(1)山水城市鎮(zhèn)江有“三山”--金山、焦山、北固山,一位游客游覽這三個景點的概率都是0.5,且該游客是否游覽這三個景點相互獨立,用ξ表示這位游客游覽的景點數(shù)和沒有游覽的景點數(shù)差的絕對值,求ξ的分布列和數(shù)學(xué)期望;
(2)某城市有n(n為奇數(shù),n≥3)個景點,一位游客游覽每個景點的概率都是0.5,且該游客是否游覽這n個景點相互獨立,用ξ表示這位游客游覽的景點數(shù)和沒有游覽的景點數(shù)差的絕對值,求ξ的分布列和數(shù)學(xué)期望.
解:(1)游客游覽景點個數(shù)為0,1,2,3,ξ可能取值為:1,3,
P(ξ=1)=
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+
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=2
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=
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,
P(ξ=3)=
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+
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=2
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=
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,
ξ的分布列為:
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所以Eξ=1×
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+3×
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=
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.
(2)當(dāng)n=2k+1,k∈N
*時,游客游覽景點個數(shù)可能為:0,1,2,…,2k+1,
ξ可能取值為:1,3,5,…,2k+1.
P(ξ=1)=
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+
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=2×
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;
P(ξ=3)=
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+
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=
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;
…
P(ξ=2k+1)=
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+
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=2×
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,
∴ξ的分布列為:
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∴Eξ=(2k+1-0)×2×
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+[(2k+1-1)-1]×2×
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+[(2k+1-2)-2]×2×
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+…+[2k+1-k)-k]×2×
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=2×
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{[(2k+1)
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+2k
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+(2k-1)
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+…+(2k+1-k)
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]-[(0×
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+1
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+2×
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+…+
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]}
=2×
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{[(2k+1)×
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+2k×
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+(2k-1)×
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+…+(k+1)
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]-[0×
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+1×
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+…+
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]},
∵
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=n
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(i=1,2,3,…,n),
Eξ=2×
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{(2k+1)×[
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]-(2k+1)×[
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]}
=2×
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×(2k+1)×[(
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)-(
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+
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)]
=2×
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×(2k+1)×
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=
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.
答:ξ的數(shù)學(xué)期望Eξ為
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.
分析:(1)游客游覽景點個數(shù)為0,1,2,3,ξ可能取值為:1,3,ξ=1表示游覽一個景點或游覽兩個景點,ξ=3表示游覽景點數(shù)為0或游覽了三個景點,根據(jù)n次獨立重復(fù)試驗中事件發(fā)生k的概率公式即可求得P(ξ=1),P(ξ=3),進(jìn)而得到分布列和期望;
(2)當(dāng)n=2k+1,k∈N
*時,游客游覽景點個數(shù)可能為:0,1,2,…,2k+1,則ξ可能取值為:1,3,5,…,2k+1.根據(jù)獨立重復(fù)試驗中事件A發(fā)生k次的概率計算公式求出ξ取各值是的概率,表示出Eξ=(2k+1-0)×2×
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+[(2k+1-1)-1]×2×
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+[(2k+1-2)-2]×2×
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+…+[2k+1-k)-k]×2×
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,分組后利用性質(zhì)
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=n
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(i=1,2,3,…,n)對上式即可進(jìn)行化簡,最后再換為n即可;
點評:本題考查離散型隨機(jī)變量的分布列、期望,考查n次獨立重復(fù)試驗中事件A發(fā)生k的概率計算公式,考查組合數(shù)性質(zhì)應(yīng)用,考查學(xué)生綜合運(yùn)用知識分析解決問題的能力,本題綜合性強(qiáng),能力要求高,屬難題.