Posted on Categories:Probability theory, 数学代写, 概率论

# 数学代写|概率论代考Probability Theory代写|A rule of products

## avatest™帮您通过考试

avatest™的各个学科专家已帮了学生顺利通过达上千场考试。我们保证您快速准时完成各时长和类型的考试，包括in class、take home、online、proctor。写手整理各样的资源来或按照您学校的资料教您，创造模拟试题，提供所有的问题例子，以保证您在真实考试中取得的通过率是85%以上。如果您有即将到来的每周、季考、期中或期末考试，我们都能帮助您！

•最快12小时交付

•200+ 英语母语导师

•70分以下全额退款

## 数学代写|概率论代考Probability Theory代写|A rule of products

Suppose $A$ is a set of $m$ elements and $B$ a set of $n$ elements. How many ways can we form ordered pairs of elements $(a, b)$ with $a \in A$ and $b \in B$ ? If $a$ and $\mathrm{b}$ may be specified independently (in our common or garden understanding of the word), then for every choice of a we may specify $m$ choices for $b$ whence there are a total of
$$\underbrace{m+m+\cdots+m}_{n \text { terms }}=m n$$
distinct choices for the pairs $(\mathrm{a}, \mathrm{b})$. Elementary, but this underscores a critical arithmetic relationship: possibilities multiply under independent selection.

ExAMPLES: 1) Coins. If a fair coin is tossed twice there are two possibilities, $\mathfrak{H}$ and $\mathfrak{T}$, for each of the tosses and $4=2 \times 2$ possible outcomes, $\mathfrak{H} \mathfrak{H}, \mathfrak{H} \mathfrak{T}, \mathfrak{T} \mathfrak{H}$, and $\mathfrak{T} \mathfrak{T}$, for the experiment. The probability of a head on the first trial is $1 / 2$, as is the probability of a tail in the second trial. The probability that we observe the outcome $\mathrm{HT}$ is $\frac{1}{4}=\frac{1}{2} \cdot \frac{1}{2}$.
2) Dice. If six dice are tossed there are $6^6$ possible outcomes. Exactly three out of the six possible outcomes of any given toss yield an even face so that the probability of an even face is $3 / 6=1 / 2$. The event that all six tosses result in an even face has $3^6$ favourable outcomes and hence probability $\frac{3^6}{6^6}=\left(\frac{1}{2}\right)^6$.
3) Urns. An urn has 1 black ball, 1 green ball, 2 blue balls, and 4 red balls, all balls considered distinguishable. If one ball is removed at random from the urn, the probability of drawing a black, green, blue, or red ball is $1 / 8,1 / 8,1 / 4$, and $1 / 2$, respectively. If four balls are drawn with replacement, the total number of possible outcomes is $8^4$; the number of outcomes favourable to the event that the draws are, in order, black, green, blue, and red is $1 \cdot 1 \cdot 2 \cdot 4=8$. The probability of this event is $\frac{8}{8^4}=\frac{1}{8} \cdot \frac{1}{8} \cdot \frac{1}{4} \cdot \frac{1}{2}$.

## 数学代写|概率论代考Probability Theory代写|What price intuition?

At the simplest levels, the definition of independence as a rule of products appears natural and unexceptionable.

EXAMPLES: 1) Cards. A card is drawn randomly from a 52-card pack. The event that it is an ace has four favourable outcomes, hence probability $4 / 52=1 / 13$. The event that the suit of the card is spades has 13 favourable outcomes and probability $13 / 52=1 / 4$. We may anticipate by symmetry that these two events are independent and, indeed, the conjoined event that the card is the ace of spades has a single favourable outcome and probability $\frac{1}{52}=\frac{1}{13} \cdot \frac{1}{4}$.
2) Relative ranks. Suppose all permutations of rankings of four students $a, b$, $c$, and $d$ are equally likely. The events $\mathrm{A}$ that $a$ ranks ahead of $d$ and B that $b$ ranks ahead of $c$ are intuitively independent. Indeed, as is easy to verify, $\mathbf{P}(A)=\mathbf{P}(B)=1 / 2$ and $\mathbf{P}(A \cap B)=1 / 4$.
3) Dice. Two dice are thrown. The event $A$ that the first die shows an ace has probability $1 / 6$ while the event $B$ that the sum of the face values of the two dice is odd has probability $1 / 2$ (as there are 18 odd-even or even-odd combinations). And as $A \cap B={(1,2),(1,4),(1,6)}$ has probability $\frac{1}{12}=\frac{1}{6} \cdot \frac{1}{2}$, the events are independent.

It is quite satisfactory in the simple situations considered above that native intuition serves in lieu of a formal calculation. This suggests that the mathematical definition of independence captures the “right” abstraction. Intuition, however, is not a reliable guide and one needs to turn to actual computation to verify whether systems of events (or random variables) are indeed, formally, independent. Cautionary examples are provided below.

# 概率论代写

## 数学代写|概率论代考Probability Theory代写|A rule of products

$$\underbrace{m+m+\cdots+m}_{n \text { terms }}=m n$$

2) 骰子。如果㚘六个骰子，则有 6 6可能的结果。任何给定的抛掷的六种可能结果中恰好有三种产生一个偶数 面，因此偶数面的概率是 $3 / 6=1 / 2$. 所有六次抛郑结果都是平脸的事件有 $3^6$ 有利的结果和概率 $\frac{3^6}{6^6}=\left(\frac{1}{2}\right)^6$. 3) 骨灰盒。一个瓮有 1 个黑球、 1 个绿球、 2 个蓝球和 4 个红球，所有球都被认为是可区分的。如果从罐子里 随机取出一个球，抽到黑色、绿色、蓝色或红色球的概率是 $1 / 8,1 / 8,1 / 4 ＼mathrm{~ ， 和 ~} 1 / 2$ ，分别。如果放回抽取四 个球，则可能的结果总数为 $8^4$ ； 有利于抽签顺序为黑色、绿色、蓝色和红色的事件的结果数是 $1 \cdot 1 \cdot 2 \cdot 4=8$. 这个事件的概率是 $\frac{8}{8^4}=\frac{1}{8} \cdot \frac{1}{8} \cdot \frac{1}{4} \cdot \frac{1}{2}$.

## 数学代写|概率论代考Probability Theory代写|What price intuition?

2) 相对等级。假设四个学生排名的所有排列 $a, b, c$ ，和 $d$ 是同样可能的。事件 $\mathrm{A}$ 那 $a$ 排在前面 $d \mathrm{~B}$ 那个 $b$ 排在前面 $c$ 在直觉上是独立的。事实上，正如很容易验证的那样， $\mathbf{P}(A)=\mathbf{P}(B)=1 / 2$ 和 $\mathbf{P}(A \cap B)=1 / 4$.
3) 骰子。掷出两个骰子。事件 $A$ 第一个骰子显示 ace 有概率 $1 / 6$ 而事件 $B$ 两个骰子的面值之和为奇数的概率 $1 / 2$ (因为有 18 种奇偶或奇偶组合) 。并作为 $A \cap B=(1,2),(1,4),(1,6)$ 有概率 $\frac{1}{12}=\frac{1}{6} \cdot \frac{1}{2}$, 事件是独立 的。

avatest.org 为您提供可靠及专业的论文代写服务以便帮助您完成您学术上的需求，让您重新掌握您的人生。我们将尽力给您提供完美的论文，并且保证质量以及准时交稿。除了承诺的奉献精神，我们的专业写手、研究人员和校对员都经过非常严格的招聘流程。所有写手都必须证明自己的分析和沟通能力以及英文水平，并通过由我们的资深研究人员和校对员组织的面试。

## 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。