Transcript 1. 5 conditional probability

Probability
theory
The department of math of central
south university
Probability and Statistics Course group
§ 1.5 probability terms, the whole formula and
Bayesian probability formula
For A, B is a randomized trial of the
two events, and P (A)> 0, said B in the event
of "A incidents have taken place" under
conditions of the probability of A in the
incident have taken place under the
conditions of the incident B of conditional
probability, known as B under the conditions
in the probability of A, recorded as P (B |
A).
Cases cited:
7 There are pockets of white balls, 3 red
and white ball in the 4-ball, 3 plastic ball;
red ball in the 2-ball, the ball is a plastic
bag from the office take a ball, assuming
that each The ball was to take the
possibility of the same. If it is known to
take the ball is the white ball, it was asked
Cricket is the probability of how many?
Set:
A term that I take the ball, made of white ball;
B says that his taking a ball, made Cricket.
Solution: List
white
ball
red
ball
subtotal
Cricket
4
2
6
plastic
ball
subtotal
3
1
4
7
3
10
4
P  B A 
7
k B A  4  k AB
n A
P( AB)

 7  k A P( A)
Consequently:
4 k AB
P B A  
7 kA
k AB

kA
n
n
4 / 10 P( AB)


P( A)
7 / 10
The definition of conditional probability
Definition:
For A, B for the two events, P (A)> 0,
Said: P (AB) / P (A) for the event A
occurred under the conditions of B
conditional probability of occurrence of the
incident, recorded as
P ( AB )
P B A 
P ( A)
Example 1
Already knows some family having 3 children ,
have one to be a girl and at least , ask be a
family's turn to probability having a boy at least
Solution:
A = (3 for children at least one of the girls)
B = (3 children at least one of the boys)
so
1 7
P  A  1  P  A   1  
8 8
therefore
6
P AB  
8
6
P AB 
6
8
PB A 


7
P  A
7
8
The nature of conditional probability
P ( B A)  0
P ( A)  1
Out to be additive

 
P   Bi A    P Bi A
 i 1
 i 1
P( B1  B2 A)  P( B1 A)  P( B2 A)  P( B1B2 A)
P ( B A)  1  P ( B A)
P ( B1  B2 A)  P ( B1 A)  P ( B1B2 A)
Multiplication formula
By the conditions of quadrature probabilities
probability events That is, a multiplication
formula
P ( AB )  P ( A) P B A ( P ( A)  0)
P ( AB )  P ( B) P  A B  ( P ( B)  0)
deduce
P ( A1 A2 A3 )  P ( A1 ) P  A2 A1  P  A3 A1 A2 
( P ( A1 A2 )  0 )
Example 2
a factory producing 1,000 hours of lamp can be a
probability of 0.8, 1500 hours can be a probability of
0.4, seek has spent 1,000 hours of bulb use to 1,500
hours of probability.
Solution: A light bulb can be made to 1000
hours ,lamp B to be 1500 hours
the probability:
P ( AB ) P ( B ) 0.4 1
P  B A 



P ( A)
P ( A) 0.8 2
B A
Example 3
A batch of products, Jiachang products accounted
for 60% of known defective products Jiachang the
rate of 10% from these products taken from a
random, and the product is defective production of
Jiachang Probability.
Solution: A set that the incident "is produced by Jia
Chang"
B said that the incident "is a defective product,"
set up by the title to know
P(A)=60%
P(B|A)=10%
Be based on multiplication formula
P ( AB )  P ( A) P ( B A)  60%  10%  6%
Example 3 A batch of 100
pieces of products


Jiachang produce 40
70 authentic
30 defective
Yichang produce 30

Jiachang produce
20
Yichang produce
30
To take a term, Hutchison A = “taking into account the
genuine”, table, and B = “Jia Chang to take the product”,
try the calculation: P(A),P(B),P(AB),P(B|A),P(A|B).
Solution:
70
P( A) 
 0.7
100
60
P( B) 
 0.6
100
40
P( AB) 
 0.4
100
40 4
P( B | A) 

70 7
40 2
P( A | B) 

60 3
Example 4 :
from a mixture of 5 counterfeit bank notes of 20 dollars in
any 2 out, which would put a test on Yanchao Ji found is
the money, seek 2 are the probability of money.
Solution:
A to "be able to get into 2 are false”
B said that "at least 2 in a false”
Then the probability is for
A B
P  A B  ( Instead of P  A )
P AB  P(A) C 52 / C 220
PB   (C  C C ) / C
2
5
1
5
1
15
2
20
therefore
PA B   P( AB) / P( B)
 C /(C  C C )  10 / 85  0.118
2
5
2
20
1
5
1
15
Example 5
in the box with 5, 3 of them first-class
products, 2 second-class goods, which do not
get back to the product, a per-seeking:
(1) take two, scored two first-class goods probability;
(2) taking two for the second time to obtain the
probability of first-class goods;
(3) take three times, third only to obtain the
probability of first-class goods;
(4) take two, known for the second time to obtain
first-class products, seek:
The first is made of the probability of secondclass goods.
Solution: Set Ai order for the first time i get
to the first-class goods
（1）
3 2 3
P( A1 A2 )  P( A1 ) P( A2 A1 )   
5 4 10
（2）
P( A2 )  P( A1 A2  A1 A2 )  P( A1 A2 )  P( A1 A2 )
2 3 3 2 3
    
5 4 5 4 5
（3） More direct answers to simple
P( A2 )  3 / 5
Question: When was the third time to obtain
first-class goods probability is
P ( A3 A1 A2 ) or P ( A1 A2 A3 ) ?
  

(3) P( A1 A2 A3 )  P A1 P A2 A1 P A3 A1 A2
2 1 3 1
   
5 4 3 10

(4)
P( A1 A2 ) P( A2 )  P( A1 A2 )
P A1 A2  

P( A2 )
P( A2 )
 1
3
10
3
5
 0.5
General
Conditional probability and the probability of
an unconditional between free to determine
the relationship between the size of
The above example
P ( A1 A2 ) 0.1 1
P ( A2 A1 ) 

  P ( A2 )
P ( A1 ) 0.6 6
if
B A
P ( AB ) P ( B )
P  B A 

 P( B)
P ( A) P ( A)
A short break to continue