计算机问题求解一论题2-7 离散概率基础 2016年04月7日
计算机问题求解 – 论题2-7 - 离散概率基础 2016年04月7日
预习检查 To compute probabilities,we assign a probability weight P(x)to each element of the sample space so that the weight represents what we believe to be the relative likelihood of that outcome.There are two rules in assigning weights.First,the weights must be nonnegative numbers,and second,the sum of the weights of all the elements in a sample space must be 1.We define the probability P(E)of the event E to be the sum of the weights of the elements of E.Algebraically,we write P(E)=∑Px (5.1) x:x∈E
预习检查
问题1： 你能否给我们讲讲书上那个关 于邮购商店的“故事”，那里 什么地方体现了“碰运气”？
If we have a table with 100 buckets and 50 keys to put in those buckets, it is possible that all 50 of those keys could be assigned (hashed)to the same bucket in the table.However,someone who is experienced with using hash functions will tell you that you'd never see this in a million years. But that same person might also tell you that neither would you ever see. in a million years,all the keys hash into different locations.In fact,it is far less likely that all 50 keys would hash into one place than that all 50 keys would hash into different places,but both events are quite unlikely.Being able to understand just how likely or unlikely such events are is a major reason for taking up the study of probability 间题2： 这里的you'd never..neither would you ever.有什么意义？
离散概率模型 间题3： 你能香以下面的过程为例解释 离散概率模型中的主要概念？ Sample space Element Event Probability A process:掷两个色子
离散概率模型 A process: 掷两个色子 Sample space Element Event Probability
Axioms for a probability space 满足下列性质的P称为一个probability distribution或者一个 probability measure. 1.P(A)≥0 for any A C S. 2.P(S)=1. 3.P(AU B)=P(A)+P(B)for any two disjoint events A and B. 记住：P是一个函数。 问题4： 为什台任何事件的概率值不会大于1？ 如果A和B相交，P(AUB)是什么？
Axioms for a probability space 满足下列性质的P 称为一个probability distribution 或者一个 probability measure。 记住：P 是一个函数
间题5： 有限样本空间中的离散 概率与计数有什么关系？ P(E)=∑Px). x:x∈E
有限样本空间 There are only finite outcomes. Each outcome individually consists an elementary event. For one coin toss,there are two outcomes-head and tail "Head"is an elementary event. The probability of an elementary event corresponds a specific outcome. ■ If all outcomes are equally likely,then the probability of an event E can be computed as: p(E)= E totalnumberof outcomesin E total number of outcomes
有限样本空间 There are only finite outcomes. Each outcome individually consists an elementary event. For one coin toss, there are two outcomes – head and tail. “Head” is an elementary event. The probability of an elementary event corresponds a specific outcome. If all outcomes are equally likely, then the probability of an event E can be computed as: totalnumber of outcomes totalnumber of outcomesin | | | | ( ) E A E p E
问题6： 你能举一个不满足equally lkey分布的例子吗？ 这种情况下，概率公式应该 如何确定？
交集非空的事件 掷均匀的色子，掷3次。出现事件“或者3次均相等，或者 没有一次是4”的概率是多少？ 合理假设：每个outcome出现的可能性是一样的 ■样本空间大小是63=216。 This is a special case of so-called inclusion- 用F表示事件“3次结果一样” exclusion principle 用G表示事件“没有一次结果是4 从集合{1,2,3,5,6}中任选3个数（可重复）的方案数) 要求的事件为F和G的并集： IFUG=lF+|G-FnG=6+125-5=126 因此，最终结果是：126/216=7/12
交集非空的事件 掷均匀的色子，掷3次。出现事件“或者3次均相等，或者 没有一次是4”的概率是多少？ 合理假设：每个outcome出现的可能性是一样的。 样本空间大小是 6 3=216。 用F表示事件“3次结果一样”，则|F|=6 (F={111,222,…,666}) 用G表示事件“没有一次结果是4”，则|G|=53=125 (G是 从集合{1,2,3,5,6}中任选3个数（可重复）的方案数) 要求的事件为F和G的并集: |FG|=|F|+|G| 因此，最终结果是：126/216 = 7/12 This is a special case of so-called inclusionexclusion principle -|FG|=6+125-5=126
