\u91cf\u5728\u4e0d\u540c\u7684\u6761\u4ef6\u4e0b\u7531\u4e8e\u5076\u7136\u56e0\u7d20\u5f71\u54cd\uff0c\u53ef\u80fd\u53d6\u5404\u79cd\u4e0d\u540c\u7684\u503c\uff0c\u6545\u5176\u5177\u6709\u4e0d\u786e\u5b9a\u6027\u548c\u968f\u673a\u6027\uff0c\u4f46\u8fd9\u4e9b\u53d6\u503c\u843d\u5728\u67d0\u4e2a\u8303\u56f4\u7684\u6982\u7387\u662f\u4e00\u5b9a\u7684\u3002
\u82e5a = 0\u5e76\u4e14b = 1\uff0c\u6240\u5f97\u5206\u5e03U\uff080,1\uff09\u79f0\u4e3a\u6807\u51c6\u5747\u5300\u5206\u5e03\u3002\u6807\u51c6\u5747\u5300\u5206\u5e03\u7684\u4e00\u4e2a\u6709\u8da3\u7684\u5c5e\u6027\u662f\uff0c\u5982\u679cu1\u5177\u6709\u6807\u51c6\u5747\u5300\u5206\u5e03\uff0c\u90a3\u4e481-u1\u4e5f\u662f\u5982\u6b64\u3002
\u7edf\u8ba1\u5b66\u4e2d\uff0c\u5f53\u4f7f\u7528p\u503c\u4f5c\u4e3a\u7b80\u5355\u96f6\u5047\u8bbe\u7684\u68c0\u9a8c\u7edf\u8ba1\u91cf\uff0c\u5e76\u4e14\u68c0\u9a8c\u7edf\u8ba1\u91cf\u7684\u5206\u5e03\u662f\u8fde\u7eed\u7684\uff0c\u5219\u5982\u679c\u96f6\u5047\u8bbe\u4e3a\u771f\uff0c\u5219p\u503c\u5747\u5300\u5206\u5e03\u57280\u548c1\u4e4b\u95f4\u3002

\u9996\u5148\uff0c\u6839\u636ex\u7684\u6982\u7387\u5bc6\u5ea6\u7b97\u51fap(X<=O.5)=1/4\uff08\u7528\u6982\u7387\u5bc6\u5ea6\u51fd\u6570\u57280\u52300.5\u79ef\u5206\uff09
P(Y=2)\uff0c\u5373\u4e09\u6b21\u72ec\u7acb\u91cd\u590d\u4e8b\u4ef6\u201cX<=O.5\u201d\u51fa\u73b0\u4e24\u6b21
\u5219P(Y=2)= C 2\uff08\u4e0a\u6807\uff093\uff08\u4e0b\u6807\uff09 *(1/4)*(1/4)*(3/4)=9/64,

解:P(X<=1/2)=1/4
Y~(3,1/4)
P(Y=1)=C(3,1)1/4*(3/4)^2=27/64
如有意见,欢迎讨论,共同学习;如有帮助,请选为满意回答!

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