概率论快速学习03:概率公理补充

简介:

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)

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))
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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