Download 1.3 classical sample

Probability
theory
The department of math of central south
university
Probability and Statistics Course group
§1.3 Classical Probability
(1) Classical Probability Model
supposeΩis the sample space of test E，if
①（The limited nature）Ωonly contains limited
sample;
②（the nature of same probability ）Each of the basic
events have same possibility;
sample.
E is said classical
(2) Estimate the probability of classical-definition, E for the
classical estimate, Ω for the
sample space of E, A for any event,.The definition of
the probability of events A
n( A) k
P( A) 

n ( ) n
NOTE:
(1) the method to determine the type of classical
sample (limited, and so on and over);
(2) the calculation of the probability of classical
steps:
① find out the test and sample point;
②count the sample points number of samples space and
the sample points number of random event ;
③list and calculate.
Example 1. throwing a dice twice, try to find
the following for the probability of events:
(1) add the twice number,is 8
(2)the twice number is 3.
Answer：
suppose A shows the event of“add the
twice is 8”
B shows the event
“ the twice number is 3。”
so
Ω  {(1,1), (1,2), , (1,6), (2,1), , (2,6), , (6,1),  (6,6)}
A  {(2,6), (3,5), (4,4), (5,3), (6,2)}
B  {(1,3), (2,3), (3,3), (4,3), (5,3), (6,3)}
so
5
P ( A) 
36
6 1
P(B)  
36 6
Example2 the box has 6 light bulbs, 2 defective 4
authentic, take twice with having taken back , each
time taking a test for the probability of the following
events:
（1）the two have get are both
defective;
（2）the two have get ,one is
fefective,one is auhentic
（3）the two have get have one
authentic at least.
Answer： supposeA ={the two have get are both
defective}，B={the two have get have one
authentic ,one defective}, C={the two have get
have one authentic at least}.
（1）the total of sample is62，the
sample number of A is 22，
so：P(A)=4/36=1/9
P(C)=32/36=8/9
（2）the sample number of B 4×2+2×4=16，
so：P(B)=16/36=4/9
（3）the sample number of C 62-2×2=32，
Thoughts: ① if not changed back to taking
it twice?
② be changed if one of the two
samples do?
Example3 A bag has 6 balls, 4 white balls, 2 red. Take
the ball from the pocket twice, each time a random
check. Taking into account both the ball:
Back a sample: for the first time taking a ball,
observed after the color back into the bag, mix after
taking a ball.
Do not back a sample: for the first time do not get a
ball back into the bag for the second time from the rest
of the ball in a ball again. For:
1）the probability of the two were taken are
both white
(2) The probability to get the ball two different
colors of;
3) the probability to take two balls at least one
white ball
8 5
1 9 4 6
7 2 3 10
Answer：Get two goals from the bag, each is
a fundamental incident.
Suppose A= “the two have get are both white
balls ”，
B= “the two have get have same colour ”，
C= “the two have get have one white ball
at least.”。
Have taken back:
42
P( A)  2  0.444
6
42  22
P( B ) 
 0.556
2
6
22
P(C )  1  P(C )  1  2  0.889
6
Without taking back:
P(A) 
C 24
C 62
C C
P(B) 
2
C6
2
4
2
2
C22
P(C)  1  P(C )  1  2
C6
Example 4 A family has three children .The probability
of having each child is male or female is same, how
many the probability of having one boy at least?
Answer：suppose A shows have one boy at
least，H shows the some child is boy。
N(S)={HHH,HHT,HTH,THH,HTT,TTH,THT,TTT}
N(A)={HHH,HHT,HTH,THH,HTT,TTH,THT}
N ( A) 7
P( A) 

N (S ) 8
Example5 take any one number from natural number 1200;
(1) the probability to obtain a number that can be divide
by 6;
(2) the probability to obtain the number can be divide by
8;
(3) the probability to obtain can be divisible by 6 and 8
answer:N(S)=200, N(1)=[200/6]=33,
N(2)=[200/8]=25, N(3)=[200/24]=8
The probability of (1),(2),(3) :33/200,1/8,1/25
Example 6 A reception center in a week, had received
12 visitors, all known to receive this 12 times in Tuesday
and Thursday. Asked whether the reception may be
inferred that the time is required?
answer：Assuming no time for the reception of the
reception center provides the visitors in the office one day
a week to go in reception center, and so is possible, then
the 12th to receive visitors are Tuesday, Thursday, with a
probability of:
212/712=0.0000003，that is 3/10000000.
Example 7 there are n persons,and everyone has the
same probabilityto be assigned to any room to a
living of N rooms. (
n  N )，Seek
the probability of the following events: (1)
designated n rooms have their own
a person living; (2) have n just rooms, all of
which lives one person .
Answer :Because everyone has N rooms
available, so the kind of living
is N n , such as they have same possibility.
In the first issue, n designated rooms have a
living person, which could total number of
individuals n the whole array of n!, Then
n!
P1 
n
N
In the second question in, n-room can be room N
arbitrary selection, which have a total
N
 ,
n
n pairs of
selected rooms, according to the foregoing discussion
we can see there are n! Species distribution, so there
are exactly n rooms All of which live in a person's
probability of
N
  n!
n
N!

P2 

Nn
N n ( N  n)!
Added: combination of notation
Additive principle: there are n-type approach to
complete one thing.for the i-type category
there are ways mi species specific methods,.The total to
n
complete this matter have
 mi
i 1
methods.
Multiplication principle: there is n steps to complete one
thing, the i steps have mi approachs, then the total
of ways to complete this matter is
n
 mi
i 1
Array:take m elements from n different
elements(without taking back ), according to a
certain degree of order lined up a different array,
A total of array is
Pnm  n( n  1)( n  2)( n  m  1)
Whole array:
Pnn  n!
Repeated array:take m elements from n different
elements ,that can repeatedly, the m elements can
.
m
line up a row.the kinds of the row is n
Portfolio: take m elements from n different
elrments(without taking back) ,the m elements
line up a row ,The kinds of the row are
n!
n
 
 m  m!( n  m )!
 n
0! 1和   1
0
Repeated Portfolio : take one elements each
time from n different elements,and take
back,then take one.taking r times line up a
portfolio,is said to repeated portfolio
.the number of repeated portfolio is
 n  r  1


r

For example: two groups have both 50 products, of
which both have 5 defective from these two
groups .Take one product from both groups.
(1) the number of species that the two products are
not defective?
(2)the number of species that only have one defective?
Answer: (1) use multiplication principle,
the results for the
1
1
C 45
.C 45
 45 2
(3) Additive combination of theory and principle of
a multiplication method for the election:
1
1
C 51 .C 45
 C 45
.C 51  2  5  45  450
A short break to continue