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概率论快速学习03:概率公理补充

 quasiceo 2017-12-26
                 原创 2014年05月25日 00:26:00

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

  “应注意到一个析取命题的对立命题是由该析取命题各部分的对立内容构成的一个合取命题” ——奥卡姆的威廉著,《逻辑学论文》


Written In The Font

  I  like maths when i was young,but I need to record them. So I am writing with some demos of Python

 

Content

  If two events, A and B are independent then the joint probability is

   P(A \mbox{ and }B) =  P(A \cap B) = P(A) P(B),\,

          Durchschnitt.png                                         

 

For example, if two coins are flipped the chance of both being heads is

\tfrac{1}{2}\times\tfrac{1}{2} = \tfrac{1}{4}.

 

In Python

A = set([1,2,3,4,5])
B = set([2,4,3,5,6])
C = set([4,6,7,4,2,1])

print(A & B & C)

Output:

{2, 4}

# & find the objects  the same in Set


         

   If either event A or event B or both events occur on a single performance of an experiment this is called the union of the events A and B denoted as:

   P(A \cup B).

  If two events are mutually exclusive then the probability of either occurring is

  P(A\mbox{ or }B) =  P(A \cup B)= P(A) + P(B).

            Vereinigung.png

 

For example, the chance of rolling a 1 or 2 on a six-sided die is

 P(1\mbox{ or }2) = P(1) + P(2) = \tfrac{1}{6} + \tfrac{1}{6} = \tfrac{1}{3}.

 

In Python

A = set([1,2,3,4,5])
B = set([2,4,3,5,6])
C = set([4,6,7,4,2,1])

print(A | B | C)

Output:

{1, 2, 3, 4, 5, 6, 7}

# | find all the objects the set has


  If the events are not mutually exclusive then

  P\left(A \hbox{ or } B\right)=P\left(A\right)+P\left(B\right)-P\left(A \mbox{ and } B\right).

Proved

  \begin{align}P(A\cup B) & =P(A\setminus B)+P(A\cap B)+P(B\setminus A)\\& =P(A)-P(A\cap B)+P(A\cap B)+P(B)-P(A\cap B)\\& =P(A)+P(B)-P(A\cap B)\end{align}

 

For example:

  Let’s use Python to show u an example about devil's bones (骰子,不是 魔鬼的骨头哈87B7B1~1_thumb)

复制代码
A = set([1,2,3,4,5,6])  # the all results of devil's bones
B = set([2,4,3])        # the A event results 
C = set([4,6])          # the B event results 

P_B =  1/2
P_C =  1/3

D = B | C
print(D)

P_D = 2/3

print(P_D == (P_B+P_C - 1/6))
复制代码

Output:

{2, 3, 4, 6}
True

 


Let me show u some others :


         P(A)\in[0,1]\, 
         P(A^c)=1-P(A)\, 
         \begin{align}P(A\cup B) & = P(A)+P(B)-P(A\cap B) \\P(A\cup B) & = P(A)+P(B) \qquad\mbox{if A and B are mutually exclusive} \\\end{align} 
         \begin{align}P(A\cap B) & = P(A|B)P(B) = P(B|A)P(A)\\P(A\cap B) &  = P(A)P(B) \qquad\mbox{if A and B are independent}\\\end{align} 
        P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{P(B|A)P(A)}{P(B)} \, 

 

If u r tired , please have a tea , or look far to make u feel better.If u r ok, Go on!


  Conditional probability is the probability of some event A, given the occurrence of some other event B. Conditional probability is written:

   P(A \mid B),

   Some authors, such as De Finetti, prefer to introduce conditional probability as an axiom of probability:

P(A \cap B) = P(A|B)P(B)

Given two events A and B from the sigma-field of a probability space with P(B) > 0, the conditional probability of A given Bis defined as the quotient of the probability of the joint of events A and B, and the probability of B:  

  P(A|B) = \frac{P(A \cap B)}{P(B)}

  the ①② expressions  are the same. Maybe u can remember one , the other will be easy to be coverted.So I am going to tell an excemple to let u remmeber it(them):

  

  “the phone has a power supply (B), the phone can be used to call others(A).”

  One →  P(A \mid B) : When the phone has a full power supply , u can call others.

  Two →P(B): has   a power supply           

  Three = One +  Two → U can call others about your love with others.

 

do u remember it?

                                                                 u=2234442768,3895906646&fm=21&gp=0_thumb[1]

 

Editor's Note

    “路漫漫其修远兮,吾将上下而求索”

 

The Next

            cya soon. We meet a big mess called The total probability and Bayes .

 

      The total probability

      P( A )=P( A | H_1) \cdot P( H_1)+\ldots +P( A | H_n) \cdot P( H_n)
      P(A)=\sum_{j=1}^n P(A|H_j)\cdot P(H_j)

      Bayes (Thomas, 1702-1761,) ;  

       P(A \vert B) = \frac {P(B \vert A) \cdot P(A)} {P(B)}

if u wanna talk with me , add the follow:

 

QQ截图20140525001523_thumb[3]

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