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1、Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,第1课时概率,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Cl
2、ient Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,2互斥事件有一个发生的概率 P(AB)P(A)P(B),Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2
3、011 Aspose Pty Ltd.,(2012山东卷)袋中有五张卡片,其中红色卡片三张,标号分别为1,2,3;蓝色卡片两张,标号分别为1,2. (1)从以上五张卡片中任取两张,求这两张卡片颜色不同且标号之和小于4的概率; (2)向袋中再放入一张标号为0的绿色卡片,从这六张卡片中任取两张,求这两种卡片颜色不同且标号之和小于4的概率,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,解析:(1)标号为1,2,3的
4、三张红色卡片分别记为A,B,C,标号为1,2的两张蓝色卡片分别记为D,E,从五张卡片中任取两张的所有可能的结果为(A,B),(A,C),(A,D),(A,E),(B,C),(B,D),(B,E),(C,D),(C,E),(D,E),共10种由于每一张卡片被取到的机会均等,因此这些基本事件的出现是等可能的 从五张卡片中任取两张,这两张卡片颜色不同且它们的标号之和小于4的结果为:(A,D),(A,E),(B,D),共3种,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyrigh
5、t 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,(1)有关古典概型的概率问题,关键是正确求出基本事件总数和所求事件包含的基本
6、事件数 (2)在用列举法把所有基本事件一一列出时,要做到不重复、不遗漏,可借助“树状图”列举,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,1(2012北京西城一模)某校高一年级开设研究性学习课程,一班和二班报名参加的人数分别是18和27.现用分层抽样的方法,从中抽取若干名学生组成研究性学习小组,已知从二班抽取了3名同学 (1)求研究性学习小组的人数; (2)规划在研究性学习的中、后期各安排1次交流活动,每次
7、随机抽取小组中1名同学发言求2次发言的学生恰好来自不同班级的概率,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Asp
8、ose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,(2012湖南卷)某超市为了解顾客的购物量及结算时间等信息,安排一名员工随机收集了在该超市购物的100位顾客的相关数据,如下表所示.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,已知这100位顾客中一次购物量超过8件的顾客占55%. (1)确
9、定x,y的值,并估计顾客一次购物的结算时间的平均值; (2)求一位顾客一次购物的结算时间不超过2分钟的概率(将频率视为概率),Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Eva
10、luation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,(1)当所求事件情况较复杂时,一般要分类计算,即用互斥事件的概率加法公式或考虑其对立事件求解 (2)在解决与互斥事件有关问题时,首先分清所
11、求事件是由哪些事件组成的,是否具备互斥的条件,一个事件是由几个互斥事件组成的,做到不重、不漏 (3)当所求事件含有“至少”“至多”或分类情况较多时,通常考虑用对立事件的概率公式求解,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2
12、004-2011 Aspose Pty Ltd.,(1)求这组志愿者的人数; (2)现在从这组志愿者中选出通晓英语的志愿者1名,通晓韩语的志愿者1名,若甲通晓英语,乙通晓韩语,求甲和乙不全被选中的概率,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,(2)设通晓中文和英语的人为A1,A2,A3,A4,A5,甲为A1,通晓中文和韩语的人为B1,B2,乙为B1,则从这组志愿者中选出通晓英语和韩语的志愿者各1名的所有
13、情况为(A1,B1),(A1,B2),(A2,B1),(A2,B2),(A3,B1),(A3,B2),(A4,B1),(A4,B2),(A5,B1),(A5,B2),共10种, 同时选中甲、乙只有(A1,B1)1种,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2
14、.0.0. Copyright 2004-2011 Aspose Pty Ltd.,答案:B,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,(1)当试验的结果构成的区域为长度、面积、体积、弧长、夹角等时,应考虑使用几何概型求解 (2)利用几何概型求概率时,关键是试验的全部结果构成的区域和事件发生的区域的寻找,有时需要设出变量,在坐标系中表示所需要的区域,Evaluation only. Created wit
15、h Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 200
16、4-2011 Aspose Pty Ltd.,探究概率应用题的新题点 高考对概率的考查由单独求事件概率,转化为与统计结合命题,而2012年高考中,全国新课标卷、福建卷把概率问题分别与函数、数列结合,出题角度新颖,将会成为今后新的命题点,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd.,(2012新课标全国卷)某花店每天以每枝5元的价格从农场购进若干枝玫瑰花,然后以每枝10元的价格出售,如果当天卖不完,剩下的玫瑰花作垃圾处理 (1)若花店一天购进17枝玫瑰花,求当天的利润y(单位:元)关于当天需求量n(单位:枝,nN)的函数解析式 (2)花店记录了100天玫瑰花的日需求量(单位:枝),整理得下表:,Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspos