概率论快速学习03：概率公理补充

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　　“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

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

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

$frac{1}{2} imes frac{1}{2} = frac{1}{4}.$

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

　　 $P(Ambox{ or }B) = P(A cup B)= P(A) + P(B).$

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

$P(1mbox{ or }2) = P(1) + P(2) = frac{1}{6} + frac{1}{6} = frac{1}{3}.$

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

　　 $Pleft(A hbox{ or } B ight)=Pleft(A ight)+Pleft(B ight)-Pleft(A mbox{ and } B ight).$

Proved

　　 $egin{align} P(Acup B) & =P(Asetminus B)+P(Acap B)+P(Bsetminus A)\ & =P(A)-P(Acap B)+P(Acap B)+P(B)-P(Acap B)\ & =P(A)+P(B)-P(Acap B) end{align}$

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)

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 :

         $P(A)in[0,1]\,$
         $P(A^c)=1-P(A)\,$
         $egin{align} P(Acup B) & = P(A)+P(B)-P(Acap B) \ P(Acup B) & = P(A)+P(B) qquadmbox{if A and B are mutually exclusive} \ end{align}$
         $egin{align} P(Acap B) & = P(A|B)P(B) = P(B|A)P(A)\ P(Acap B) & = P(A)P(B) qquadmbox{if A and B are independent}\ end{align}$
        $P(A mid B) = frac{P(A cap B)}{P(B)} = frac{P(B|A)P(A)}{P(B)} \,$

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:

$P(A cap B) = P(A|B)P(B)$ ^①

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:　　

　　 $P(A|B) = frac{P(A cap B)}{P(B)}$ ^②

　　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?

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,) ;

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

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