概率论高速学习04:概率公理 全概率 贝叶斯 事件独立性

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概率论高速学习04:概率公理 全概率 贝叶斯 事件独立性

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    数学和生活是技术之本, 有了数学,加上生活,才会开心.

    今天继续概率论:

  • 全概率
  • 贝叶斯
  • 事件独立性

Content

The total probability


In the Set :
    image imageimage

                                                             image

The law of total probability is the proposition that if \\left\\{{B_n : n = 1, 2, 3, \\ldots}\\right\\} is a finite or countably infinitepartition of a sample space (in other words, a set of pairwise disjoint events whose union is the entire sample space) and each event B_n is measurable, then for any event A of the same probability space:

             \\Pr(A)=\\sum_n \\Pr(A\\mid B_n)\\Pr(B_n),\\,

example:

例. 甲、乙两家工厂生产某型号车床,当中次品率分别为20%, 5%。已知每月甲厂生产的数量是乙厂的两倍,现从一个月的产品中随意抽检一件,求该件产品为合格的概率?

A表示产品合格,B表示产品来自甲厂

image

 

Bayes


for some partition {Bj} of the event space, the event space is given or conceptualized in terms of P(Bj) and P(A|Bj). It is then useful to compute P(Ausing the law of total probability:        

                               image

 

example:

An entomologist spots what might be a rare subspecies of beetle, due to the pattern on its back. In the rare subspecies, 98% have the pattern, or P(Pattern|Rare) = 98%. In the common subspecies, 5% have the pattern. The rare subspecies accounts for only 0.1% of the population. How likely is the beetle having the pattern to be rare, or what is P(Rare|Pattern)?

From the extended form of Bayes\' theorem (since any beetle can be only rare or common),

\\begin{align}P(\\text{Rare}|\\text{Pattern}) &=\\frac{P(\\text{Pattern}|\\text{Rare})P(\\text{Rare})} {P(\\text{Pattern}|\\text{Rare})P(\\text{Rare}) \\, + \\, P(\\text{Pattern}|\\text{Common})P(\\text{Common})} \\\\[8pt]&= \\frac{0.98 \\times 0.001} {0.98 \\times 0.001 + 0.05 \\times 0.999} \\\\[8pt]&\\approx 1.9\\%. \\end{align}

 

One more example:

image

 

Independence


Two events

Two events A and B are independent if and only if their joint probability equals the product of their probabilities:

\\mathrm{P}(A \\cap B) = \\mathrm{P}(A)\\mathrm{P}(B).

Why this defines independence is made clear by rewriting with conditional probabilities:

\\begin{align}\\mathrm{P}(A \\cap B) = \\mathrm{P}(A)\\mathrm{P}(B) &\\Leftrightarrow \\mathrm{P}(A) = \\frac{\\mathrm{P}(A \\cap B)}{\\mathrm{P}(B)} \\\\&\\Leftrightarrow \\mathrm{P}(A) = \\mathrm{P}(A\\mid B)\\end{align}

how about Three events

           image

 

sometimes , we will see the Opposition that can be used to make the mess done. We will use the rule of independence such as : P(A^c)=1-P(A)\\,

 

Editor\'s Note

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