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assignmentutor-lab™ 为您的留学生涯保驾护航 在代写线性代数linear algebra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。我们的专家在代写线性代数linear algebra代写方面经验极为丰富，各种代写线性代数linear algebra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|线性代数代写linear algebra代考|Conditional Probability and Bayes’ Theorem

A probability topic that easily confuses people is the concept of conditional probability, which is the probability of an event A occurring given event $\mathrm{B}$ has occurred. It is typically expressed as $P(\mathrm{~A}$ GIVEN B) or $P(\mathrm{~A} \mid \mathrm{B})$.

Let’s say a study makes a claim that $85 \%$ of cancer patients drank coffee. How do you react to this claim? Does this alarm you and make you want to abandon your favorite morning drink? Let’s first define this as a conditional probability $P$ (Coffee given Cancer) or $P$ (Coffee|Cancer). This represents a probability of people who drink coffee given they have cancer.

Within the United States, let’s compare this to the percentage of people diagnosed with cancer $(0.5 \%$ according to cancer.gov) and the percentage of people who drink coffee ( $65 \%$ according to statista.com):
\begin{aligned} &P(\text { Coffee })=.65 \ &P(\text { Cancer })=.005 \ &P(\text { Coffee } \mid \text { Cancer })=.85 \end{aligned}
Hmmmm…study these numbers for a moment and ask whether coffee is really the problem here. Notice again that only $0.5 \%$ of the population has cancer at any given time. However $65 \%$ of the population drinks coffee regularly. If coffee contributes to cancer, should we not have much higher cancer numbers than $0.5 \%$ ? Would it not be closer to $65 \%$ ?

This is the sneaky thing about proportional numbers. They may seem significant without any given context, and media headlines can certainly exploit this for clicks: “New Study Reveals $85 \%$ of Cancer Patients Drink Coffee” it might read. Of course, this is silly because we have taken a common attribute (drinking coffee) and associated it with an uncommon one (having cancer).

## 数学代写|线性代数代写linear algebra代考|Joint and Union Conditional Probabilities

Let’s revisit joint probabilities and how they interact with conditional probabilities. I want to find the probability somebody is a coffee drinker AND they have cancer. Should I multiply $P$ (Coffee) and $P$ (Cancer)? Or should I use $P$ (Coffee|Cancer) in place of $P$ (Coffee) if it is available? Which one do I use?
Option 1:
$P($ Coffee $) \times P($ Cancer $)=.65 \times .005=.00325$
Option 2:
$P($ Coffee $\mid$ Cancer $) \times P($ Cancer $)=.85 \times .005=.00425$

If we already have established our probability applies only to people with cancer, does it not make sense to use $P$ (Coffee|Cancer) instead of $P$ (Coffee)? One is more specific and applies to a condition that’s already been established. So we should use $P$ (Coffee|Cancer) as $P$ (Cancer) is already part of our joint probability. This means the probability of someone having cancer and being a coffee drinker is $0.425 \%$ :
$P($ Coffee and Cancer $)=P($ Coffee $\mid$ Cancer $) \times P($ Cancer $)=.85 \times .005=.00425$
This joint probability also applies in the other direction. I can find the probability of someone being a coffee drinker and having cancer by multiplying $P$ (Cancer|Coffee) and $P$ (Coffee). As you can observe, I arrive at the same answer:
$$P(\text { Cancer } \mid \text { Coffee }) \times P(\text { Coffee })=.0065 \times .65=.00425$$
If we did not have any conditional probabilities available, then the best we can do is multiply $P$ (Coffee Drinker) and $P($ Cancer) as shown here:
$$P(\text { Coffee Drinker }) \times P(\text { Cancer })=.65 \times .005=.00325$$

# 线性代数代考

## 数学代写|线性代数代写linear algebra代考|Conditional Probability and Bayes’ Theorem

$$P(\text { Coffee })=.65 \quad P(\text { Cancer })=.005 P(\text { Coffee } \mid \text { Cancer })=.85$$

## 数学代写|线性代数代写linear algebra代考|Joint and Union Conditional Probabilities

$P$ (咖啡 $) \times P($ 㾬症 $)=.65 \times .005=.00325$

$P($ 咖啡与痹症 $)=P$ (咖啡癌症 $) \times P($ 癌症 $)=.85 \times .005=.00425$

$$P(\text { Cancer } \mid \text { Coffee }) \times P(\text { Coffee })=.0065 \times .65=.00425$$

$$P(\text { Coffee Drinker }) \times P(\text { Cancer })=.65 \times .005=.00325$$

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## 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。

assignmentutor™您的专属作业导师
assignmentutor™您的专属作业导师