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Li Jie
Probability and Statistics
For Engineering and the Sciences
Fifth Edition
JAY L.DEVORE
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Birthday
• How many students in your class?
• The same birthday (at least two classmates)?
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What Is Probability
• Probability is a way of expressing knowledge or
belief that an event will occur or has occurred. In
mathematics the concept has been given an exact
meaning in probability theory, that is used
extensively in such areas of study as mathematics,
statistics, finance, gambling, science, and philosophy
to draw conclusions about the likelihood of potential
events and the underlying mechanics of complex
systems.
• http://en.wikipedia.org/wiki/Probability
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2 Probability
Introduction
The term probability refers to the study of randomness and
uncertainty. In any situation in which one of a number of
possible outcomes may occur, the theory of probability
provides methods for quantifying the chances, or likelihoods,
associated with the various outcomes. The language of
probability is constantly used in an informal manner in both
written and spoken contexts.
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2 Probability
Introduction
In this chapter, we introduce some elementary probability
concepts, indicate how probabilities can be interpreted, and
show how the rules of probability can be applied to compute
the probabilities of many interesting events. The methodology
of probability will then permit us to express in precise language
such informal statements as those given above.
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2.1
Sample Spaces and Events
Random Experiment
An experiment is any action or process that generates
observations.
For examples
1.tossing a coin once or several times
2.selecting a card or cards from a deck
3.weighing a loaf of bread
4.ascertaining the commuting time from home to work on
particular morning
5.obtaining blood types from a group of individuals, etc.
throw six-sided die
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Sample Space and Sample Point
Definition : The sample space of an experiment, denoted
by S (Ω), is the set of all possible outcomes of that
experiment, and sample point of the sample space,
denoted by ω, is a outcome of the experiment.
8 5
1 9 4 6
7 2 3 10
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Example 2.1 : The simplest experiment to which
probability applies is one with two possible outcomes. One
such experiment consists of examining a single fuse to see
whether it is defective. The sample space for this
experiment can be abbreviated as Ω={N,D}, where N
represents not defective, D represents defective, and the
braces are used to enclose the elements of a set.
Another such experiment would involve tossing a
thumbtack and noting whether it landed point up or point
down, with sample space Ω={U,D}, and yet another would
consist of observing the sex of the next child born at the
local hospital, with Ω={M,F}.
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Example 2.2 : driving to work , a commuter passes
through a sequence of three intersections with traffic
lights. At each light, she either stops, s, or continues, c, the
sample space is the set of all possible outcomes :
Ω ={ccc, ccs, css, csc, sss, ssc, scc, scs }
Where csc for example denotes the outcomes that the
commuter continues through the first light,stops at the
second light,and continues through the third light.
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• Example 2.3 Two gas stations are located at a certain
intersection. Each one has six gas pumps. Consider
the experiment in which the number of pumps in use
at a particular time of day is determined for each of
the stations.
• An experimental outcome specifies how many pumps
are in use at the first station and how many are in use
at the second one. One possible outcome is (2,2),
another is (4,1).
• The 49 outcomes is Ω are displayed in the
companying table. The sample space for the
experiment in which a six-sided die is thrown twice
results from deleting the 0 row and 0 column from the
table, given 36 outcomes
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Example 2.4 The number of jobs in a print queue of a
mainframe computer may be modeled as random . Here the
sample space can be taken as Ω= 0,1,2,3,...
That is ,all the nonnegative integers. In practice there is
probably an upper limit,N,on how large the print queue can
be ,so instead the sample space might be defined as
Ω= 0,1,2,..., N 
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Example 2.5
Earthquakes exhibited very erratic
behavior,which is sometimes modeled as random. For
example, the length of time between successive
earthquakes in a particular region that are greater in
magnitude than a given threshold may be regarded as an
experiment. Here Ω is the set of all nonnegative numbers:

Ω= t t  0
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Random Event
Definition : An event, are usually denoted by italic uppercase
letters, is any collection (subset) of outcomes contained in the
sample space Ω. An event is said to be simple if it consists of
exactly one outcome and compound if it consists of more than
one outcome.
Remark: When an experiment is performed, a particular event A
is said to occur if the resulting experimental outcome is
contained in A. In general, exactly one simple event will occur,
but many compound events will occur simultaneously.
The event that the commuter stops at the first light is the
subset of Ω denoted by
A ={sss, ssc, scc, scs}
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Example 2.5 : Consider an experiment in which each of
three automobiles taking a particular freeway exit turns
left (L) or right (R) at the end of the exit ramp. The eight
possible outcomes that comprise the sample space are
LLL,RLL,LRL,LLR,LRR,RLR, RRL, and RRR. Thus,
there are eight simple events, among which are E1={LLL}
and E5={LRR}. Some compound events include
A={RLL,LRL,LLR}=the event that exactly one of the
three cars turns right
B={LLL,RLL,LRL,LLR}=the event that at most one of
the cars turns right
C={LLL,RRR}=the event that all three cars turn in the
same direction
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Suppose that when the experiment is performed, the
outcome is LLL. Then the simple event E1 has occurred
and so also have the events B and C (but not A).
Some Relations From Set Theory
An event is nothing but a set, so that relationships and
results from elementary set theory can be used to study
events. The following concepts from set theory will be
used to construct new events from given events.
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Union(∪)
The algebra of set theory carries over directly into
probability theory. The union of two events , A and B,is
the event C that either A occurs or B occurs or both
occur:C=A∪B.
For example,if A is the event that commuter stops at the
first light and if B is the event that she stops at the third
light,
B= sss , scs, ccs, css
Then C is the event that he stops at the first light or stops
at the third light and consists of the outcomes that are in A
or in B or in both :
C= sss, ssc, scc, scs, ccs, css


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Intersection (∩)
The intersection of two events, C=A∩ B ,is the event that
both A and B occur.
If A and B are as listed previously,then C is the event that
the commuter stops at the first light and stops at the third
light and thus consists of those outcomes that are common
to both A and B:
C= sss, scs
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Complement(
A ) A, A
c
The complement of an event , A ,is the event that A does
not occur and thus consists of all those elements in the
sample space that are not in A ,the complement of the event
that the commuter stops at the first light is the event that she
continues at the first light:
A = ccc, ccs, css, csc


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You may recall from previous exposure to set theory that
rather mysterious set called the empty set,usually denoted
by Ф. The empty set with no elements; it is the event with
no outcomes. For example,if A is the event that the
commuter stops at the first light and C is the event that she
continues through all three lights , c=CCC then A and C
have no outcomes in common, and we can write:
A C  
In such cases ,A and C are said to be mutually exclusive or
disjoint.
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Venn diagrams
B
A
Ω
A
A B
B
A
A
A
B
A
B
A
B
A∩B
A
B
A－B
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The following are some lows of set theory.
Commutative laws: A  B  B  A
A B  B  A
Associative laws:
 A  B   C  A  B  C 
 A  B   C  A  B  C 
Distributive laws:  A  B   C   A  C   B  C 
 A  B   C   A  C   B  C 
De Morgan laws:
A B  A  B
A B  A  B
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2.2 Axioms, Interpretations And Properties
Of Probability
Given an experiment and a sample spaceΩ, the objective of
probability is to assign to each event A a number P(A),
called the probability of the event A, which will give a
precise measure of the chance that A will occur. To ensure
that the probability assignments will be consistent with our
intuitive notions of probability, all assignments should
satisfy the following axioms :
1.ΡΩ   1
2.If A  Ω, then Ρ A  0
3.If A1 and A2 are disjoint ,then
Ρ A1  A2   Ρ A1   Ρ A2 
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More generally ,if
then
1 , 2 , n  are mutually disjoint ,
  
Ρ  Αi    Ρ Αi 
 i 1  i 1
The first two axioms are rather obvious. Since  consists of all
possible outcomes.
ΡΩ   1
The second axiom simply states that a probability is nonnegative.
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The third axiom states that if A and B are disjoint that is have
no outcomes in common then
Ρ Α  Β   Ρ Α  ΡΒ 
and also that this property extends to limits. For example the
probability that the print queue contains either one or three jobs
is equal to the probability that it contains one plus the
probability that it contains three.
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Properties Of Probability
The following properties of probability measures are consequences
of the axioms.
Property 1
ΡΑ   1  Ρ Α
this property follows since A and are disjoint with Α  Α  Ω
and thus ,by the first and third axioms, Ρ Α  Ρ Α  1
 
in words this property says that the probability that an event does
not occur equals one minus the probability that it does occur.
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Property 2
Ρφ  0
this property follows from property 1 since
φΩ
In words this says that the probability that there is no
outcome at all is zero.
if A  B  Φ then P( A  B)  0
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Property 3 if
Α  Β then
Ρ Α  ΡΒ 
this property follows since B can be expressed as
the union of two disjoint sets: Β  Α  Β  Α

Then from the third axiom,
And thus

ΡΒ   Ρ Α  ΡΒ  Α 
Ρ Α  ΡΒ   ΡΒ  Α   ΡΒ 
This property states that if B occurs whenever A occurs,
then Ρ Α  ΡΒ
.
For example if whenever it rains (A) it is cloudy (B), then the
probability that it rains is less than or equal to the probability that
it is cloudy.
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Property 4 addition law
Ρ Α  Β  Ρ Α  ΡΒ  Ρ Α  Β
to see this ,we decompose
Α Β
into three disjoint subsets, as shown in figure2.1 :
C  ΑΒ
D  ΑΒ
E  ΒΑ
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FIGURE 2.1
Venn diagram illustrating the addition
law.
A
B
C
D
E
We then have from the third axiom,
Ρ Α  Β  ΡC   ΡD  ΡE 
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Also , Α  C  D and C and D are disjoint ; so
Ρ Α  ΡC   ΡD
similarly, ΡΒ  ΡD  ΡE 
putting these results together ,we see that
Ρ Α  ΡΒ  ΡC   ΡE   2ΡD  Ρ Α  Β  ΡD
or
Ρ Α  Β  Ρ Α  ΡΒ  ΡD
this property is easy to see from the venn
diagram 2.1. if P(A) and P(B) are added
together , Ρ Α  Β  is counted twice.
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The probability of a union of more than two events can
be computed analogously. For three events A, B, and C,
the result is
P( A  B  C )  P( A)  P( B)  P(C )  P( A  B)
 P( A  C )  P( B  C )  P( A  B  C ).
P(( A  B)  C )  P( A  B)  P(C )  P(( A  B)  C )
 P( A)  P( B)  P(C )  P( A  B)
 P( A  C )  P( B  C )  P( A  B  C ).
P( A  B)  P( A  AB)  P( A)  P( AB)
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Equally Likely Outcomes
In many experiments consisting of N outcomes, it is reasonable
to assign equal probabilities to all N simple events. With p=P(Ei)
for every i,
N
N
1
1   P( Ei )   p  pN so p 
N
i 1
i 1
That is, if there are N possible outcomes, then the probability
assigned to each is 1/N. Now consider an event A, with N(A)
denoting the number of outcomes contained in A. Then
P( A) 
 P( Ei ) 
Ei in A
number of ways A can occur
1 N ( A)



total number of outcomes
N
Ei in A N
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Example 2.6 : When two dice are rolled separately, there are
N=36 outcomes. If both the dice are fair, all 36 outcomes are
equally likely, so P(Ei)=1/36. Then the event A={sum of two
numbers=7} consists of the six outcomes (1,6), (2,5), (3,4), (4,3),
(5,2), and (6,1), so
N ( A) 6 1
P ( A) 


N
36 6
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2.3
Counting Techniques
When the various outcomes of an experiment are equally likely,
then the task of computing probabilities reduces to counting. In
particular, if N is the number of outcomes in a sample space
and N(A) is the number of outcomes contained in an event A,
then
P( A) 
number of ways A can occur
N ( A)

total number of outcomes
N
(2.1)
If a list of the outcomes is available or easy to construct and
N is small, then the numerator and denominator of Equation
(2.1) can be obtained without the benefit of any general
counting principles.
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Example 2.7: a black urn(瓮、缸) contains 5 red and 6
green balls and a white urn contains 3 red and 4 green balls.
You are allowed to choose an urn and then choose a ball at
random from the urn. If you choose a red ball ,you get a
prize. Which urn should you choose to draw from?
If you draw from the black urn ,the probability of choosing
a red ball is 0.455. If you choose to draw from the white
urn, the probability of choosing a red ball is 0.429,so you
should choose to draw from the black urn.
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Now consider another game in which a second black urn has 6
red and 3 green balls and a second white urn has 9 red and 5
green balls. If you draw from the black urn, the probability of
red ball is 6/9. whereas if you choose to draw from the white
urn ,the probability is 9/14. so ,again you should choose to
draw from the black urn.
In the final game, the contents of the second black urn are added
to the first black urn and the contents of the second white urn
are added to the first white urn .again you can choose which urn
you can draw from .which should you choose? Intuition says
choose the black urn, but let’s calculate the probabilities.
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The black urn now contains 11 red and 9green balls, so the
probability of drawing a red ball from it is 11/20=0.55 .The white
urn now contains 12 red and 9 green balls, so the probability of
drawing a red ball from it is 12/21=0.571 .
so, you should choose the white urn. This counterintuitive result
is an example of Simpson’s paradox.
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The Product Rule For Ordered Pairs
If one experiment has m outcomes and another experiment has n
outcomes ,then there are m×n possible outcomes for the two
experiments.
Proof: Denote the outcomes of the first experiment by a1 , a2 ,, am
and The outcomes of the second experiment by b1 , b2 ,bn the
outcomes for the two experiments are the ordered pairs ai , b j 
these pairs can be exhibited as the entries of an m × n
rectangular array, in which the pair ai , b j is in the ith row and the
jth column. There are mn entries in this array.
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Example 2.8: A family has just moved to a new city and
requires the services of both an obstetrician and a
pediatrician. There are two easily accessible medical clinics,
each having two obstetricians and three pediatricians. The
family will obtain maximum health insurance benefits by
joining a clinic and selecting both doctors from that clinic. In
how many ways can this be done? Denote the obstetricians by
O1,O2,O3, and O4 and the pediatricians by P1,…,P6. Then we
wish the number of pairs (Oi, Pj) for which Oi and Pj are
associated with the same clinic. Because there are four
obstetricians, n1=4, and for each there are three choices of
pediatrician, so n2=3. Applying the product rule gives N= n1
n2=12 possible choices.
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Tree diagrams
P1
P2
O1
O2
O3
O4
P3
P1
P2
P3
P4
P5
P6
P4
P5
P6
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A More General Product Rule
If there are p experiments and the first has n1 possible outcomes,
the second n2 , and the pth n p possible outcomes ,then there
are a total of n1  n2    n p possible outcomes for the p
experiments.
Example 2.9: an 8-bit binary word is a sequence of 8 digits, of
which each may be either 0 or 1. How many different 8-bit words
are there?
There are two choices for first bit, two for the second, etc, and
thus there are
2 × 2 × 2 × 2 × 2 × 2 × 2 × 2= 28=256
such words.
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Permutations(排列）
Definition : Any ordered sequence of k objects taken from a
set of n distinct objects is called a permutation of size k of
the objects. The number of permutations of size k that can be
constructed from the n objects is denoted by Pnk .
Pnk  nn  1n  2 n  k  1
Definition : For any positive integer m, m! is read “m factorial”
and is defined by
m! mm 1m  22 1
so
n!
P 
(n  k )!
k
n
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Combinations (组合）
Definition : Given a set of n distinct objects, any unordered
subset of size k of the objects is called a combination. The
number of combinations of size k that can be formed from the
n distinct objects will be denoted by C nk .
k
P
n!
k
n
Cn 

k! k!(n  k )!
Cnm  Cnn  m
Cnm  Cnm1  Cnm11
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Example A suppose that a fair coin is thrown twice and the
sequence of heads and tails is recorded. The sample space is
  hh, ht , th, tt
As in Example A of previous section, we assume that each
outcome in  has probability 0.25. let A denote the event that at
least one head is throw. Then
  hh, ht , th
And P(A)=.75
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Example Consider the experiment of selecting five cards from
a deck of 52 cards. Let
A={the five cards number different each other}
B={two cards have the same number, and other three cards also
have homology number}
C={two cards have the same number, the other three cards have
different number}
D={five cards have four kinds of designs}
5
C13 45
P( A) 
5
C52
1
2
1
C C C C
P( B)  13 4 512 4
C52
1
2
3
C13 C4 C12 43
P(C ) 
5
C52
3
1
2
1
1
1
C C C C C
P( D)  4 13 13 5 13 13
C52
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Example There are m boys and n girls, they stand randomly
to one row, Let
A={girls stand together}
B={girls stand together, and boys stand together,too}
C={there is at least one boy in any two girls}
Solution:
(m  1)!n!
P( A) 
(m  n)!
m!n! P22
P( B) 
(m  n)!
m! Pmn1
P(C ) 
(m  n)!
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Li Jie
2.4
Conditional probability
Example 2.10: Complex components are assembled in a
plant that uses two different assembly lines, A and A . Line
A uses older equipment than A , so it is somewhat slower
and less reliable. Suppose on a given day line A has
assembled 8 components, of which 2 have been identified
as defective (B) and 6 as nondefective (B), whereas A has
produced 1 defective and 9 nondefective components. This
information is summarized in the accompanying table.
Condition
Line
A
A
B
2
1
B
6
9
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Unaware of this information, the sales manager randomly
selects 1 of these 18 components for a demonstration. Prior to
the demonstration
N ( A) 8
P(line A component selected )  P( A) 

N
18
However, if the chosen component turns out to be defective,
then the event B has occurred, so the component must have
been 1 of the 3 in the B column of the table. Since these 3
components are equally likely among themselves after B has
occurred,
2 2 / 18 P( A  B)
P( A | B)  

3 3 / 18
P( B)
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The Definition Of Conditional Probability
Definition : For any two events A and B with P(B)>0, the
conditional probability of A given that B has occurred is
defined by
P( A  B)
P( A | B) 
P( B)
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Example 2.11: Suppose that of all individuals buying a certain
personal computer, 60% include a word processing program in
their purchase, 40% include a spreadsheet program, and 30%
include both types of programs. Consider randomly selecting a
purchaser and let A={word processing program included}
and B={spreadsheet program included}. Then P(A)=0.6,
P(B)=0.4, and P(both included)=P(A∩B)=0.3. Given that the
selected individual included a spreadsheet program, the
probability that a word processing program was also included is
P( A  B) 0.3
P( A | B) 

 0.75
P( B)
0.4
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Example In one year, if the first city has 20% rainy days, the
second city has 18% rainy days, raining in both cities at the
same time has 12%. Determine the probability of A,B
A={at least one city has raining}
B={if known one city has raining, then the other city has raining
too}
Solution: A1={the first city is raining}, A2={the second city is
raining}
P( A1  A2 )  P( A1 )  P( A2 )  P( A1 A2 )  0.20  0.18  0.12  0.26
P(A2|A1)=P(A1A2)/P(A1) =0.12/0.2=0.6
P(A1|A2)=P(A1A2)/P(A2)
=0.12/0.18=0.67
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Example A batch of accessories is 100, in which there are 5
defectives. Each time select one accessory to inspect, then
put it aside. Select two accessories continuously to inspect.
Determine the probabilities of the following events.
a. the first inspected accessory is good.
b. If the first inspected accessory is good, the second
inspected accessory is good too.
c. the accessories inspected in two times are all good.
Solution: A1={the first inspected accessory is good}, A2={the
second inspected accessory is good }
95
(a) P(A ) 
 0.95
100
94
(b) P(A | A ) 
 0.9495
99
1
2
1
(c ) P(A A )  P(A )  P(A | A )  0.9020
1
2
1
2
1
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.Multiplication Law
Let A and B be events and assume P(B) 0.Then
P(A  B)= P(A|B)P(B)
The multiplication law is often useful in finding the
probabilities of intersections, as the following
examples illustrate.
Li Jie
Example An urn contains three red balls and one blue
ball. Two balls are selected without replacement.
What is the probability that they are both red?
Let R 1and R 2 denote the events that a red ball
is drawn on the first trial and on the second trial,
respectively. From the multiplication law,
P(R1  R 2 )  P(R1 )P(R2 |R1 )
P(R )
1
is clearly 3 , and if a red ball has been
4
removed on the first trial, there are two red balls and one
blue ball left. Therefore,
1
2
P( R2 | R1 )  . Thus, P( R1  R2 ) 
.
3
2
Li Jie
Example Suppose that if it is cloudy (B), the probability
that it is raining (A) is .3, and that the probability it is
cloudy is
P( B)  .2
The probability that it is cloudy and raining is
P( A  B)  P( A | B) P( B)  .3  .2  .06
Another useful tool for computing probabilities is
provided by the following law.
Li Jie
Corollary:
P( A1 A2 A 3)
 P( A1 ) P( A2 | A1 ) P( A3 | A1 A2 )
P( A1 A2  An )
 P( A1 ) P( A2 | A1 ) P( A3 | A1 A2 )  P( An | A1 A2  An 1 )
Li Jie
Example A fair coin is tossed three times in an honest
manner. Given that the head shows at least once, what
is the probability that it shows all three times?
Solution: suppose A={the head shows all three times}
B={the head shows at least once}
P( AB ) P( A)
1/ 8
1
P( A | B) 



P( B)
P( B) 1  1 / 8 7
Li Jie
Example There are ten same size coins in one bag, in
which 7 coins is 1 jiao, 3 coins is 5 jiao. Sampling
with non-replacement. Please determine:
(1) The probability of 5 jiao coin only selected in the
third time.
(2) The probability of 5 jiao coin at least selected once
in three selections.
Solution: suppose Ai={5 jiao coin be selected in ith
time}, i=1,2,3
5 jiao coin only selected in the third time A A A
1
2
3
Li Jie
5 jiao coin at least selected once in three selections
A1  A2  A3
(1) Due to multiplication law
P( A1 A2 A3 )  P( A1 )P( A2 | A1 )P( A3 | A1 A2 )
7 6 3 7

  
 0.175
10 9 8 40
(2) Due to dual law
A1  A2  A3  A1 A2 A3
Li Jie
P( A1  A2  A3 )  1  P ( A1 A2 A3 )
P( A1  A2  A3 )  1  P ( A1 A2 A3 )
 1  P( A1 )P( A2 | A1 )P( A3 | A1 A2 )
7
 1
10
5 17
 
8 24
Li Jie
Li Jie
The law of total probability
Definition: if A1,A2,…,An satisfied:
(1) P( Ai )  0, i  1,2,n
(2) A1 , A2 ,, An be mutually exclusive
(3) A1  A2    An  
exhaustive events
A1
An
B
A2
A3
A4
Then we named A1,A2,…,An is a dividing of sample space
Li Jie
Theorem: suppose A1,A2,…,An is a dividing of sample space, then
n
P( B)   P( Ai ) P( B | Ai )
i 1
Proof:
P( B | A1 )
B  B  B( A1  A2    An )
 BA1  BA2    BAn
BA1 , BA2 ,, BAn are mutex
and P( BAi )  P( Ai ) P( B | Ai )
B
P( A1 )
P( A2 )
P( A3 )
P( An )
P( B)  P( BA1 )  P( BA2 )    P( BAn )
n
  P( Ai ) P( B | Ai )
A1
B
P( B | A2 )
A2
B
P( B | A3 )
A3
B

An
P( B | An )

B
Tree diagram
i 1
Li Jie
Example2.13: Suppose that occupations are grouped into
upper (U), middle (M), and lower (L) levels. U1will denote the
event that a father’s occupation is upper-level; U 2 will denote
the event that a son’s occupation is upper-level, etc.
U2
M2
L2
U1
.45
.48
.07
M1
.05
.70
.25
L1
.01
.50
.49
Li Jie
Such a table, which is called a matrix of transition
probabilities, is to be read in the following way: If a father is
in U, the probability that his son is in U .45, the probability
that his son is in M is .48, Examination of the table reveals
that there is more upward mobility from L into M than from
M into U. Suppose that of the father’s generation, 10% are in
U, 40% in M, and 50% in L. What is the probability that a son
in the next generation is in U?
Applying the law of total probability, we have
P(U 2 )  P(U 2 | U1 ) P(U1 )  P(U 2 | M 1 ) P( M 1 )  P(U 2 | L1 ) P( L1 )
 0.45  0.10  0.05  0.40  0.01  0.50  0.07
Li Jie
Example: Urn A has three red balls and two white balls.
and urn B has two red balls and five white balls. A fair
coin is tossed; if it lands heads up, a ball is drawn from
urn A and otherwise a ball is drawn from urn B.
a. What is the probability that a red ball is drawn?
b. If a red ball is drawn, what is the probability that the
coin landed heads up?
Li Jie
Bayes’ theorem
In the discussion of conditional probability, we indicated that
revising probabilities when new information phase of
probability analysis. Often, we begin the analysis with initial
or prior probability estimates for specific events if interest.
Then, from sources such as a sample, a special report, or a
product test, we obtain additional information about the
events. Given this new information, we update the prior
probability values by calculating revised probabilities,
referred to as posterior probabilities. Bayes’ theorem
provides a means for making these probability calculations.
Li Jie
Example2.14: As an application of Bayes’ theorem, consider a
manufacturing firm that receives shipments of parts from two
different suppliers A1 and A2. Let P(A1)=0.65, P(A2)=0.35.
The quality of the purchased parts varies with the source of
supply. We let G denote the event that a part is good and B
denote the event that a part is bad,
P(G|A1)=0.98 P(B|A1)=0.02, P(G|A2)=0.95, P(B|A2)=0.05
Suppose now that the parts from the two suppliers are used in
the firm’s manufacturing process and that a machine breaks
down because it attempts to process a bad part. Given
information that the part is bad , what is the probability that
it came from supplier 1 and what is the probability that it came
from supplier 2?
Li Jie
P(G|A1)=0.98
P(A1)=0.65
P(A2)=0.35
P(B|A1)=0.02
P(G|A2)=0.95
P(B|A2)=0.05
P(A1G)=P(A1)P(G|A1)=0.637
P(A1B)=P(A1)P(B|A1)=0.013
P(A2G)=P(A2)P(G|A2)=0.3325
P(A2B)=P(A2)P(B|A2)=0.0175
Solution: we will determine P(A1|B), P(A2|B)
P( A1B)
P( A1B)  P( B | A1 ) P( A1 )
P( A1 | B) 
P( B)
P( B)  P( B | A1 ) P( A1 )  P( B | A2 ) P( A2 )
P( A1B)
P( B | A1 ) P( A1 )
P( A1 | B) 

P( B)
P( B | A1 ) P( A1 )  P( B | A2 ) P( A2 )
Li Jie
Baye’s theorem (Two-event case)
P( A1 B )
P( B | A1 )P( A1 )
P( A1 | B ) 

P( B ) P( B | A1 )P( A1 )  P( B | A2 )P( A2 )
P( A2 B )
P( B | A2 )P( A2 )
P( A2 | B ) 

P( B ) P( B | A1 )P( A1 )  P( B | A2 )P( A2 )
Corollary: if A1,A2,….,An are mutually exclusive and their
union is the entire sample space. Then
P( Ai B ) P( Ai )P( B| Ai )
P( Ai | B ) 

P( B )
P( B )

P( Ai )P( B| Ai )
n
 P( A )P( B| A )
k 1
k
k
Li Jie
Example 2.15:
In digit communication, on account of
random factors, the signal received would not be the same
signal send out. If the digits send out are ‘0’ and ‘1’, then the
receiving digits will ‘0’, ‘1’, and ‘*’. if ‘0’ would be send out
by probability 0.6, then the probability of receiving ‘0’, ‘1’,
‘*’are 0.8, 0.05,0.15, if ‘1’would send out by probability 0.4,
then the probability of receiving ‘0’, ‘1’, ‘*’are 0.9, 0.05,0.05.
If now we have received signal ‘*’, what are the probability
of original send signal ‘0’, and ‘1’.
receive
send
0
0.8
0.05
0.15
0
1
*
receive
send
1
0
0.05
0.9
0.05
1
*
Li Jie
Solution: A={send ‘0’}, AC={send ‘1’}, B={receive ‘*’}
receive
0
C|B)
We
asked
to
determine
P(A|B),
P(A
1
send 0.8
0
0.05
0.15
*
P( AB )
P( B | A) P( A)
P( A | B) 

C
C
P
(
B
)
P
(
B
|
A
)
P
(
A
)

P
(
B
|
A
)
P
(
A
)
receive
send
0.6  0.15
0
0.05

 0.8182
1
0.6  0.15  0.4  0.05
0.9
1
0.05
C
C
C
P
(
A
B
)
P
(
B
|
A
)
P
(
A
)
C
* P( A | B) 

P( B)
P ( B | A) P( A)  P ( B | AC ) P( AC )
0.4  0.05

 0.1818
0.6  0.15  0.4  0.05
Li Jie
Example: A factory runs three shifts. In a given day, 1% of
the items produced by the first shift are defective, 2% of
the second shift’s items are defective, 5% of the third shift’s
items are defective. If the shifts all have the same
probability, what percentage of the items produced in a day
are defective? If an item is defective, what is the probability
that it was produced by the third shift?
Example: There are 12 new ping-pong balls in one bag. We
take 3 balls at each match, then replace its to the bag after
play. What is the probability that the 3 balls are all new at
the third match.
Li Jie
Example: There are 12 new ping-pong balls in one bag. We
take 3 balls at each match, then replace its to the bag after
play. What is the probability that the 3 balls are all new at
the third match.
Bi  {i new balls at the 2nd match}, i  0,1,2,3
i
9
3 i
3
3
12
CC
P( Bi ) 
C
, i  1,2,3
C93i
P(C3 Bi )  3 , i  1,2,3
C12
3
P(C3 )   P( Bi ) P(C3 Bi )
i 0
Li Jie
2.5
Independence
Definition: Two events A and B are independent if
P(A|B)=P(A) and are dependent otherwise.
Remark: The definition of independence might seem
“unsymmetric” because we do not demand that P(B|A)= P(B)
also. However, using the definition of conditional probability
and the multiplication rule, we have
P(A|B)=P(A) if and only if P(B|A)=P(B)
P(B|A)=P(AB)/P(A) = P(A|B) P(B)/P(A)
so P(B|A)=P(B)
Li Jie
P(A∩B) When Events Are Independent
Proposition: A and B are independent if and only if
P(A∩B)=P(A)·P(B)
and
P( A  B)  P( A)  P( B)
 P( A  B)  P( A )  P( B)
 P( A  B )  P( A)  P( B )
 P( A  B )  P( A )  P( B )
If A and B are independent, then so are the
following pairs of events:
(1) A' and B
(2) A and B ' (3) A' and B '
Li Jie
If two events are mutually exclusive , they
cannot be independent.
Example 2.32 Let A and B be any two mutually exclusive
events with P(A)>0. For example, for a randomly chosen
automobile, let A={the car has four cylinders}, B={the car
has six cylinders}. A and B mutually exclusive,
P(A︱B)=0≠P(A), The message here is that if two events are
mutually exclusive , they cannot be independent.
Li Jie
Example2.16: A card selected randomly from a deck. Let A
denote the event that it is an ace and D the event that it is a
diamond. Knowing that the card is an ace gives no information
about its suit.
Checking formally that the events are independent,
we have P(A)  4  1 and P ( D )  1 Also, A  D is the
52 13
4
event that the card is the ace of diamonds
1
and P(A  D) 52 . Since P(A)P(D) ( 1 )  ( 1 )  1 ,the
4
13 52
events are in fact independent.
Li Jie
Example2.17: Suppose that a system consists of components
connected in series,so the system fails if any one component
fails. If there are n mutually independent components and
each fails with probability p, what is the probability that the
system will fail?
It is easier to find the probability of the complement of this
event;the system works if and only if all the components
work, and this situation has probability (1-p)n .The probability
that the system fails is then 1-(1-p)n. For example, if n=10 and
p=0.05,the probability that the system work is only 0.9510=0.6,
and the probability that the system fails is 0.40.
Suppose, instead, that the components are connected in
parallel, so the system fails only when they all fail. In this
case, the probability that the system fails is only
0.0510=9.8×10-14.
Li Jie
Independence Of More Than Two Events
Definition: Events A1,…,An are mutually independent if for
every k (k=2,3,…,n) and every subset of indices i1,i2,…ik,
P( Ai1  Ai2    Aik )  P( Ai1 )  P( Ai2 )  P( Aik )
ABC mutually independent , Then
P(AB)=P(A)P(B), P(AC)=P(A)P(C), P(BC)=P(B)P(C),
P(ABC)=P(A)P(B)P(C)
Li Jie
Normal tetrahedron
regular tetrahedron
1 red 2 white 3 black 4 red white black
regular octahedron
1234 red 1235 white 1678 black
A={ red }
B={white}
C={ black}
Li Jie
Example B A system is designed so that fails only if
a unit and a backup unit both fail. Assuming that
these failures are independent and that each unit fails
with probability p .the system fails with probability
p2 .If, for example ,the probability that any unit fails
during a given year is .1, then the probability that the
system fails is .01, which represent a considerable
improvement in reliability.
Li Jie
Things become more complicated when we consider
more than two events. For example, suppose we know that
events A, B, and C are pairwise independent (any two are
independent ). We would like to be able to say that they
are all independent based on the assumption that knowing
something about two of the events does not tell us anything
about the third, for example, P(C | A  B)  P(C)
.
But
as the following example shows, pairwise independence
does not guarantee mutual independence.
Li Jie
Example C
A fair coin is tossed twice. Let A denote the
event of heads on the first toss, B the event of heads on the
second toss, and C the event that exactly one head is thrown.
A and B are clearly independent, and P(A)=P(B)=P(C)=.5.
To see that A and C are independent, we observe that
P(C|A)=.5. But
P(A  B  C)  0  P(A)P(B)P(C)
Li Jie
Example A fair die is repeatedly rolled in an honest manner
until one spot shows for the first time. Determine the
probability that ten rolls are required.
Solution:
Li Jie