### 物理代写|统计力学代写Statistical mechanics代考|PHYC30017

statistics-lab™ 为您的留学生涯保驾护航 在代写统计力学Statistical mechanics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。我们的专家在代写统计力学Statistical mechanics代写方面经验极为丰富，各种代写统计力学Statistical mechanics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

## 物理代写|统计力学代写Statistical mechanics代考|EXAMPLES INVOLVING DISCRETE PROBABILITIES

• Two cards are drawn from a 52-card deck, with the first being replaced before the second is drawn. What is the probability that both cards are spades? Let $A$ be the event of drawing a spade, with $B$ the event of drawing another spade after the first has been replaced in the deck. This is an “and” kind of problem: What is the probability of a spade being drawn and another spade being drawn. $P(A)=P(B)=13 / 52=1 / 4$. The two events are independent, and thus from Eq. (3.5), $P(A \cap B)=P(A) P(B)=1 / 16$.
• What is the probability of at least one spade in drawing two cards, when the first is replaced? The slick way to work this problem is to calculate the probability of not drawing a spade-the probability of at least one spade is the complement of the probability of no spades in two draws. The probability of no spades (not drawing a spade and not drawing another one) is $(39 / 52)^{2}=9 / 16$ (independent events). The probability of at least one spade is then $1-P$ (no spades) $=7 / 16$. The direct approach is to treat this as an “or” problem: What is the probability of drawing one or two spades? Let $A$ be the event of drawing a spade and not drawing a spade on the other draw, with $B$ the event of drawing two spades. The probability of at least one spade is $P(A$ or $B)=P(A)+P(B)$ (mutually exclusive). $P(A)=P$ (spade on one draw and not a spade on the other $)=(1 / 4)(3 / 4)=3 / 16$ (independent). There are two ways to realize the first experiment, however, draw a spade and then not, or not draw a spade and then a spade, so we add the probabilities: The probability of one spade is $2 \times(3 / 16)$. The probability of two spades, $P(B)=(1 / 4)^{2}=1 / 16$. The probability of at least one spade is $2 \times(3 / 16)+(1 / 16)=7 / 16$, in agreement with the first answer.
• Two cards are drawn from a deck, but now suppose the first is not put back. What is the probability that both are spades? This is an “and” problem, the probability of drawing a spade and drawing another one. The events are independent. Thus, $P=(13 / 52) \times(12 / 51)=1 / 17$.
• What is the probability that the second card is a spade, when it’s not known what the first card was? Let $B$ be the event of drawing a spade on the second draw. All we know about the first event is that a card was drawn and not replaced. There are two mutually exclusive possibilities: The first card was a spade or not, call these events $A$ and $\bar{A}$. Then, $P(A \cap B)+P(\bar{A} \cap B)=$ $P(A) P(B)+P(\bar{A}) P(B)=(P(A)+P(\bar{A})) P(B)=P(B)$. Thus, $P(B)=1 / 4$. The probability of a spade on the second draw, when the result of the first draw is unknown, is the probability of a spade on the first draw.

## 物理代写|统计力学代写Statistical mechanics代考|Probability distributions on discrete sample spaces

The collection of probabilities associated with the range of values of a random variable is known as a probability distribution. ${ }^{10}$ For each value $x_{j}$ of a random variable $x$, the aggregate of sample points associated with $x_{j}$ form the event for which $x-x_{j}$; its probability is denoted $P\left(x-x_{j}\right)$. From Fig. 3.4, for example, $f(1)=1 / 2$ is associated with the event $T H$ or $H T$.
Definition. A function $f(x)$ such that $f\left(x_{j}\right)=P\left(x=x_{j}\right)$ is the probability distribution of $x$.
For the range of values $\left{x_{j}\right}$ of $x, f\left(x_{j}\right) \geq 0$ and $\sum_{j} f\left(x_{j}\right)=1$; see Figs. $3.4$ and 3.5.
There can be more than one random variable defined on the same sample space. Consider random variables $x$ and $y$ that take on the values $x_{1}, x_{2}, \ldots$ and $y_{1}, y_{2}, \ldots$, and let the corresponding probability distributions be $f\left(x_{j}\right)$ and $g\left(y_{k}\right)$. The aggregate of events for which the two conditions $x=x_{j}$ and $y=y_{k}$ are satisfied forms the event having probability denoted $P\left(x=x_{j}, y=y_{k}\right)$.
Definition. A function $p(x, y)$ for which $p\left(x_{j}, y_{k}\right)=P\left(x=x_{j}, y=y_{k}\right)$ is called the joint probability distribution of $x$ and $y$.

Clearly, $p\left(x_{j}, y_{k}\right) \geq 0$ and $\sum_{j k} p\left(x_{j}, y_{k}\right)=1$. Moreover, for fixed $x_{j}$,
$$\sum_{k} p\left(x_{j}, y_{k}\right)=f\left(x_{j}\right),$$
while for fixed $y_{k}$
$$\sum_{j} p\left(x_{j}, y_{k}\right)=g\left(y_{k}\right) .$$
That is, adding the probabilities for all events $y_{k}$ for fixed $x_{j}$ produces the probability distribution for $x_{j}$, and adding the probabilities for all events $x_{j}$ produces the probability distribution for $y_{k}$.

## 物理代写|统计力学代写Statistical mechanics代考|EXAMPLES INVOLVING DISCRETE PROBABILITIES

• 从一副 52 张牌中抽出两张牌，在抽出第二张牌之前先替换第一张牌。两张牌都是黑桃的概率是多少? 让 $A$ 是 绘制黑桃的事件，与 $B$ 在第一个黑桃被替换在甲板上之后绘制另一个黑桃的事件。这是一个”和”类型的问题: 一个铲子被抽出和另一个铲子被抽出的概率是多少。 $P(A)=P(B)=13 / 52=1 / 4$. 这两个事件是独立 的，因此来自方程式。(3.5), $P(A \cap B)=P(A) P(B)=1 / 16$.
• 当第一张被替换时，至少有一张黑桃抽两张牌的概率是多少? 解决这个问题的巧妙方法是计算没有抽到黑桃的 概率一一至少有一个黑桃的概率是两次抽到没有黑桃的概率的补数。没有黑桃的概率 (没有画黑桃，也没有画 另一个) 是 $(39 / 52)^{2}=9 / 16$ (独立事件) 。那么至少有一把铁锹的概率是 $1-P$ (没有黑桃) $=7 / 16$. 直 接的方法是将其视为一个”或”问题: 抽到一两个黑桃的概率是多少? 让 $A$ 是画黑挑而不是在另一张画上画黑桃 的事件，与 $B$ 绘制两个黑桃的事件。至少有一把铲子的概率是 $P(A$ 或者 $B)=P(A)+P(B)$ (互斥) 。 $P(A)=P($ (一平局是铁锹，另一个不是铁锹 $)=(1 / 4)(3 / 4)=3 / 16$ (独立的)。实现第一个实验有两 种方法，但是，画一个黑桃然后不画，或者不画一个黑桃然后一个黑桃，所以我们添加概率: 一个黑桃的概率 是 $2 \times(3 / 16)$. 两个黑桃的概率， $P(B)=(1 / 4)^{2}=1 / 16$. 至少有一把铲子的概率是 $2 \times(3 / 16)+(1 / 16)=7 / 16$ ，与第一个答案一致。
• -从一副牌中抽出两张牌，但现在假设第一张牌没有放回。两者都是黑桃的概率是多少? 这是一个”与”问题，即 画出一把铁锹再画另一个的概率。事件是独立的。因此， $P=(13 / 52) \times(12 / 51)=1 / 17$.
• 当不知道第一张牌是什么时，第二张牌是黑桃的概率是多少? 让 $B$ 是在第二次抽奖时抽到黑桃的事件。关于第 一个事件，我们所知道的只是一张牌被抽出来而不是被替换。有两种相互排斥的可能性: 第一张牌是否是黑 桃，调用这些事件 $A$ 和 $\bar{A}$. 然后， $P(A \cap B)+P(\bar{A} \cap B)=$ $P(A) P(B)+P(\bar{A}) P(B)=(P(A)+P(\bar{A})) P(B)=P(B)$. 因此， $P(B)=1 / 4$. 当第一次平局的 结果末知时，第二次平局出现黑桃的概率是第一次平局出现黑桃的概率。

## 物理代写|统计力学代写Statistical mechanics代考|Probability distributions on discrete sample spaces

$$\sum_{k} p\left(x_{j}, y_{k}\right)=f\left(x_{j}\right),$$

$$\sum_{j} p\left(x_{j}, y_{k}\right)=g\left(y_{k}\right) .$$

## 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。