解:(1)小李獨立參加每次考核合格的概率依次組成一個公差為
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的等差數(shù)列,
且他直到第二次考核才合格的概率為
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.
得(1-p
1)(p
1+
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)=
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,
解得p
1=
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或p
1=
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.
∵p
1≤
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,∴p
1=
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,
即小李第一次參加考核就合格的概率為
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(2)由(1)的結(jié)論知,ξ的可能取值是1,2,3,4
小李四次考核每次合格的概率依次為
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,
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,
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,
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,
∴P(ξ=1)=
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,P(ξ=2)=
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,
P(ξ=3)=(1-
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)(1-
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)
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=
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P(ξ=4)=(1-
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)•(1-
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)•(1-
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)•1=
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∴小李參加測試的次數(shù)的數(shù)學期望為Eξ=1•
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+2•
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+3•
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+4•
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=
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分析:(1)小李獨立參加每次考核合格的概率依次組成一個公差為
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的等差數(shù)列,他直到第二次考核才合格表示他第一次不合格第二次才合格,這兩個事件是相互獨立的,寫出概率的關(guān)系式,列出方程,得到結(jié)果.
(2)小李參加考核的次數(shù)ξ,ξ的可能取值是1,2,3,4,小李四次考核每次合格的概率依次為
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,
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,
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,
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,根據(jù)相互獨立事件同時發(fā)生的概率,得到分布列和期望.
點評:本題考查離散型隨機變量的分布列和期望,考查相互獨立事件同時發(fā)生的概率,考查利用概率知識解決實際問題的能力,是一個綜合題目.