高二数学概率题!
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P\uff5b\u4e24\u4eba\u5728\u4e0d\u540c\u5c42\u79bb\u5f00\u7684\u6982\u7387\uff5d\uff1d1\uff0dP\uff5b\u4e24\u4eba\u5728\u540c\u4e00\u5c42\u79bb\u5f00\u7684\u6982\u7387\uff5d
\uff1d1\uff0d6/(6\u00d76)
=5/6
\u7b54\uff1a\u4e24\u4eba\u5728\u4e0d\u540c\u5c42\u79bb\u5f00\u7684\u6982\u7387\u662f5/6.
\u4f1a\u52a0\u5206\u5417\uff1f
0 1 2 3
\u6982\u7387 0.008 0.096 0.384 0.512
E=0*0.008+1*0.096+2*0.384+3*0.512
=2.4
\uff082\uff09\u81f3\u5c11\u6709\u4e00\u6b21\u51fb\u4e2d\u76ee\u6807\u7684\u6982\u7387=1-0.008=0.992
最后是甲方获胜,说明最后一场(第9场)一定甲方获胜,而且前面的8场双方各赢4场.
这就相当于在前面的8场(独立重复试验8次)中,甲需要获得其中任意4场胜利,所以概率是:
[(1/2)^4]*[(1/2)^(8-4)]*C(8,4)=35/128.
而最后一场甲一定要胜,概率为1/2,所以总的概率是:
(35/128)*(1/2)=35/256.
这一共有5种情况,即用一人,二人,三人,四人,五人打完比赛,而4名队员被淘汰的概率就是1/5,因为甲,乙胜的概率都是1/2,所以甲方有 4 名队员被淘汰且最后战胜乙方的概率是1/5*1/2=1/10.
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绛旓細姒傜巼涓篶5/1*3*c4/2/3^5=30/81 鏁呮鐜囧拰涓簆=20/81+30/81=50/81
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