## 数学代写|概率论代写Probability theory代考|POPH90148

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写概率论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|概率论代写Probability theory代考|Addition and Multiplication Rules

There are two helpful rules for counting, phrased in terms of “jobs” which are to be done.

1. The Addition Rule: Suppose we can do job 1 in $p$ ways and job 2 in $q$ wavs. Then we can do either job I OR job 2 (but not both), in $p+q$ ways. instructor can pick one student to answer a question. If there are 5 vowels and 20 consonants on a list and I must pick one letter, this can be done in $5+20 \text { ways. }$
2. The Multiplication Rule: Suppose we can do job I in p ways and, for each of these ways. we can do job 2 in $q$ ways. Then we can do both job I AND job 2 in $p \times q$ ways.

For example, if there are 5 vowels and 20 consonants and I must choose one consonant followed by one vowel for a two-letter word, this can be done in $20 \times 5$ ways (there are 100 such words). To ride a bike, you must have the chain on both a front sprocket and a rear sprocket. For a 21 speed bike there are 3 ways to select the front sprocket and 7 ways to select the rear sprocket, which gives $3 \times 7=21$ such combinations.

This interpretation of “OR” as addition and “AND” as multiplication evident in the addition and multiplication rules above will occur throughout probability, so it is helpful to make this association in your mind. Of course questions do not always have an AND or an OR in them and you may have to play around with re-wording the question to discover implied AND’s or OR’s.

Example: Suppose we pick 2 numbers from digits 1, 2, 3, 4, 5 with replacement. (Note: “with replacement” means that after the first number is picked it is “replaced” in the set of numbers, so it could be picked again as the second number.) Assume a uniform distribution on the sample space, that is, assume that every pair of numbers has the same probability. Let us find the probability that one number is even. This can be reworded as: “The first number is even AND the second is odd (this can be done in $2 \times 3$ ways) OR the first is odd AND the second is even (done in $3 \times 2$ ways).” Since these are connected with the word OR, we combine them using the addition rule to calculate that there are $(2 \times 3)+(3 \times 2)=12$ ways for this event to occur. Since the first number can be chosen in 5 ways AND the second in 5 ways, $S$ contains $5 \times 5=25$ points and since each point has the same probability, they all have probability $\frac{1}{25}$. Therefore
$$P \text { (one number is even })=\frac{12}{25}$$
When objects are selected and replaced after each draw, the addition and multiplication rules are generally sufficient to find probabilities. When objects are drawn without being replaced, some special rules may simplify the solution.

## 数学代写|概率论代写Probability theory代考|Counting Subsets or Combinations

In some problems, the outcomes in the sample space are subsets of a fixed size. Here we look at counting such subsets. Again, it is useful to write a short list of the subsets you are counting.

Example: Suppose we randomly select a subset of three digits from the set ${0,1,2,3,4,5,6,7,8,9}$ so that the sample space is
$$S={{1,9,3},{0,1,3},{0,1,4}, \ldots, 8,9}}$$
All the digits in each outcome are unique, that is, we do not consider ${1,1,2}$ to be a subset of $S$. Also, the order of the elements in a subset is not relevant. This is true in general for sets; the subsets ${1,2,3}$ and ${3,1,2}$ are the same. To count the number of outcomes in $S$, we use what we have learned about counting arrangements. Suppose there are $m$ such subsets. Using the elements of any subset of size 3 , we can form 3 ! arrangements of length 3 . For example, the subset ${1,2,3}$ generates the $3 !=6$ arrangements $123,132,213,231,312,321$ and any other subset generates a different $3 !$ arrangements so that the total number of arrangements of 3 digits taken without replacement from the set ${0,1,2,3,4,5,6,7,8,9}$ is $3 ! \times m$. But we know the total number of arrangements is $10^{(3)}$ so $3 ! \times m=10^{(3)}$. Solving we get
$$m=\frac{10^{(3)}}{3 !}=120$$
Number of subsets of size $k$ : We use the combinatorial symbol $\left(\begin{array}{l}n \ k\end{array}\right)$ (” $n$ choose $k$ “) to denote the number of subsets of size $k$ that can be selected from a set of $n$ objects. By an argument similar to that above, if $m$ denotes the number of subsets of size $k$ that can be selected from $n$ things, then $m \times k !=n^{(k)}$ and so we have
$$m=\left(\begin{array}{l} n \ k \end{array}\right)=\frac{n^{(k)}}{k !}$$
In the example above we selected the subset at random so each of the $\left(\begin{array}{c}10 \ 3\end{array}\right)=120$ subsets has the same probability $\frac{1}{120}$. We now find the probability of the following events:
$A$ : the digit 1 is included in the selected subset
$B$ : all the digits in the selected subset are even
$C$ : at least one of the digits in the selected subset is less than or equal to 5
To count the outcomes in event $A$, we must have 1 in the subset and we can select the other two elements from the remaining 9 digits in $\left(\begin{array}{l}9 \ 2\end{array}\right)$ ways. And so
$$P(A)=\frac{\left(\begin{array}{c} 9 \ 2 \end{array}\right)}{\left(\begin{array}{c} 10 \ 3 \end{array}\right)}=\frac{9^{(2)} / 2 !}{10^{(3)} / 3 !}=\frac{3}{10}$$

# 概率论代考

## 数学代写|概率论代写Probability theory代考|Addition and Multiplication Rules

1. 加法规则：假设我们可以在 $p$ 方法和工作 $2 q$ 波浪。然后我们可以做工作 $\mid$ 或工作 2 （但不能同时 做），在 $p+q$ 方法。讲师可以挑选一名学生回答问题。如果列表中有 5 个元音字母和 20 个辅音 字母，我必须选择一个字母，这可以用 $\$ 5+20 \backslash$Itext${$方式完成。$} \$$2. 乘法规则：假设我们可以用 p 种方式完成工作 I ，并且对于其中的每一种方式。我们可以做工作 2 q 方法。然后我们可以同时完成工作 I 和工作 2 p \times q 方法。 例如，如果有 5 个元音和 20 个辅音，我必须选择一个辅音后跟一个元音来表示一个两个字母的单词， 这可以在 20 \times 5 方式 (有 100 个这样的词) 。要骑自行车，您必须在前链轮和后链轮上都安装链条。对 于 21 速自行车，有 3 种方式选择前链轮和 7 种方式选择后链轮，这给出了 3 \times 7=21 这样的组合。 在上述加法和乘法规则中，将“或”解释为加法，将“与”解释为乘法，这种解释在整个概率过程中都会出 现，因此在您的脑海中形成这种联想是有帮助的。当然，问题中并不总是包含 AND 或 OR，您可能不得 不尝试重新措辞问题以发现隐含的 AND 或 O R 。 示例：假设我们从数字 1、2、3、4、5 中选择 2 个数字并进行替换。（注： “with replacement”是指第 一个数被选出后，它在数集中被“替换”，所以它可以作为第二个数再次被选出。）假设在样本空间上均 匀分布，即假设每对数字都有相同的概率。让我们找出一个数是偶数的概率。这可以改写为：“第一个数 字是偶数，第二个是奇数（这可以在 2 \times 3 方式）或第一个是奇数，第二个是偶数（完成 3 \times 2 方 法)。”由于这些与单词OR相关联，因此我们使用加法规则将它们组合在一起以计算出有 (2 \times 3)+(3 \times 2)=12 此事件发生的方式。由于第一个数字有 5 种选择方式，第二个有 5 种选择方 式， S 包含 5 \times 5=25 点，因为每个点都有相同的概率，所以它们都有概率 \frac{1}{25}. 因此 \ \$$
$P \backslash \operatorname{text}{($ 一个数是偶数 $})=\backslash \operatorname{frac}{12} 25}$ $\$ \$$在每次抽取后选择和替换对象时，加法和乘法规则通常足以找到概率。在不替换对象的情况下绘制对象 时，一些特殊规则可能会简化解决方案。 ## 数学代写|概率论代写Probability theory代考|Counting Subsets or Combinations 在某些问题中，样本空间中的结果是固定大小的子集。在这里，我们着眼于对此类子集进行计数。同 样，写下您正在计数的子集的简短列表很有用。 示例：假设我们从集合中随机选择三个数字的子集 0,1,2,3,4,5,6,7,8,9 使得样本空间为$$
\mathrm{S}={{1,9,3},{0,1,3},{0,1,4}, \backslash \text { Idots, } 8,9}}
$$每个结果中的所有数字都是唯一的，也就是说，我们不考虑 1,1,2 成为一个子集 S. 此外，子集中元素的 顺序无关紧要。对于集合来说，这通常是正确的；子集 1,2,3 和 3,1,2 是相同的。计算结果的数量 S ， 我们使用我们学到的关于计数安排的知识。假设有 m 这样的子集。使用大小为 3 的任何子集的元素，我 们可以形成 3 ! 长度的安排 3 . 例如，子集 1,2,3 生成 3 !=6 安排 123,132,213,231,312,321 和任 何其他子集生成一个不同的 3 !安排，使 3 位数字的安排总数从集合中取出而无需替换 0,1,2,3,4,5,6,7,8,9 是 3 ! \times m. 但是我们知道总的排列数是 10^{(3)} 所以 3 ! \times m=10^{(3)}. 解决我们 得到$$
m=\frac{10^{(3)}}{3 !}=120
$$大小的子集数 k : 我们使用组合符号 (n k) (” n 选择 k^{\prime \prime} ) 表示大小的子集数 k 可以从一组中选择 n 对象。通 过与上述类似的论证，如果 m 表示大小的子集数 k 可以从中选择 n 事情，那么 m \times k !=n^{(k)} 所以我们 有$$
m=(n k)=\frac{n^{(k)}}{k !}
$$在上面的示例中，我们随机选择了子集，因此每个 (103)=120 子集具有相同的概率 \frac{1}{120}. 我们现在找 到以下事件的概率: A : 数字 1 包含在所选子集中 B : 所选子集中的所有数字都是偶数 C : 所选子集中至少有一位数字小于或等于 5 来统计事件的结果 A ，我们必须在子集中有 1 并且我们可以从中的剩余 9 位中选择其他两个元素 (92) 方 法。所以$$
P(A)=\frac{(92)}{(103)}=\frac{9^{(2)} / 2 !}{10^{(3)} / 3 !}=\frac{3}{10}
$$统计代写请认准statistics-lab™. statistics-lab™为您的留学生涯保驾护航。 ## 金融工程代写 金融工程是使用数学技术来解决金融问题。金融工程使用计算机科学、统计学、经济学和应用数学领域的工具和知识来解决当前的金融问题，以及设计新的和创新的金融产品。 ## 非参数统计代写 非参数统计指的是一种统计方法，其中不假设数据来自于由少数参数决定的规定模型；这种模型的例子包括正态分布模型和线性回归模型。 ## 广义线性模型代考 广义线性模型（GLM）归属统计学领域，是一种应用灵活的线性回归模型。该模型允许因变量的偏差分布有除了正态分布之外的其它分布。 术语 广义线性模型（GLM）通常是指给定连续和/或分类预测因素的连续响应变量的常规线性回归模型。它包括多元线性回归，以及方差分析和方差分析（仅含固定效应）。 ## 有限元方法代写 有限元方法（FEM）是一种流行的方法，用于数值解决工程和数学建模中出现的微分方程。典型的问题领域包括结构分析、传热、流体流动、质量运输和电磁势等传统领域。 有限元是一种通用的数值方法，用于解决两个或三个空间变量的偏微分方程（即一些边界值问题）。为了解决一个问题，有限元将一个大系统细分为更小、更简单的部分，称为有限元。这是通过在空间维度上的特定空间离散化来实现的，它是通过构建对象的网格来实现的：用于求解的数值域，它有有限数量的点。边界值问题的有限元方法表述最终导致一个代数方程组。该方法在域上对未知函数进行逼近。[1] 然后将模拟这些有限元的简单方程组合成一个更大的方程系统，以模拟整个问题。然后，有限元通过变化微积分使相关的误差函数最小化来逼近一个解决方案。 tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。 ## 随机分析代写 随机微积分是数学的一个分支，对随机过程进行操作。它允许为随机过程的积分定义一个关于随机过程的一致的积分理论。这个领域是由日本数学家伊藤清在第二次世界大战期间创建并开始的。 ## 时间序列分析代写 随机过程，是依赖于参数的一组随机变量的全体，参数通常是时间。 随机变量是随机现象的数量表现，其时间序列是一组按照时间发生先后顺序进行排列的数据点序列。通常一组时间序列的时间间隔为一恒定值（如1秒，5分钟，12小时，7天，1年），因此时间序列可以作为离散时间数据进行分析处理。研究时间序列数据的意义在于现实中，往往需要研究某个事物其随时间发展变化的规律。这就需要通过研究该事物过去发展的历史记录，以得到其自身发展的规律。 ## 回归分析代写 多元回归分析渐进（Multiple Regression Analysis Asymptotics）属于计量经济学领域，主要是一种数学上的统计分析方法，可以分析复杂情况下各影响因素的数学关系，在自然科学、社会和经济学等多个领域内应用广泛。 ## MATLAB代写 MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 ## 数学代写|概率论代写Probability theory代考|STAT4061 如果你也在 怎样代写概率论Probability theory这个学科遇到相关的难题，请随时右上角联系我们的24/7代写客服。 概率论是与概率有关的数学分支。虽然有几种不同的概率解释，但概率论以严格的数学方式处理这一概念，通过一套公理来表达它。 statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写概率论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。 我们提供的概率论Probability theory及其相关学科的代写，服务范围广, 其中包括但不限于: • Statistical Inference 统计推断 • Statistical Computing 统计计算 • Advanced Probability Theory 高等概率论 • Advanced Mathematical Statistics 高等数理统计学 • (Generalized) Linear Models 广义线性模型 • Statistical Machine Learning 统计机器学习 • Longitudinal Data Analysis 纵向数据分析 • Foundations of Data Science 数据科学基础 ## 数学代写|概率论代写Probability theory代考|Sample Spaces and Probability Consider some phenomenon or process which is repeatable, at least in theory, and suppose that certain events or outcomes A_1, A_2, A_3, \ldots are defined. We will often term the phenomenon or process an “experiment” and refer to a single repetition of the experiment as a “trial”. The probability of an event A, denoted P(A), is a number between 0 and 1 . For probability to be a useful mathematical concept, it should possess some other properties. For example, if our “experiment” consists of tossing a coin with two sides, Head and Tail, then we might wish to consider the two events A_1= “Head turns up” and A_2= “Tail turns up”. It does not make much sense to allow P\left(A_1\right)=0.6 and P\left(A_2\right)=0.6, so that P\left(A_1\right)+P\left(A_2\right)>1. (Why is this so? Is there a fundamental reason or have we simply adopted 1 as a convenient scale?) To avoid this sort of thing we begin with the following definition. Definition 1 A sample space S is a set of distinct outcomes for an experiment or process, with the property that in a single trial. one and only one of these outcomes occurs. The outcomes that make up the sample space may sometimes be called “sample points” or just “points” on occasion. A sample space is defined as part of the probability model in a given setting but it is not necessarily uniquely defined, as the following example shows. Example: Roll a six-sided die, and define the events$$
a_i=\text { there are } i \text { pips on the top face, for } i=1,2, \ldots, 6
$$Then we could take the sample space as S=\left{a_1, a_2, \ldots, a_6\right}. (Note we use the curly brackets ” {\ldots} ” to indicate the elements of a set). Instead of using this definition of the sample space we could instead define the events E : the event that there are an even number of pips on the top face Q : the event that there are an odd number of pips on the top face and take S={E, O}. Both sample spaces satisfy the definition. Which one we use depends on what we wanted to use the probability model for. If we expect never to have to consider events like “there are less than three pips on the top face” then the space S={E, O} will suffice, but in most cases, if possible, we choose sample points that are the smallest possible or “indivisible”. Thus the first sample space is likely preferred in this example. ## 数学代写|概率论代写Probability theory代考|SAMPLE SPACES AND PROBABILITY Definition 4 The probability P(A) of an event A is the sum of the probabilities for all the simple events that make up A or P(A)=\sum_{a \in A} P(a). For example, the probability of the compound event A=\left{a_1, a_2, a_3\right} is P\left(a_1\right)+P\left(a_2\right)+P\left(a_3\right). Probability theory does not say what numbers to assign to the simple events for a given application, only those properties guaranteeing mathematical consistency. In an actual application of a probability model, we try to specify numerical values of the probabilities that are more or less consistent with the frequencies of events when the experiment is repeated. In other words we try to specify probabilities that are consistent with the real world. There is nothing mathematically wrong with a probability model for a toss of a coin that specifies that the probability of heads is zero, except that it likely won’t agree with the frequencies we obtain when the experiment is repeated. Example: Suppose a six-sided die is rolled, and let the sample space be S={1,2, \ldots, 6}, where i represents the simple event that there are i pips on the top face, i=1,2, \ldots, 6. If the die is an ordinary one, (a fair die) we would likely define probabilities as$$
P(i)=\frac{1}{6} \text { for } i=1,2, \ldots, 6

## 数学代写|概率论代写Probability theory代考|SAMPLE SPACES AND PROBABILITY

$$P(i)=\frac{1}{6} \text { for } i=1,2, \ldots, 6$$

(1) 指定样本空间 $S$.
(2) 为中的简单事件分配概率分布 $S$.
(3) 对于任何复合事件 $A$ ，寻找 $P(A)$ 通过添加构成的所有简单事件的概率 $A$.

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|概率论代写Probability theory代考|STAT4028

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写概率论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|概率论代写Probability theory代考|The GenERAL CONSTRUCTION OF THE PROBABILITY SPACE

We have considered the variants of probabilistic spaces in situations where the number of outcomes of some experiment is finite or even countable. It should be noted that such schemes are very popular. Elementary events in such situations may be, for example, the following:
“the appearance of the six when throwing a die,”
“getting a ticket with the number 7 during a random selection of 24 examination tickets,”
“three defeats of a football team before its first victory in the championship,”
“the five-time appearance of the letter ” $s$ ” on the first page of a readable newspaper,”
“the winning combination of numbers $(2,8,11,22,27,31)$ falls out in the draw of a lotttery.”
However, many experiments do not fit into these discrete schemes. For example, the result of some experiment may be the coordinate of a randomly thrown point on a real line or the coordinates of a randomly thrown point on a unit square. Therefore, a further generalization of our construction of probability spaces must be useful.

Now let $\Omega={\omega}$ be an arbitrary (not necessarily, finite or countable) set of elementary events. When moving from $\Omega$ to a set of random events, problems may arise of the type,” which combinations of elementary outcomes can be taken as elements of $F$ ?..” The examples from the previous paragraph suggest that this choice is sufficiently arbitrary. The only condition is that the elements (random events) contained in $F$ must present some kind of configurations which could be called $\sigma$-algebra. The “poorest” and very exotic will be the $\sigma$-algebra, which includes only two elements – an impossible event $\theta$ and the authentic event $\Omega$. The next in simplicity but already actually used there may be an $\sigma$-algebra composed of 4 events $A, \bar{A}, \theta$ and, where as the event $A$ one can take an arbitrary union of elementary outcomes. Naturally, to solve any specific problems we must work with some more eventful set $F$. The only condition, as already was noted, is that this set must form an $\sigma$-algebra. For example, if $\Omega$ contains all the points of the real axis, then it is convenient (but not at all necessary!) to take the Borel $\sigma$-algebra containing all segments and their various combinations.

## 数学代写|概率论代写Probability theory代考|RANDOM VARIABLES AND DISTRIBUTION FUNCTIONS

Any probabilistic space constructed is inherently a card file in which a certain set of events and a set of probabilities corresponding to them are located, which determine the degrees of opportunities for the appearance of these events. In many cases, these probabilities could be found without building such heavy construction, which is a probabilistic space, but it turns out that this construction is necessary for defining and working with such important probabilistic object, which is a random variable.

The fact is that very often random outcomes of some experiment completely unrelated to any numbers or numbering can determine certain numerical characteristics depending on these outcomes.

Let’s give the simplest example. The International Football Federation (FIFA) is going to use a lot to determine where the qualifying match of the world championship between the teams of Russia and Finland will take place. The drum contains three cards with the names of the stadiums in St. Petersburg, Helsinki and the neutral field in Berlin. A randomly selected card must determine the city in which this match will take place. A fan from St. Petersburg who is going to visit this game without fail assesses his future expenses (depending on the choice of one of these three stadiums), respectively, as 3,000,10,000 and 20,000 rubles. For him, before the draw, the future cost is a random variable, taking one of these three values with equal probabilities $1 / 3$.

Let’s consider another example. A symmetric coin must be thrown three times. The possible outcomes of this experiment are expressed in terms of the appearance of the reverse or the face in each of these three tosses:
\begin{aligned} & \omega_1={r, r, r}, \omega_2={r, r, f}, \omega_3={r, f, r}, \omega_4={f, r, r}, \ & \omega_5={r, f, f}, \omega_6={f, r, f}, \omega_7={f, f, r}, \omega_8={f, f, f} . \end{aligned}
On the set, represented by these 8 elementary outcomes, you can specify various real functions.

# 概率论代考

## 数学代写|概率论代写Probability theory代考|The GenERAL CONSTRUCTION OF THE PROBABILITY SPACE

“掷骰子时出现 6″、
“在随机抽取 24 张考试票时得到一张数字为 7 的票”、
“足球输了 3 次”首胜前的球队”、
“五次出场的信” $s$ ” 在可读报纸的第一页上，”

## 数学代写|概率论代写Probability theory代考|RANDOM VARIABLES AND DISTRIBUTION FUNCTIONS

(取决于选择这三个球场之一) 分别为 3000、10000和20000卢布。对他来说，在抽签之前，末来的成 本是一个随机变量，以相等的概率取这三个值之一 $1 / 3$.

$$\omega_1=r, r, r, \omega_2=r, r, f, \omega_3=r, f, r, \omega_4=f, r, r, \quad \omega_5=r, f, f, \omega_6=f, r, f, \omega_7=f, f$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|概率论代写Probability theory代考|STAT4061

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写概率论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|概率论代写Probability theory代考|THE SIMPLEST PROBABILISTIC MODELS

For the simplest situations, discussed before methods were proposed to rate the chances of the occurrences of events. It would be quite natural to introduce some characteristic, which makes it possible to compare the chances of the success in carrying out various experiments. Such sufficiently convenient characteristic is a certain measure of the success of the experiment (the probability of occurrence of the desired event) turned out to be the ratio $m / n$, where $n$ is the possible number of outcomes of this experiment, and $\mathrm{m}$ is the number of outcomes that suit us.

In order to consider more complex situations in which this measure of the success can be evaluated for various events of interest to us, we will try to give some scientific form to the classical model already considered before, in which this measure is determined by the ratio $m / n$.

So, we are conducting some experiment, the result of which can be (with equal chances for any of them!) $n$ outcomes. These outcomes we treat as elementary events and denote them $\omega_1, \omega_2, \ldots, \omega_n$. Thus, we define the first element of the probabilistic space – the so-called set of elementary events
$$\Omega=\left{\omega_1, \omega_2, \ldots, \omega_n\right}$$

For example, under the single throwing of a dice we have $n=6$ and $\Omega=$ $\left{\omega_1, \omega_2, \ldots, \omega_6\right}$, where $\omega_k$ means the appearance of the face with the digit $k, k=1,2,3,4,5,6$. If the coin is thrown three times, then
$$n=2^3=8, \omega_1={r, r, r}, \omega_2={r, r, f}, \ldots, \omega_8={f, f, f}$$
where the symbol ” $r$ ” corresponds to the appearance of its reverse on the first place, on the second place or on the third place, and ” $f$ ” indicates the appearance of its face during the first, second or third coin toss.

Along with the elementary situations, we may be interested in more complex outcomes of the experiment. For example, it may be important for us to have exactly an even face when throwing a dice or to get the event consisting in the appearance of at least one of three possible reverses of the coins when one deals with the throwing of three coins. What types of the cumbersome structures can be built from the original “bricks” – the elementary outcomes that we have already fixed? To construct these complex events, we can take the different groups
$$\left{\omega_{\alpha(1)}, \omega_{\alpha(2)}, \ldots, \omega_{\alpha(r)}\right}, r=1,2, \ldots, n$$
which are composed from our “bricks.” The number of such possible groups is $C_n^1+C_n^2+\cdots+C_n^n$.

## 数学代写|概率论代写Probability theory代考|DISCRETE GENERALIZATIONS

In the classical probabilistic model considered above, we are dealing with $n$ outcomes of some experiment having equal chances for their appearances. The simplest examples of such classical schemes are connected, for example, with throwing of the “correct” dices or some symmetrical coins, as well as with the random selection of one or several playing cards from a well-mixed deck. However, there are substantially more situations when the possible outcomes of the carrying out experiment are not equally probable. For example, imagine that two “correct” dices are throwing, but we are interested in the sum of the readings of the two fallen faces only, then the outcomes of this experiment $\omega_2, \omega_3, \ldots, \omega_{12}$, where $\omega_k$ corresponds to the sum, which is equal to $\mathrm{k}$, no longer will be equally probable. Therefore, the first simplest generalization of the classical probability model presented above is fairly obvious.

Now let us consider the set of the elementary outcomes $\Omega=$ $\left{\omega_1, \omega_2, \ldots, \omega_n\right}$ in the case, when each outcome $\omega_k$ has its own (not necessarily equal to $1 / \mathrm{n}$ ) weight $p_k$ and the sum of all these $n$ nonnegative weights is equal to one. Then the total weight (probability)
$$P(A)=p_{\alpha(1)}+p_{\alpha(2)}+\cdots+p_{\alpha(m)}$$
corresponds to event A, which is formed from the “bricks” (elementary outcomes)
$$\left{\omega_{\alpha(1)}, \omega_{\alpha(2)}, \ldots, \omega_{\alpha(m)}\right}$$
We note that the probabilities of the impossible event and the reliable event remain equal, respectively, to zero and to one.

If we go further along the path of generalizations, we can start with the following example. Let’s return to our symmetrical coin, when the chances of falling out of the obverse or the reverse are the same and the corresponding probabilities are equal to $1 / 2$. We will now throw the coin until the appearance of the first reverse and calculate the number of obverses that fell out. It is evident that one can no longer confine ourselves to a finite number of $n$ elementary outcomes. Suppose that $\omega_k, k=0,1,2, \ldots$, is the outcome of this experiment, as a result of the situation, when a series of $k$ obverses was obtained. Note, a little ahead of the time, that the probability $p_k$, corresponding to the elementary event $\omega_k$ is equal to $1 / 2^{k+1}, k=$ $0,1,2, \ldots$

# 概率论代考

## 数学代写|概率论代写Probability theory代考|THE SIMPLEST PROBABILISTIC MODELS

$$n=2^3=8, \omega_1=r, r, r, \omega_2=r, r, f, \ldots, \omega_8=f, f, f$$

## 数学代写|概率论代写Probability theory代考|DISCRETE GENERALIZATIONS

$$P(A)=p_{\alpha(1)}+p_{\alpha(2)}+\cdots+p_{\alpha(m)}$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|概率论代写Probability theory代考|STAT7614

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写概率论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|概率论代写Probability theory代考|Probability as a mathematical tool

From the result (3.75) one may obtain a number of identities obeyed by the binomial coefficients. For example, we may decide not to distinguish between colors 1 and 2 ; i.e. a ball of either color is declared to have color ‘ $a$ ‘. Then from (3.75) we must have, on the one hand,
$$h\left(r_a, r_3, \ldots, r_k \mid N_a, N_3, \ldots, N_k\right)=\frac{\left(\begin{array}{c} N_a \ r_a \end{array}\right)\left(\begin{array}{c} N_3 \ r_3 \end{array}\right) \cdots\left(\begin{array}{c} N_k \ r_k \end{array}\right)}{\left(\begin{array}{c} \sum N_i \ \sum r_i \end{array}\right)}$$
with
$$N_a=N_1+N_2, \quad r_a=r_1+r_2$$

But the event $r_a$ can occur for any values of $r_1, r_2$ satisfying (3.77), and so we must have also, on the other hand,
$$h\left(r_a, r_3, \ldots, r_k \mid N_a, N_3, \ldots, N_k\right)=\sum_{r_1=0}^{r_a} h\left(r_1, r_a-r_1, r_3, \ldots, r_k \mid N_1, \ldots, N_k\right)$$
Then, comparing (3.76) and (3.78), we have the identity
$$\left(\begin{array}{c} N_a \ r_a \end{array}\right)=\sum_{r_1=0}^{r_a}\left(\begin{array}{c} N_1 \ r_1 \end{array}\right)\left(\begin{array}{c} N_2 \ r_a-r_1 \end{array}\right)$$
Continuing in this way, we can derive a multitude of more complicated identities obeyed by the binomial coefficients. For example,
$$\left(\begin{array}{c} N_1+N_2+N_3 \ r_a \end{array}\right)=\sum_{r_1=0}^{r_a} \sum_{r_2=0}^{r_1}\left(\begin{array}{l} N_1 \ r_1 \end{array}\right)\left(\begin{array}{c} N_2 \ r_2 \end{array}\right)\left(\begin{array}{c} N_3 \ r_a-r_1-r_2 \end{array}\right)$$
In many cases, probabilistic reasoning is a powerful tool for deriving purely mathematical results; more examples of this are given by Feller (1950, Chap. 2 \& 3) and in later chapters of the present work.

## 数学代写|概率论代写Probability theory代考|The binomial distribution

Although somewhat complicated mathematically, the hypergeometric distribution arises from a problem that is very clear and simple conceptually; there are only a finite number of possibilities and all the above results are exact for the problems as stated. As an introduction to a mathematically simpler, but conceptually far more difficult, problem, we examine a limiting form of the hypergeometric distribution.

The complication of the hypergeometric distribution arises because it is taking into account the changing contents of the urn; knowing the result of any draw changes the probability for red for any other draw. But if the number $N$ of balls in the urn is very large compared with the number drawn $(N \gg n)$, then this probability changes very little, and in the limit $N \rightarrow \infty$ we should have a simpler result, free of such dependencies. To verify this, we write the hypergeometric distribution (3.22) as
$$h(r \mid N, M, n)=\frac{\left[\frac{1}{N^r}\left(\begin{array}{c} M \ r \end{array}\right)\right]\left[\frac{1}{N^{n-r}}\left(\begin{array}{c} N-M \ n-r \end{array}\right)\right]}{\left[\frac{1}{N^n}\left(\begin{array}{l} N \ n \end{array}\right)\right]} .$$
The first factor is
$$\frac{1}{N^r}\left(\begin{array}{c} M \ r \end{array}\right)=\frac{1}{r !} \frac{M}{N}\left(\frac{M}{N}-\frac{1}{N}\right)\left(\frac{M}{N}-\frac{2}{N}\right) \cdots\left(\frac{M}{N}-\frac{r-1}{N}\right)$$ and in the limit $N \rightarrow \infty, M \rightarrow \infty, M / N \rightarrow f$, we have
$$\frac{1}{N^r}\left(\begin{array}{c} M \ r \end{array}\right) \rightarrow \frac{f^r}{r !}$$
Likewise,
$$\begin{gathered} \frac{1}{N^{n-r}}\left(\begin{array}{c} M-1 \ n-r \end{array}\right) \rightarrow \frac{(1-f)^{n-r}}{(n-r) !} \ \frac{1}{N^n}\left(\begin{array}{l} N \ n \end{array}\right) \rightarrow \frac{1}{n !} \end{gathered}$$

# 概率论代考

## 数学代写|概率论代写Probability theory代考|Probability as a mathematical tool

$$h\left(r_a, r_3, \ldots, r_k \mid N_a, N_3, \ldots, N_k\right)=\frac{\left(N_a r_a\right)\left(N_3 r_3\right) \cdots\left(N_k r_k\right)}{\left(\sum N_i \sum r_i\right)}$$

$$N_a=N_1+N_2, \quad r_a=r_1+r_2$$

$$h\left(r_a, r_3, \ldots, r_k \mid N_a, N_3, \ldots, N_k\right)=\sum_{r_1=0}^{r_a} h\left(r_1, r_a-r_1, r_3, \ldots, r_k \mid N_1, \ldots, N_k\right)$$

$$\left(N_a r_a\right)=\sum_{r_1=0}^{r_a}\left(N_1 r_1\right)\left(N_2 r_a-r_1\right)$$

$$\left(N_1+N_2+N_3 r_a\right)=\sum_{r_1=0}^{r_a} \sum_{r_2=0}^{r_1}\left(N_1 r_1\right)\left(N_2 r_2\right)\left(N_3 r_a-r_1-r_2\right)$$

## 数学代写|概率论代写Probability theory代考|The binomial distribution

$$h(r \mid N, M, n)=\frac{\left[\frac{1}{N^r}(M r)\right]\left[\frac{1}{N^{n-r}}(N-M n-r)\right]}{\left[\frac{1}{N^n}(N n)\right]} .$$

$$\frac{1}{N^r}(M r)=\frac{1}{r !} \frac{M}{N}\left(\frac{M}{N}-\frac{1}{N}\right)\left(\frac{M}{N}-\frac{2}{N}\right) \cdots\left(\frac{M}{N}-\frac{r-1}{N}\right)$$

$$\frac{1}{N^r}(M r) \rightarrow \frac{f^r}{r !}$$

$$\frac{1}{N^{n-r}}(M-1 n-r) \rightarrow \frac{(1-f)^{n-r}}{(n-r) !} \frac{1}{N^n}(N n) \rightarrow \frac{1}{n !}$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|概率论代写Probability theory代考|MATHS2103

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写概率论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|概率论代写Probability theory代考|Expectations

Another way of looking at this result appeals more strongly to our intuition and generalizes far beyond the present problem. We can hardly suppose that the reader is not already familiar with the idea of expectation, but this is the first time it has appeared in the present work, so we pause to define it. If a variable quantity $X$ can take on the particular values $\left(x_1, \ldots, x_n\right)$ in $n$ mutually exclusive and exhaustive situations, and the robot assigns corresponding probabilities $\left(p_1, p_2, \ldots, p_n\right)$ to them, then the quantity
$$\langle X\rangle=E(X)=\sum_{i=1}^n p_i x_i$$ is called the expectation (in the older literature, mathematical expectation or expectation value ) of $X$. It is a weighted average of the possible values, weighted according to their probabilities. Statisticians and mathematicians generally use the notation $E(X)$; but physicists, having already pre-empted $E$ to stand for energy and electric field, use the bracket notation $\langle X\rangle$. We shall use both notations here; they have the same meaning, but sometimes one is easier to read than the other.

Like most of the standard terms that arose out of the distant past, the term ‘expectation’ seems singularly inappropriate to us; for it is almost never a value that anyone ‘expects’ to find. Indeed, it is often known to be an impossible value. But we adhere to it because of centuries of precedent.

Given $R_{\text {later }}$, what is the expectation of the number of red balls in the urn for draw number one? There are three mutually exclusive possibilities compatible with $R_{\text {later }}$ :
$$R_2 W_3, W_2 R_3, R_2 R_3$$
for which $M$ is $(1,1,0)$, respectively, and for which the probabilities are as in (3.64) and $(3.65)$
$$\begin{gathered} P\left(R_2 W_3 \mid R_{\text {later }} B\right)=\frac{P\left(R_2 W_3 \mid B\right)}{P\left(R_{\text {later }} \mid B\right)}=\frac{(1 / 2) \times(2 / 3)}{(5 / 6)}=\frac{2}{5}, \ P\left(W_2 R_3 \mid R_{\text {later }} B\right)=\frac{2}{5}, \ P\left(R_2 R_3 \mid R_{\text {later }} B\right)=\frac{1}{5} . \end{gathered}$$
So
$$\langle M\rangle=1 \times \frac{2}{5}+1 \times \frac{2}{5}+0 \times \frac{1}{5}=\frac{4}{5}$$

## 数学代写|概率论代写Probability theory代考|Other forms and extensions

The hypergeometric distribution (3.22) can be written in various ways. The nine factorials can be organized into binomial coefficients also as follows:
$$h(r \mid N, M, n)=\frac{\left(\begin{array}{l} n \ r \end{array}\right)\left(\begin{array}{l} N-n \ M-r \end{array}\right)}{\left(\begin{array}{l} N \ M \end{array}\right)} .$$
But the symmetry under exchange of $M$ and $n$ is still not evident; to see it we must write out (3.22) or (3.73) in full, displaying all the individual factorials.

We may also rewrite ( $3.22)$, as an aid to memory, in a more symmetric form: the probability for drawing exactly $r$ red balls and $w$ white ones in $n=r+w$ draws, from an urn containing $R$ red and $W$ white, is
$$h(r)=\frac{\left(\begin{array}{l} R \ r \end{array}\right)\left(\begin{array}{l} W \ w \end{array}\right)}{\left(\begin{array}{c} R+W \ r+w \end{array}\right)},$$
and in this form it is easily generalized. Suppose that, instead of only two colors, there are $k$ different colors of balls in the urn, $N_1$ of color $1, N_2$ of color $2, \ldots, N_k$ of color $k$. The probability for drawing $r_1$ balls of color $1, r_2$ of color $2, \ldots, r_k$ of color $k$ in $n=\sum r_i$ draws is, as the reader may verify, the generalized hypergeometric distribution:
$$h\left(r_1 \cdots r_k \mid N_1 \cdots N_k\right)=\frac{\left(\begin{array}{c} N_1 \ r_1 \end{array}\right) \cdots\left(\begin{array}{l} N_k \ r_k \end{array}\right)}{\left(\begin{array}{c} \sum N_i \ \sum r_i \end{array}\right)}$$

# 概率论代考

## 数学代写|概率论代写Probability theory代考|Expectations

$$\langle X\rangle=E(X)=\sum_{i=1}^n p_i x_i$$

$$R_2 W_3, W_2 R_3, R_2 R_3$$

$$P\left(R_2 W_3 \mid R_{\text {later }} B\right)=\frac{P\left(R_2 W_3 \mid B\right)}{P\left(R_{\text {later }} \mid B\right)}=\frac{(1 / 2) \times(2 / 3)}{(5 / 6)}=\frac{2}{5}, P\left(W_2 R_3 \mid R_{\text {later }} B\right)$$

$$\langle M\rangle=1 \times \frac{2}{5}+1 \times \frac{2}{5}+0 \times \frac{1}{5}=\frac{4}{5}$$

## 数学代写|概率论代写Probability theory代考|Other forms and extensions

$$h(r \mid N, M, n)=\frac{(n r)(N-n M-r)}{(N M)} .$$

$$h(r)=\frac{(R r)(W w)}{(R+W r+w)},$$

$$h\left(r_1 \cdots r_k \mid N_1 \cdots N_k\right)=\frac{\left(N_1 r_1\right) \cdots\left(N_k r_k\right)}{\left(\sum N_i \sum r_i\right)}$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|概率论代写Probability theory代考|STAT4528

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写概率论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|概率论代写Probability theory代考|Qualitative properties

Now let us check to see how the theory based on (2.63) and (2.64) is related to the theory of deductive logic and the various qualitative syllogisms from which we started in Chapter 1. In the first place it is obvious that in the limit as $p(A \mid B) \rightarrow 0$ or $p(A \mid B) \rightarrow 1$, the sum rule (2.64) expresses the primitive postulate of Aristotelian logic: if $A$ is true, then $\bar{A}$ must be false, etc.

Indeed, all of that logic consists of the two strong syllogisms (1.1), (1.2) and all that follows from them; using now the implication sign (1.14) to state the major premise:
$$\begin{array}{cc} A \Rightarrow B & A \Rightarrow B \ \frac{A \text { is true }}{B \text { is true }} & \frac{B \text { is false }}{A \text { is false }} \end{array}$$
and the endless stream of their consequences. If we let $C$ ‘ stand for their major premise:
$$C \equiv A \Rightarrow B$$
then these syllogisms correspond to our product rule (2.63) in the forms
$$p(B \mid A C)=\frac{p(A B \mid C)}{p(A \mid C)}, \quad p(A \mid \bar{B} C)=\frac{p(A \bar{B} \mid C)}{p(\bar{B} \mid C)},$$
respectively. But from (2.68) we have $p(A B \mid C)=p(A \mid C)$ and $p(A \bar{B} \mid C)=0$, and so $(2.70)$ reduces to
$$p(B \mid A C)=1, \quad p(A \mid \bar{B} C)=0$$
as stated in the syllogisms (2.68). Thus the relation is simply: Aristotelian deductive logic is the limiting form of our rules for plausible reasoning, as the robot becomes more and more certain of its conclusions.

## 数学代写|概率论代写Probability theory代考|Numerical values

We have found so far the most general consistent rules by which our robot can manipulate plausibilities, granted that it must associate them with real numbers, so that its brain can operate by the carrying out of some definite physical process. While we are encouraged by the familiar formal appearance of these rules and their qualitative properties just noted, two evident circumstances show that our job of designing the robot’s brain is not yet finished.
In the first place, while the rules (2.63), (2.64) place some limitations on how plausibilities of different propositions must be related to each other, it would appear that we have not yet found any unique rules, but rather an infinite number of possible rules by which our robot can do plausible reasoning. Corresponding to every different choice of a monotonic function $p(x)$, there seems to be a different set of rules, with different content.

Secondly, nothing given so far tells us what actual numerical values of plausibility should be assigned at the beginning of a problem, so that the robot can get started on its calculations. How is the robot to make its initial encoding of the background information into definite numerical values of plausibilities? For this we must invoke the ‘interface’ desiderata (IIIb), (IIIC) of (1.39), not yet used.

The following analysis answers both of these questions, in a way both interesting and unexpected. Let us ask for the plausibility $\left(A_1+A_2+A_3 \mid B\right)$ that at least one of three propositions $\left{A_1, A_2, A_3\right}$ is true. We can find this by two applications of the extended sum rule (2.66), as follows. The first application gives
$$p\left(A_1+A_2+A_3 \mid B\right)=p\left(A_1+A_2 \mid B\right)+p\left(A_3 \mid B\right)-p\left(A_1 A_3+A_2 A_3 \mid B\right)$$
where we first considered $\left(A_1+A_2\right)$ as a single proposition, and used the logical relation
$$\left(A_1+A_2\right) A_3=A_1 A_3+A_2 A_3 .$$

Applying (2.66) again, we obtain seven terms which can be grouped as follows:
\begin{aligned} p\left(A_1+A_2+A_3 \mid B\right) & =p\left(A_1 \mid B\right)+p\left(A_2 \mid B\right)+p\left(A_3 \mid B\right) \ & -p\left(A_1 A_2 \mid B\right)-p\left(A_2 A_3 \mid B\right)-p\left(A_3 A_1 \mid B\right) \ & +p\left(A_1 A_2 A_3 \mid B\right) \end{aligned}

# 概率论代考

## 数学代写|概率论代写Probability theory代考|Qualitative properties

$$A \Rightarrow B \quad A \Rightarrow B \frac{A \text { is true }}{B \text { is true }} \quad \frac{B \text { is false }}{A \text { is false }}$$

$$C \equiv A \Rightarrow B$$

$$p(B \mid A C)=\frac{p(A B \mid C)}{p(A \mid C)}, \quad p(A \mid \bar{B} C)=\frac{p(A \bar{B} \mid C)}{p(\bar{B} \mid C)},$$

$$p(B \mid A C)=1, \quad p(A \mid \bar{B} C)=0$$

## 数学代写|概率论代写Probability theory代考|Numerical values

$$p\left(A_1+A_2+A_3 \mid B\right)=p\left(A_1+A_2 \mid B\right)+p\left(A_3 \mid B\right)-p\left(A_1 A_3+A_2 A_3 \mid B\right)$$

$$\left(A_1+A_2\right) A_3=A_1 A_3+A_2 A_3 .$$

$$p\left(A_1+A_2+A_3 \mid B\right)=p\left(A_1 \mid B\right)+p\left(A_2 \mid B\right)+p\left(A_3 \mid B\right) \quad-p\left(A_1 A_2 \mid B\right)$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|概率论代写Probability theory代考|STAT4061

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写概率论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|概率论代写Probability theory代考|The product rule

We first seek a consistent rule relating the plausibility of the logical product $A B$ to the plausibilities of $A$ and $B$ separately. In particular, let us find $A B \mid C$. Since the reasoning is somewhat subtle, we examine this from several different viewpoints.

As a first orientation, note that the process of deciding that $A B$ is true can be broken down into elementary decisions about $A$ and $B$ separately. The robot can
(1) decide that $B$ is true;
$(B \mid C)$
(2) having accepted $B$ as true, decide that $A$ is true.
$(A \mid B C)$
Or, equally well,
(1′) decide that $A$ is true;
$(A \mid C)$
(2′) having accepted $A$ as true, decide that $B$ is true.
$(B \mid A C)$
In each case we indicate above the plausibility corresponding to that step.
Now let us describe the first procedure in words. In order for $A B$ to be a true proposition, it is necessary that $B$ is true. Thus the plausibility $B \mid C$ should be involved. In addition, if $B$ is true, it is further necessary that $A$ should be true; so the plausibility $A \mid B C$ is also needed. But if $B$ is false, then of course $A B$ is false independently of whatever one knows about $A$, as expressed by $A \mid \bar{B} C$; if the robot reasons first about $B$, then the plausibility of $A$ will be relevant only if $B$ is true. Thus, if the robot has $B \mid C$ and $A \mid B C$ it will not need $A \mid C$. That would tell it nothing about $A B$ that it did not have already.

Similarly, $A \mid B$ and $B \mid A$ are not needed; whatever plausibility $A$ or $B$ might have in the absence of information $C$ could not be relevant to judgments of a case in which the robot knows that $C$ is true. For example, if the robot learns that the earth is round, then in judging questions about cosmology today, it does not need to take into account the opinions it might have (i.e. the extra possibilities that it would need to take into account) if it did not know that the earth is round.

Of course, since the logical product is commutative, $A B=B A$, we could interchange $A$ and $B$ in the above statements; i.e. knowledge of $A \mid C$ and $B \mid A C$ would serve equally well to determine $A B|C=B A| C$. That the robot must obtain the same value for $A B \mid C$ from either procedure is one of our conditions of consistency, desideratum (IIIa).

We can state this in a more definite form. $(A B \mid C)$ will be some function of $B \mid C$ and $A \mid B C$ :
$$(A B \mid C)=F[(B \mid C),(A \mid B C)]$$

## 数学代写|概率论代写Probability theory代考|The sum rule

Since the propositions now being considered are of the Aristotelian logical type which must be either true or false, the logical product $A \bar{A}$ is always false, the logical sum $A+\bar{A}$ always true. The plausibility that $A$ is false must depend in some way on the plausibility that it is true. If we define $u \equiv w(A \mid B), \quad v \equiv w(\bar{A} \mid B)$, there must exist some functional relation
$$v=S(u)$$
Evidently, qualitative correspondence with common sense requires that $S(u)$ be a continuous monotonic decreasing function in $0 \leq u \leq 1$, with extreme values $S(0)=1, S(1)=0$. But it cannot be just any function with these properties, for it must be consistent with the fact that the product rule can be written for either $A B$ or $A \bar{B}$ :
$$\begin{gathered} w(A B \mid C)=w(A \mid C) w(B \mid A C) \ w(A \bar{B} \mid C)=w(A \mid C) w(\bar{B} \mid A C) . \end{gathered}$$
Thus, using (2.36) and (2.38), Eq. (2.37) becomes
$$w(A B \mid C)=w(A \mid C) S[w(\bar{B} \mid A C)]=w(A \mid C) S\left[\frac{w(A \bar{B} \mid C)}{w(A \mid C)}\right]$$

Again, we invoke commutativity: $w(A B \mid C)$ is symmetric in $A, B$, and so consistency requires that
$$w(A \mid C) S\left[\frac{w(A \bar{B} \mid C)}{w(A \mid C)}\right]=w(B \mid C) S\left[\frac{w(B \bar{A} \mid C)}{w(B \mid C)}\right]$$
This must hold for all propositions $A, B, C$; in particular, (2.40) must hold when
$$\bar{B}=A D$$
where $D$ is any new proposition. But then we have the truth values noted before in (1.13):
$$A \bar{B}=\bar{B}, \quad B \bar{A}=\bar{A}$$
and in $(2.40)$ we may write
\begin{aligned} & w(A \bar{B} \mid C)=w(\bar{B} \mid C)=S[w(B \mid C)] \ & w(B \bar{A} \mid C)=w(\bar{A} \mid C)=S[w(A \mid C)] \end{aligned}

# 概率论代考

## 数学代写|概率论代写Probability theory代考|The product rule

(1) 决定 $B$ 是真的;
$(B \mid C)$
(2) 已接受 $B$ 为真，决定 $A$ 是真的。
$(A \mid B C)$

(1′) 决定 $A$ 是真的；
$(A \mid C)$
(2) 已接受 $A$ 为真，决定 $B$ 是真的。
$(B \mid A C)$

$$(A B \mid C)=F[(B \mid C),(A \mid B C)]$$

## 数学代写|概率论代写Probability theory代考|The sum rule

$$v=S(u)$$

$$w(A B \mid C)=w(A \mid C) w(B \mid A C) w(A \bar{B} \mid C)=w(A \mid C) w(\bar{B} \mid A C) .$$

$$w(A B \mid C)=w(A \mid C) S[w(\bar{B} \mid A C)]=w(A \mid C) S\left[\frac{w(A \bar{B} \mid C)}{w(A \mid C)}\right]$$

$$w(A \mid C) S\left[\frac{w(A \bar{B} \mid C)}{w(A \mid C)}\right]=w(B \mid C) S\left[\frac{w(B \bar{A} \mid C)}{w(B \mid C)}\right]$$

$$\bar{B}=A D$$

$$A \bar{B}=\bar{B}, \quad B \bar{A}=\bar{A}$$

$$w(A \bar{B} \mid C)=w(\bar{B} \mid C)=S[w(B \mid C)] \quad w(B \bar{A} \mid C)=w(\bar{A} \mid C)=S[w(A \mid C)]$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|概率论代写Probability theory代考|STAT4528

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写概率论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|概率论代写Probability theory代考|The Ball and Urn (cell) model

Tn some combinatorial analysis problems, the goal is to count the number of states 1 of putting some objects (balls) into some containers (cells or urns). In this type of problems, the term “indistinguishable objects” means that only the number of each container’s objects matters. Moreover, the displacement of an object from a container with the other object from another container does not create a new state (the value of objects is the same). On the contrary, the term “distinguishable objects” means that both the number of objects and their values in each container are important. In other words, if the values of objects are assumed to be different, the displacement of one object from a container with the other object from another container creates a new state.

Furthermore, the term “indistinguishable containers” means that groupmates of each object matter. Nonetheless, “the different containers” means that in addition to the groupmates, the urn of each object is important as well.

In this book, when distributing the people into physical places (such as a class, a room, and an avenue, to name but a few), we consider individuals to be different unless otherwise stated in the problem (For instance, it is explicitly stated in the problem that the value of people is the same or only the number of individuals lying in each place is important.). Furthermore, we consider physical places to be distinct urns unless otherwise stated in the problem (For instance, it is explicitly stated in the problem that only the number of each person’s groupmates matters, not their place.) If we want to group the people, we consider the groups to be identical urns, unless otherwise stated in the problem.

Now, we investigate some well-known cases in the ball and urn model. Note that the classification scheme of this book is merely a suggestion for classifying the ball and urn model’s problems, which is not necessarily adopted in all reference books.

## 数学代写|概率论代写Probability theory代考|Random trial, sample space, and event

Oonsider a trial with an unknown prior result, yet known possible results. Such a trial is called a random trial, and the set of its possible results is called sample space, usually denoted by the letter $S$. For more clarification, consider the following examples:
In the trial of tossing one coin, the sample space is as follows:
$$S={H, T}^1$$
In the trial of tossing two coins, the sample space is defined as:
$$S={(H, H),(T, H),(H, T),(T, T)}$$

In the trial of tossing two dice, the sample space consists of 36 states and is defined as:
$$S={(i, j): i, j=1,2,3,4,5,6}$$
In the trial of measuring the lifetime of a particular light bulb (in hours), the sample space is defined as:
$$S={x: x \geq 0}$$
Each subset of a sample space with possible outcomes belonging to a trial is called the sample space event.

For instance, consider the trial of tossing two coins. If the event $E$ denotes at least one heads appears, the event is expressed as follows:
$$E={(H, T),(T, H),(H, H)}$$
Alternatively, consider the trial of tossing two dice. If the event $E$ denotes the sum of the results of two dice is equal to 4 , the event is expressed as:
$$E={(1,3),(2,2),(3,1)}$$
Also, in the trial of measuring the lifetime of a particular light bulb, the event $E$ is defined as the lifetime of the light bulb with a maximum value of 10 hours. This event is represented as follows:
$$E={x: 0 \leq x \leq 10}$$
Note that we say the event $E$ has occurred when one of its results has occurred. Namely, in the trial of tossing two dice, assume that the event $E$ denotes the sum of the results of two dice is equal to 4 . Then, if one of the results $(1,3),(2,2)$, or $(3,1)$ occurs, we say that the event $E$ has occurred.

# 概率论代考

## 数学代写|概率论代写Probability theory代考|Random trial, sample space, and event

$$S=H, T^1$$

$$S=(H, H),(T, H),(H, T),(T, T)$$

$$S=(i, j): i, j=1,2,3,4,5,6$$

$$S=x: x \geq 0$$

$$E=(H, T),(T, H),(H, H)$$

$$E=(1,3),(2,2),(3,1)$$

$$E=x: 0 \leq x \leq 10$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

## 数学代写|概率论代写Probability theory代考|STAT406

statistics-lab™ 为您的留学生涯保驾护航 在代写概率论Probability theory方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写概率论Probability theory代写方面经验极为丰富，各种代写概率论Probability theory相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 数学代写|概率论代写Probability theory代考|The Basic Principle of Counting

A ll methods of counting rely on the Basic Principle of Counting or the Principle of Multiplication, which is expressed as follows:
Suppose that two trials are to be done. If the first trial can obtain one out of the $n$ possible results and each of those results correspond with the $m$ possible results of the second trial, then altogether there are $n \times m$ possible results for performing the two trials.

A noteworthy point in applying the multiplication principle is to pay attention to the phrase “each of those results”. Even though it seems obvious, many mistakes occurring in usage of the multiplication principle result from disregarding the very point. Note the following examples:
Example $2.1$
There are 12 coaches, each of whom has 4 athletes participating in a ceremony. If one coach and one of his athletes are to be chosen as the coach and athlete of the year, respectively, how many different choices are possible to do so?

Solution. We define the first and second trials to be choosing the coach and athlete of the year, respectively. The first trial can be done in 12 states, and given the selection of each coach in the first trial, choosing his athlete can be done in 4 states. Hence, the trials can be performed in $12 \times 4=48$ states.
Example $2.2$
Suppose that five coaches have two athletes each and the other seven coaches have three athletes each. Now, if we want to choose one coach and one of his athletes as the coach and athlete of the year, how many different choices are possible to do so?

## 数学代写|概率论代写Probability theory代考|Permutation of ” n ” distinct elements at a round table

The number of states that ” $n$ ” distinct elements can be arranged at a round table is equal to $(n-1)$ !. To prove it, we should know that the only difference between the problem of arranging people at a round table and in a row is that the location of people does not matter in the former case, which the only important point is the way of arranging the people. We are now trying to establish a relationship between the number of states of this problem and the number of states of seating people in a row. Also, it is intended to show that every ” $n$ ” states of seating people in a row are equivalent to one state of seating people at a round table.

As mentioned previously, in the problem of arranging people at a round table, the only important issue is the order of sitting. Hence, the states shown below are considered indistinguishable: Therefore, there is a relationship between the states of seating people in a row and at a round table as follows:

Hence, the number of states of seating people at a round table can be written as follows:
The number of states of seating $n$ people at a round table
$$=(\text { The number of states of seating } n \text { people in a row }) \times \frac{1}{n}=n ! \times \frac{1}{n}=(n-1) \text { ! }$$
There is also another way to justify the formula of arranging people around a round table. Since different possible places of the round table do not create a new state for the first person, there is only one state for him. However, after he sits, since the way of sitting relative to the first person is important for the other ones, the value of places turns out to be different, and the number of states of seating them relative to the first person equals:
$$1 \times(n-1) \times(n-2) \times(n-3) \times \ldots \times 1=(n-1) !$$

How many ways can ” $n “$ people be seated at a round table such that person $\mathrm{A}$ sits between person $B$ and person $C$ ?

Solution 1. There is one state for person A. Then, there are two states for person B to sit on the left or right side of the person A. In this status, there is one state for person C. Finally, the other $(n-3)$ people can sit on the remaining places in $(n-3)$ ! states. Therefore, the number of states equals:
$$1 \times 2 \times 1 \times(n-3) !=2 \times(n-3) !$$

# 概率论代考

## 数学代写|概率论代写Probability theory代考|Permutation of ” n ” distinct elements at a round table

: $n$ 圆桌会议上的人们
$=($ The number of states of seating $n$ people in a row $) \times \frac{1}{n}=n ! \times \frac{1}{n}=(n-1) !$

$$1 \times(n-1) \times(n-2) \times(n-3) \times \ldots \times 1=(n-1) !$$

$$1 \times 2 \times 1 \times(n-3) !=2 \times(n-3) !$$

## 有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

## MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。