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  • 概率论快速学习04:概率公理 全概率 贝叶斯 事件独立性

    原创地址: http://www.cnblogs.com/Alandre/ (泥沙砖瓦浆木匠),需要转载的,保留下! Thanks

           加油! 大牛总是不断努力,你却更需要加倍努力.

    Written In The Font

        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}
ight} 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(Amid B_n)Pr(B_n),\,

    example:

    例. 甲、乙两家工厂生产某型号车床,其中次品率分别为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任意抽检一件,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(A) using 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),

    egin{align}P(	ext{Rare}|	ext{Pattern}) &=
frac{P(	ext{Pattern}|	ext{Rare})P(	ext{Rare})} {P(	ext{Pattern}|	ext{Rare})P(	ext{Rare}) \, + \, P(	ext{Pattern}|	ext{Common})P(	ext{Common})} \[8pt]
&= frac{0.98 	imes 0.001} {0.98 	imes 0.001 + 0.05 	imes 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:

    egin{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}(Amid 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

    “学吧,至少不亏.”一句良言 终身受用.

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  • 原文地址:https://www.cnblogs.com/Alandre/p/3764119.html
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