题目内容
设集合
,
,分别从集合
和
中随机取一个数
和
.
(1)若向量
,
,求向量
与
的夹角为锐角的概率;
(2) 记点
,则点
落在直线
上为事件
,
求使事件
的概率最大的
.
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(1)若向量
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(2) 记点
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求使事件
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(1)
;(2)3或4.
解:(1) 设向量
与
的夹角为
因为
为锐角 ∴
,且向量
与
不共线,因为
,
,
显然
与
不共线,所以,
,
分别从集合
和
中随机取一个数
和
的基本事件有;
所以向量
与
的夹角为锐角的概率
(2)由(Ⅰ)知;当
时,满足条件的概率
当
时,满足条件的概率
当
时,满足条件的概率
当
时,满足条件的概率
所以使事件
的概率最大的
值为
或
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解:(1) 设向量
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因为
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显然
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分别从集合
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所以向量
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(2)由(Ⅰ)知;当
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当
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当
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当
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所以使事件
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本试题主要考查了古典概型的概率的求解,以及运用分类讨论的思想求解概率的最值。
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