Transcript 1. 8 The first chapter Exercise

Probability
theory
The department of math of central south
university
Probability and Statistics Course group
The first chapter Exercise
Example 1
Assuming there are three boxes of similar models
of parts, which were equipped with 50, 30 and 40,
and first-class goods were 20, 12 and 24. Now choose
from a case of random samples of all has a spare
parts (for the first time to take back the parts do not)
try to seek access to spare parts is the probability of
first-class goods and out calculating both the
probability of first-class goods.
Solution: set
B1 , B2 , B3 respectively in the first and
second election of the first-class products with pieces of
the box for 20,12,24, respectively for the first time, for
the second out of first-class products, according to
Italian title,
3
P ( A1 )   P ( Bi ) P ( A1 | Bi )
i 1
1 20 19 1 12 11 1 24 23
          0.22
3 50 49 3 30 29 3 40 39
3
P ( A1 A2 )   P ( Bi ) P ( A1 A2 | Bi )
i 1
Example 2
A, B, C at the same time of three aircraft fire,
hitting three of the probability of 0.4,0.5,0.7.
Planes were shot down and one hit a probability
of 0.2 and was hit by two shot down by a
probability of 0.6 If all three hit, the aircraft must
have been shot down, and the probability of the
aircraft was shot down.
Solution: Let B = ( plane was shot down),
Ai = (aircraft was hit i people), i = 1,2,3
By the whole probability formula
P (B) = P (A1) P (B | A1) + P (A2) P (B | A2)
+ P (A3) P (B | A3)
= 0.36 × 0.2 +0.41 × 0.6 +0.14 × 1
= 0.458
That the aircraft was shot down by a
probability of 0.458.
Example 3
From the assumption that there are three regions of the
10, 15 and 25 candidates of the nomination form, the
application forms which girls were 3, 7 and 5. Are
randomly out of a region of the entry form, and from
one after another at random Two out.
(1) to seek out a female form of probability;
(2) known after the draw is a table of boys,to seek out a
female form of probability.
solution:introduction of events：
H j {application forms is the first region j} ；( j  1,2,3)
Ai ={I is the first time be able to get into the boys
table} (i  1,2,3) . by the conditions of knowledge:
1
P ( H1 )  P ( H 2 )  P ( H 3 )  ，
3
7
8
20
P( A1 | H1 )  ，P( A1 | H 2 )  ，P( A1 | H 3 )  ．
10
15
25
by the probability of the whole formula, we have
3
1 3 7
5  29
  P( A1 )  P( H j ) P( A1 | H j )       ．
3  10 15 25  90
j 1

Easy to see P( A A｜H )  3  7  7 ;
1 2
1
10  9
P( A1 A｜
2 H2 ) 
30
78
8
5  20
5
 ; P( A1 A｜
H
)


2
3
15 14 30
25  24 30
by the probability of the whole formula:
P( A2 ) 
P( A1 A2 ) 
3
1  7 8 20  61
P( H j ) P( A2 | H j )       ；
3  10 15 25  90
j 1

3
1 7
8
5 2
P( H j ) P( A1 A2 | H j )   
  ．
3  30 30 30  9
j 1

by the definition of conditional probability
P( A1 A2 )
29
20
  P( A1 | A2 ) 


P( A2 )
61 90 61
Example 4 set A, B, C the probability of the same
events and independent of each other, P( ABC )  7 8
seek P ( A)
Solution
set p  P( A)  P( B)  P(C．and
the formula by
)
adding event
A, B, C independent ,we have
7
 P( A  B  C )  1  P( A B C )
8
 1  P( A )P( B )P(C )  1  (1  p)3，
P ( A)  1 2
Example 5
Prove events A, B, C that the full
independence and a necessary condition is that they
are pairwise independent and any of the events and
the remaining two events of the independent pay.
Proof: (1) need
Events for A, B, C independent, any two of them
independent of each other; In addition, by
P (A [BC]) P (ABC) = P (A) P (B) P (C) = (A) P (BC),
A visible with the independent BC. The same evidence, B
and AC independent, C and AB independence.
(2) Adequacy events A, B, C pairwise independent
and any of the events and the rest of the second
incident of cross-independent. Pairwise independent by
the independent, to know
P( AB)  P( A)P( B )，P( AC )  P( A)P(C )，P( BC )  P( B )P(C )．
A result of BC and independence, can be seen
P ( ABC )  P ( A[ BC ])  P ( A) P ( BC )
 P ( A) P ( B ) P (C )
So events A, B, C independent．
A short break to continue