Based on the probability of the problem (independent and conditional probabilities) caused by infectious diseases

Overview:

Yesterday a working mass infection infection probability questions about infectious diseases, the whole group were bombed, planning, back-end, front-end, test each with their own thinking to try to give this issue a correct answer. Of course, I joined them, but the result is, emmm, I began to suspect that he is not is not suitable for a programmer.

problem:

Result of discussion:

I think the final outcome of the discussion seems more reliable about two:

  • 52.734375%

Computing ideas

  • 50%

Computing ideas

Finally, we discuss to find out which is the correct answer, we want to solve this problem we must understand some basic concepts of probability, the following is a brief look at some things you need to know some of probability.

Probability

definition

Probability, also known as "probability", which is a reflection of random events likelihood of occurrence (likelihood) size. Random event is under the same conditions, the event may not appear likely. For example, from a number of genuine and defective merchandise, the one at random, "pits an authentic" it is a random event. A random set of n trials phenomena observed, wherein A is m times the event occurs, i.e., their frequency is m / n. After much trial and error, often m / n closer and closer to a determined constant (see prove this assertion Bernoulli law of large numbers). This constant is the probability that event A occurs, commonly used P (A) represented.

 Probability theory discussed in separate incidents

When before the start, we have to clearly describe the probability of the problem is a problem, we must accurately grasp the "sample space", the probability of this book is generally known as the set of all possible outcomes constitute as "sample space." If A in the sample space is a description of the problem A, which is based on the probability that an A, and B in a further different sample derived a probability space B, and then discuss the relationship caution is necessary or else donkey lip not the horse's mouth.

1. conditional probability

When learning the conditional probability conditional probability will get the following formula:

           P(E|F)=\frac{P(EF)}{P(F)}, P(F)>0           

This formula means that under the conditions of the incident F, E probability events. Literally, subjective feelings feel this is very easy to be understood by a formula. Understanding the probability of this condition formula, then there is a deep understanding of the need to switch when discussing the problem of probability "sample space". With "Venn diagram" to understand more easily grasp the essence:

 

P(E|F)=\frac{S_{red}}{S_{F}},即条件概率P(E|F)等于红色部分面积(EF相交部分面积)除以F事件面积(绿色+红色面积)

也就是说求P(E|F)时候的样本空间是以F事件的样本空间为参考的,这与P(E)=\frac{S_{E}}{S_{A}}(即F事件面积除以A原始样本空间面积)。

也就是说P(E|F)P(E)两个概率所参考的样本空间完全不一样:

 

P(E)是基于原始样本空间A,P(E|F)是基于新的样本空间F。

2. 事件独立

定义:对于事件E和事件F,如果满足下面的公式,那么称它们是独立的。若两个事件E和F不独立,则称它们是相依的,或者相互不独立。

P(EF)=P(E)P(F)

因此如果事件E和事件F独立,那么肯定满足下满的式子:

P(E)=P(E|F)

观察上面的维恩图,可知:

\frac{S_{red}}{S_{F}}=\frac{S_{E}}{S_{A}}

这也就说明了事件F的样本空间对事件E样本空间的切割后这部分(即形成维恩图中红色部分空间)在F中的比例和事件E在原来总体样本空间A中的比例是一致的,实际上这种"同比例切割"的特性,是确定F与E是否独立的一个标志,如果F事件样本空间同比例切割E事件空间,那么E和F就是独立的。这样子的描述和"F事件的发生并不影响E发生的概率,那么E和F就是独立的", 事实上这样子的描述在主观上有时候不是特别容易判断的。用"同比例切割"有时候更容易判断两个事件是否是独立的。相反的,不能同比例切割的话,那可以判断E和F是不独立的。

利用这个结论,观察上面这个维恩图,它告诉我们,E和F事件没有相交的部分,按照"同比例切割"的观点,F事件和E事件是"不独立"的! 当然也可以利用是否满足公式P(EF)=P(E)P(F)方式去验证独立性。 这个图告诉我们:

两个不相交的事件,反而是"相互不独立"的。除了一种情况,事件E不可能出现,也就是P(E)=0。

这给我们一种新的认识:世界上两个没有任何交集的人,却相互不独立。除非你不存在。

造成这种错觉的原因是,讨论问题的角度不一样,相交讨论的是两个事件的集合,而"独立性"与否讨论的是比例(也就是概率)的问题。另外,概率论中的"独立"都是特别针对概率值的影响的,而人的独立性讨论的是人格特征。概率论中只是借用了"独立"这个词,概念上被赋予了严格的数学意义。

例1. 从一副洗好的52张扑克牌里随机抽取一张牌,令E表示事件"抽取的牌为一张A",令F表示事件"抽取的牌为一张黑桃",那么E和F就是独立的。因为P(EF)=1/52,而P(E)=4/52且P(F)=13/52。

这个例子也可以用"同比例分割"的方法来判断。原始样本空间大小为52,事件E空间大小有4(因为有4张牌A),因此事件E在原来空间中的分割比例时4/52。 相交事件EF样本空间(既是牌A又是黑桃)1,事件F的样本空间很明显是13(因为有13张黑桃),因此,EF在F中的分割比例为1/13。4/52=1/13,因此是独立的。

例2. 掷两枚均匀的骰子,令表示事件"骰子点数和为6",令F表示事件"第一枚骰子点数为4",那么

P(E_{1}F)=P({4,2})=\frac{1}{36}

P(E_{1})P(F)=\frac{5}{36}\times \frac{1}{6}=\frac{5}{216}

因此,和F不独立。也可以用"同比例分割"法。E1F相交事件在F中分割的比例为1/6,而E1事件在原来空间的比例是5/36。

例2, 如果令表示事件"骰子点数和为7",那么F和是独立的。请自证。

需要强调的是,两个事件独立并不代表两个事件之间没有影响,影响这个词太笼统了,因为,影响这个词没有说具体什么影响。而概率论中,事件之间是否独立,它强调的是事件F的出现与否对事件E原来发生的概率是否有影响!它明确了影响什么!即便事件F对事件E产生其他影响,只要不影响E的概率,那就是"独立"!

结果

最终结果是:d被感染的概率是50%

讨论结果一错误的原因是:

当bc互相握手后,b和c感染的概率都是62.5%(50%的概率是bc全都感染,12.5%的概率是bc只感染一个),此时62.5%中的50%和12.5%的采样空间是不一致的,必须分来计算与d握手时,d被感染的概率。

所以讨论结果一的第三步是错的。
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版权声明:本文为CSDN博主「皮皮君」的原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接及本声明。
原文链接:https://blog.csdn.net/qq_16587307/article/details/89415118

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