Transcript Document

Probability and Random
Process for Electrical
Engineering
主講人 : Huan Chen
Office : 427
E-Mail : [email protected]
助教 : 黃政偉 ([email protected])
鄭志川 ([email protected])
Lab : 223
Outline



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2.1 Specifying Random Experiments
2.2 The Axioms of Probability
2.3 Computing Probabilities Using
Counting Methods
2.4 Conditional Probability
2.5 Independence of Events

2.1 Specifying Random Experiments
A random experiment is specified by stating an experimental procedure and
a set of one or more measurements or observations.
EXAMPLE 2.1 :
Experiment E1 : Select a ball form an urn containing balls numbered 1
to 50. Note the number of the ball.
Experiment E2 : Select a ball form an urn containing balls numbered 1
to 4. Suppose that balls 1 and 2 are black and that balls 3 and 4 are
white. Note number and color of the ball you select.
Experiment E3 : Toss a coin three times and note the sequence of
heads and tails.
Experiment E4 : Toss a coin three times and note the number of
heads.
Experiment E7 : Pick a number at random between zero and one.
Experiment E12 : Pick two numbers at random between zero and one.
Experiment E13 : Pick a number X at random between zero and one,
then pick a number Y at random between zero and X.
The sample space
We define an outcome or sample point of a random experiment as a
result that cannot be decomposed into other results.
The sample space S of a random experiment is defined as the
set of all possible outcomes.
We will denote an outcome of an experiment by ζ, where ζ is
an element or point in S.
EXAMPLE 2.2 :
The sample spaces corresponding to the experiments in Example
2.1 are given below using set notation :
S1  1,2,......,50
S 2  1, b , 2, b , 3, w, 4, w
S3  HHH , HHT , HTH , THH , TTH , THT , HTT , TTT 
S 4  0,1,2,3
S 7  x : 0  x  1  0,1

See Fig. 2.1 (a).

S12   x, y  : 0  x  1 and 0  y  1
S13   x, y  : 0  y  x  1
See Fig. 2.1 (c).
See Fig. 2.1 (d).
We will call S a discrete sample space if S is countable; that is, its
outcomes can be put into one-to-one correspondence with the
positive integers.
We will call S a continuous sample space if S is not
countable.
Events
We define an event as a subset of S.
Two events of special interest are the certain event, S, which
consists of all outcomes and hence always occurs, and the
impossible or null event, φ, which contains no outcomes and
hence never occurs.
An event from a discrete sample space that consists of a single
outcome is called an elementary event.
Set Operations
We can combine events using set operations to obtain other events.
The union of two events A and B is denoted by A  B .
The intersection of two events A and B is denoted by A  B
Two events are said to be mutually exclusive if their
intersection is the null event, A  B   .
The complement of an event A is denoted by Ac
If an event A is a subset of an event B, that is, A  B .
The events A and B are equal, A  B , if they contain the same
outcomes.
The following properties of set operations and their
combinations are useful :
Commutative Properties :
A B  B  A
and
A B  B  A
(2.1)
Associative Properties :
A  B  C    A  B  C and A  B  C    A  B  C (2.2)
Distributive Properties :
A  ( B  C )  ( A  B)  ( A  C )
and
A  ( B  C )  ( A  B)  ( A  C )
(2.3)
DeMorgan’s Rules :
( A  B)c  Ac  B c
and
( A  B)c  Ac  B c
(2.4)
A
A
B
(a) A  B
B
(b) A  B
A
Ac
A
B
(d ) A  B  
(c) Ac
A
B
(e) A  B
EXAMPLE 2.4
For Experiment E10, let the events A, B, and C be defined by
A  v : v  10 ,
“magnitude of v is greater than 10 volts,”
B  v : v  5 , ” v is less than -5 volts,” and
C  v : v  0 , “v is positive.”
You should then verify that


A  B  v : v  5 or v  10 ,
A  B  v : v  10,
C c  v : v  0,
( A  B)  C  v : v  10,
A  B  C   , and
( A  B) c  v : 5  v  10.

2.2 The Axioms of Probability
Let E be a random experiment with sample spaces S. A probability
law for the experiment E is a rule that assigns to each event A a
number PA, called the probability of A, that satisfies the following
axioms :
Axiom Ⅰ
0  PA ,
Axiom Ⅱ
PS   1 ,
Axiom Ⅲ
Axiom Ⅲ’
if A  B   , then PA  B  PA  PB
If A1 , A2 ..... is a sequence of events such
that Ai  A j   for all i  j, then
 
P      P Ak 
 k 1 k 1
 
COROLLARY 1. P Ac  1  PA
Proof : Since an event A and its complement Ac are mutually
exclusive, A  Ac   , we have form Axiom Ⅲ that


 
P A  Ac  PA  P Ac
Since S  A  Ac, by Axiom Ⅱ,


 
1  PS   P A  Ac  PA  P Ac
 
c
The corollary follows after solving for P A
COROLLARY 2. PA  1
Proof : From Corollary 1,
 
PA  1  P Ac ,1
 
since P Ac  0
COROLLARY 3. P   0
Proof : Let A  S and Ac   in Corollary 1 :
P   1  PS   0
COROLLARY 4. If A1 , A2 ,...., An are pairwise mutually exclusive,
then
n
 n
P   Ak    PAk 
k 1  k 1
for n  2 .
Proof :
Suppose that the result is true for some
; that is ,
n
 n
P   Ak    PAk 
k 1  k 1
(2.5)
and consider the n + 1 case
 n 

n 1 
n 
P   Ak   P   Ak   An 1   P   Ak   PAn 1 
k 1 
k 1 
k 1 

(2.6)
where we have applied Axiom Ⅲ to the second expression after
noting that the union of events A1 to An is mutually exclusive with
An+1. The distributive property then implies
n
n
n

  Ak   An 1   Ak  An 1      .
k 1
k 1
k 1 
Substitution of Eq. (2.5) into Eq. (2.6) gives the n + 1 case
n
 n 1
P   Ak    PAk 
k 1  k 1
COROLLARY 5.
PA  B  PA  PB  PA  B
Proof : First we decompose A  B , A, and B as unions of disjoint
events. From the Venn diagram in Fig. 2.4,

 

PA  B   P A  B c  P B  Ac  PA  B 


PB   PB  A  PA  B 
By substituting PA  B  and PB  A  from the two lower
PA  P A  B c  PA  B 
c
c
c
equations into the top equation, we obtain the corollary.
Corollary 5 is easily generalized to three events,
PA  B  C   PA  PB  PC   PA  B
 PA  C  PB  C  PA  B  C,
(2.7)
FIGURE 2.4
Decomposition of A  B into three disjoint sets.
A  Bc
A
A B
Ac  B
B
Ac  B
A
B
COROLLARY 6.
 


 n  n
P  Ak    P Aj   P Aj  Ak  
 k 1  j 1
j k
  1 PA1  An .
n 1
Proof is by induction (See Problems 18 and 19).
Since probabilities are nonnegative, Corollary 5 implies that
the probability of the union of two events is no greater than the sum
of the individual event probabilities
PA  B  PA  PB
(2.8)
COROLLARY 7. If
A  B, then
PA  PB .
Proof : In Fig. 2.5, B is the union of A and


PA  B  0 .
Ac  ,Bthus
PB  PA  P Ac  B  PA ,
since
c
Discrete Sample Spaces

First, suppose that the sample space is finite, S  a1' , a2'  am'
is given by


 Pa  Pa    Pa 

PB   P a1' , a2' ,, am'
'
1
'
2
'
m
(2.9)
If S is countably infinite, then Axiom Ⅲ’implies that the probability
of an event such as D  b1' ,b2'  is given by


PD  Pb  Pb  
'
1
'
2
(2.10)
If the sample space has n elements, S  a1 ,an , a
probability assignment of particular interest is the case of equally
likely outcomes. The probability of the elementary events is
1
Pa1  Pa2     Pan  
(2.11)
n
The probability of any event that consists of k outcomes, say
B  a1' , ak' , is


k
PB   Pa    Pa  
n
'
1
'
k
2.12 
Thus, if outcomes are equally likely, then the probability of an event is
equal to the number of outcomes in the event divided by the total number of
outcomes in the sample space.
EXAMPLE 2.6
An urn contains 10 identical balls numbered 0,1,…, 9. A random
experiment involves selecting a ball from the urn and noting the
number of the ball. Find the probability of the following events :
A = “number of ball selected is odd,”
B = “number of ball selected is a multiple of 3,”
C = “number of ball selected is less than 5,”
and of A  B and A  B  C .
The sample space is S  0,1,,9, so the sets of outcomes
corresponding to the above events are
A  1,3,5,7,9,
B  3,6,9 , and C  0,1,2,3,4.
If we assume that the outcomes are equally likely, then
PA  P1  P3  P5  P7  P9 
PB   P3  P6  P9 
5
10
3
10.
PC   P0  P1  P2  P3  P4 
5
.
10
From Corollary 5.
5 3 2
6
PA  B   PA  PB   PA  B      .
10 10 10 10
where we have used the fact that A  B  3,9, so PA  B  2 10.
From Corollary 6,
PA  B  C   PA  PB  PC   PA  B
 PA  CPB  CPA  B  C
5 3 5 2 2 1 1
      
10 10 10 10 10 10 10
9

10
EXAMPLE 2.7
Suppose that a coin is tossed three times. If we observe the
sequence of heads and tails, then there are eight possible outcomes
. If we
S  HHH , HHT , HTH , THH , TTH , THT , HTT , TTT 
assume that the outcomes of S3 are equiprobable, then the
probability of each of the eight elementary events is 1/8. This
probability assignment implies that the probability of obtaining two
heads in three tosses is, by Corollary 3,


P "2 heads in 3 tosses"  PHHT , HTH , THH 
 PHHT   PHTH   PTHH  
3
8
This second probability assignment predicts that the probability of
obtaining two heads in three tosses is
1
P "2 heads in 3 tosses"  P2 
4


Continuous Sample Spaces
Probability laws in experiments with continuous sample spaces specify a
rule for assigning numbers to intervals of the real line and rectangular
regions in the plane.
EXAMPLE 2.9
Consider the random experiment “pick a number x at random
between zero and one.” The sample space S for this experiment is
the unit interval [0,1], which is uncountably infinite. If we suppose that
all the outcomes S are equally likely to be selected, then we would
guess that the probability that the outcome is in the interval [0,1/2] is
the same as the probability that the outcome is in the interval [1/2,1].
We would also guess that the probability of the outcome being
exactly equal to ½ would be zero since there are an uncountably
infinite number of equally likely outcomes.
Consider the following probability law : “The probability that the
outcome falls in a subinterval of S is equal to the length of the
subinterval,” that is,
Pa, b  (b  a)
for 0  a  b  1
(2.14)
where by Pa, b we mean the probability of the event corresponding
to the interval [a,b].
We now show that the probability law is consistent with the
previous guesses about the probabilities of the events[0,1/2],[1/2,1],
and {1/2} :
P0,0.5  0.5  0  0.5
P0.5,1  1  0.5  0.5
In addition, if x0 is any point in S, then Px0 , x0   0 since individual
points have zero width.
A  0,0.2 0.8,1 Since the two intervals are disjoint, we have
by Axiom Ⅲ
PA  P0,0.2  P0.8,1  .4
EXAMPLE 2.10
Suppose that the lifetime of a computer memory chip is measured,
and we find that “the proportion of chips whose lifetime exceeds t
decreases exponentially at a rate α” Find an appropriate probability
law.
Let the sample space in this experiment be S  (0, ). If we
interpret the above finding as “the probability that a chip’s lifetime
exceeds t decreases exponentially at a rate α,” we then obtain the
following assignment of probabilities to events of the form (t,∞):
P(t , )  e t
for t  0,
where α > 0. Axiom Ⅱis satisfied since
PS   P(0, )  1.
The probability that the lifetime is in the interval (r, s] is found by
noting in Fig.27 that r, s s,   r, , so by Axiom Ⅲ,
Pr,   Pr, s  Ps, .
By rearranging the above equation we obtain
Pr , s   Pr ,   Ps,   e r  e s .
We thus obtain the probability of arbitrary intervals in S
Figure 2.7
r,   r, s s, 
(
](
r
s
EXAMPLE 2.11
Consider Experiment E12, where we picked two number x and y at
random between zero and one. The sample space is then the unit
square shown in Fig. 2.8(a). If we suppose that all pairs of numbers
in the unit square are equally likely to be selected, then it is
reasonable to use a probability assignment in which the probability of
any region R inside the unit square is equal to the area of R. Find the
probability of the following events: A  x  0.5, B  y  0.5, and
C  x  y.
Figures 2.8(b) through 2.8(c) show the regions corresponding to
the events A, B, and C. Clearly each of these regions has area ½.
Thus
1
1
1
PC  
PA 
PB  
2
2
2
FIGURE 2.8
a Two-dimensional sample space and three events.
y
y
x
S
x
x
0
1
(a) Sample space
0
1
(b) Event
y
1
2
1

x  
2

y
y
1
2
x y
0
(c) Event
1
1

y  
2

x
x
0
1
(d) Event x  y
2.4 CONDITINAL PROBABILITY
The conditional probability, PA | B, of event A given that event B has
occurred. The conditional probability is defined by
PA  B
PA | B 
PB
for PB  0
(2.24)
FIGURE 2.11
If B is known to have occurred,
then A can occur only if A  B
occurs.
S
B
A B
A
Suppose that the experiment is performed n times, and suppose
that event B occurs nB times, and that event A  B occurs nAB
times. The relative frequency of interest is then
n A B n A  B n
P A  B 


,
nB
nB n
PB 
where we have implicitly assumed that PB  0 . This is an
agreement with Eq. (2.24)
EXAMPLE 2.21
A ball is selected from an urn containing two black balls, numbered 1
and 2, and two white balls, numbered 3 and 4. The number and color
of the ball is noted, so the sample space is 1, b, 2, b, 3, w, 4, w.
Assuming that the four outcomes are equally likely, find PA | B and
PB | C  , where A, B, and C are the following events:
A  1, b , 2, b  ,
B  2, b , 4, w,
C  3, w
. 4, w,
" black ball selected, "
"even  numbered ball selected, " and
" number of ball is greater then 2"
Since PA  B  P2, b and PA  C   P   0, Eq. (2.21) gives
PA | B  
PA  B  .25

 .5  PA
PB 
.5
PA | C  
PA  C  0
  0  PA .
PC 
.5
If we multiply both sides of the definition of PA | B by PB we
obtain
PA  B  PA | BPB .
(2.25a)
Similarly we also have that
PA  B  PB | APA .
(2.25b)
EXAMPLE 2.23
Many communication systems can be modeled in the following way.
First, the user inputs a 0 or a 1 into the system, and a corresponding
signal is transmitted. Second, the receiver makes a decision about
what was the input it the system, based on the signal it received.
Suppose that the user sends 0s with probability 1-p and 1s with
probability p, and suppose that the receiver makes random decision
errors with probability ε. For i  0, 1,let Abe
the event “input was ”
i
and let
be the
i,
Bi event “receiver decision was ” Find ithe
.
probabilities P Ai  B j for i  0, 1 and j  0, 1
The tree diagram for this experiment is shown in Fig. 2.13. We
then readily obtain the desired probabilities


PA0  B0   1  p 1    ,
PA0  B1   1  p  ,
PA1  B0   p , and
PA1  B1   p1    .
FIGURE 2.13
Probabilities of input-output
pairs in a binary transmission
system.
0
1-p
1-ε
(1-p)(1-ε)
p
1
0
Input into binary channel
1
0
ε
(1-p)ε
pε
Output from binary channel
ε
1-ε
1
p(1-ε)
Let B1 , B2 ,, Bn be mutually exclusive events whose union
equals the sample space S as shown in Fig. 2.14. We refer to these
sets as a partition of S. Any event A can be represented as the union
of mutually exclusive events in the following way :
A  A  S  A  B1  B2    Bn 
  A  B1    A  B2      A  Bn  .
FIGURE 2.14
A partition of S into n
disjoint sets.
B1

B3
Bn1
A
B2
Bn

(See Fig. 2.14.) By Corollary 4, the probability of A is
PA  PA  B1   PA  B2     PA  Bn .
By applying Eq. (2.25a) to each of the terms on the right-hand side,
we obtain the theorem on total probability :
PA  PA | B1 PB1   PA | B2 PB2     PA | Bn PBn . (2.26)
EXAMPLE 2.25
A manufacturing process produces a mix of “good” memory chips
and “bad” memory chips. The lifetime of good chips follows the
exponential law introduced in Example 2.10, with a rate of failure α.
The lifetime of bad chips also follows the exponential law, but the rate
of failure is 1000 α.
Suppose that the fraction of good chips is 1-p and of bad chips, p.
Find the probability that a randomly selected chip is still functioning
after t seconds.
Let C be the event “chip still functioning after t seconds,” and let
G be the event “chip is good,” and B the event “chip is bad.” By the
theorem on total probability we have
PC  PC | GPG  PC | BPB
 PC | G1  p  PC | Bp
 1  p et  pe 1000t ,
where we used the fact that PC | G  e t and PC | B  e 1000t
Bayes’ Rule
Let B1 , B2 ,, Bn be a partition of a sample space S. Suppose that
event A occurs; what is the probability of event B1? By the definition
of conditional probability we have


PA | B PB 
PB | A 

,
PA
 PA | B PB 
P A  Bj
j
j
j
n
k 1
k
(2.27)
k
where we used the theorem on total probability to replace PA .
Eq.(2.27) is called Bayes’ rule.
EXAMPLE 2.26 Binary Communication System
In the binary communication system in Example 2.23, find which
input is more probable given that the receiver has output a 1. Assume
that, a priori, the input is equally likely to be 0 or 1.
Let Ak
be the event that the input was k , k = 0, 1, then A0 and A1 are a
partition of the sample space of input-output pairs. Let B1 be the
event “receiver output was a 1.” The probability of B1 is
PB1   PB1 | A0 PA0   PB1 | A1 PA1 
1
1 1
     1      .
2
2 2
Applying Bayes’ rule, we obtain the a posteriori probabilities
PA0 | B1  
PB1 | A0 PA0   2


PB1 
12
PA1 | B1  
PB1 | A1 PA1  1    2

 1    .
PB1 
12
Thus, if εis less than 1/2, then input 1 is more likely than input 0 when
a 1 is observed at the output of the channel.
EXAMPLE 2.27 Quality Control
Consider the memory chips discussed in Example 2.25. Recall that a
fraction p of the chips are bad and tend to fail much more quickly
than good chips. Suppose that in order to “weed out” the bad chips,
every chip is tested for t seconds prior to leaving the factory. The
chips that fail are discarded and the remaining chips are sent out to
customers. Find the value of t for which 99% of the chips sent out to
customers are good.
Let C be the event “chip still functioning after t seconds,” and let
G be the event “chip is good,” and B the event “chip is bad,” The
problem requires that we find the value of t for which
PG | C   .99 .
We find PG | C  by applying Bayes’ rule :
PG | C  
PC | G PG 
PC | G PG   PC | B PB

1  p e t

1  p et  pe1000t

1
 .99
1000t
pe
1
1  p e t
The above equation can then be solved for t :
t
 99 p 
1
 .
ln 
999  1  p 
For example, if 1/α=20,000 hours and p = .10, then t = 48 hours.
2.5 INDEPENDENCE OF EVENTS
We will define two events A and B to be independent if
PA  B  PAPB.
(2.28)
Equation (2.28) then implies both
PA | B  PA
(2.29a)
PB | A  PB
(2.29b)
Note also that Eq. (2.29a) implies Eq. (2.28) when PB  0 and
Eq. (2.29b) implies Eq. (2.28) when PA  0 .
EXAMPLE 2.28
A ball is selected from an urn containing two black balls, numbered 1
and 2, and two white balls, numbered 3 and 4. Let the events A, B,
and C be defined as follows:
A  1, b , 2, b  ,
B  2, b , 4, w,
C  3, w
. 4, w,
" black ball selected, "
"even  numbered ball selected, " and
" number of ball is greater then 2"
Are events A and B independent? Are events A and C independent?
1
PA  PB   ,
2
1
PA  B   P2, b  .
4
1
PA  B    PA PB  . the events A and B are independent
4
by Equation (2.29b),
P A  B 
P2, b 
14 1
P A | B  



PB 
P2, b , 4, w 1 2 2
PA 
PA
P1, b , 2, b 
12


PS  P1, b , 2, b , 3, w, 4, w 1
These two equations imply that PA  PA | B because the proportion
of outcomes in S that lead to the occurrence of A is equal to the proportion
of outcomes in B that lead to A.
Events A and C are not independent since PA  C   P   0
so PA  C   0  PA  .5
EXAMPLE 2.29
Two numbers x and y are selected at random between zero and one.
Let the events A, B, and C be defined as follows:
A  x  0.5,
B  y  0.5,
and C  x  y.
Are the events A and B independent? Are A and C independent?
Using Eq.(2.29a),
PA | B  
PA  B  1 4 1

  PA ,
PB 
12 2
so events A and B are independent.
Using Eq.(2.29b),
P A | C  
P A  C  3 8 3 1

   PA ,
PC 
12 4 2
so events A and C are not independent.
FIGURE 2.15
Examples of independent and
nonindependent events.
y
y
A
B
A
0
1/2
1
x
(a) Events A and B are independent
C
0 1/2
1
x
(b) Events A and C are not
independent
What conditions should three events A, B, and C satisfy in order
for them to be independent? First, they should be pairwise
independent, that is,
PA  B  PAPB, PA  C  PAPC , and
PB  C   PBPC .
In addition, knowledge of the joint occurrence of any two, say A and B,
should not affect the probability of the third, that is
PC | A  B  PC .
and PC | A  B 
PA  B  C 
 PC .
PA  B
Then PA  B  C   PA  BPC  PAPBP C  ,
Thus we conclude that three events A, B, and C are independent if the
probability of the intersection of any pair or triplet of events is equal to the
product of the probabilities of the individual events.
EXAMPLE 2.30
Consider the experiment discussed in Example 2.29 where two
numbers are selected at random from the unit interval. Let the events
B, D, and F be defined as follows:
1
1


B  y   ,
D  x  
2
2


1
1 
1
1

F   x  and y     x  and y   .
2
2 
2
2

It can be easily verified that any pair of these events is independent :
1
PB  D    PB PD  ,
4
1
PB  F    PB PF  , and
4
1
PD  F    PD PF . since B  D  F   , so
4
1
PB  D  F   P   0  PB PD PF  .
8
FIGURE 2.16
Events B, D, and F are pairwise
independent, but the triplet B, D,
F are not independent events.
y
1
y
B
D
|
1
2
x
0
x
1
1

(a) B   y  
2

0
1/2
1
1

(b) D   x  
2

y
1
F
|
1
2
F
x
0
1/2
1
1
1 
1
1

(c) F   x  and y   x  and y  
2
2 
2
2

The events A1, A2, …, An are said to be independent if for k = 2, …, n

    
P Ai1  Ai2    Ain  P Ai1 P Ai2 P Ain ,
(2.30)
where 1  i1  i2    ik  n .
EXAMPLE 2.31
Suppose a fair coin is tossed three times and we observe the
resulting sequence of heads and tails. Find the probability of the
elementary events.
Sample space S  HHH , HHT , HTH , THH , TTH , THT , HTT , TTT .
The assumption that the coin is fair, PH   PT   1 2 .
If we assume that the outcomes of the coin tosses are independent,
PHHH   PH PH PH   1 8 ,
PHHT   PH PH PT   1 8 ,
PHTH   PH PT PH   1 8 ,
PTHH   PT PH PH   1 8 ,
PTTH   PT PT PH   1 8 ,
PTHT   PT PH PT   1 8 ,
PHTT   PH PT PT   1 8 ,
PTTT   PT PT PT   1 8 ,
2.6 SEQUENTIAL EXPERIMENTS
Sequences of Independent Experiments
Let A1, A2, …, An be events such that Ak concerns only the outcome of
the kth subexperiment. If the subexperiments are independent, then
PA1  A2    An   PA1 PA2  PAn .
(2.31)
This expression allows us to compute all probabilities of events of the
sequential experiment
EXAMPLE 2.33
Suppose that 10 numbers are selected at random from the interval
[0,1]. Find the probability that the first 5 numbers are less than 1/4
and the last numbers are greater than 1/2 .
Let x1, x2,…, xn be the sequence of 10 numbers, then
1

Ak   xk  
4

1

Ak   xk  
2

for k  1,  , 5
for k  6,  , 10 .
If we assume that each selection of a number is independent of the
other selections, then
PA1  A2    A10   PA1 PA2  PA10 
5
5
1  1 
    .
4 2
The Binomial Probability Law
EXAMPLE 2.34
Suppose that a coin is tossed three times. If we assume that the
tosses are independent and the probability of heads is p, then the
probability for the sequences of heads and tails is
PHHH   PH PH PH   p 3 ,
PHHT   PH PH PT   p 2 1  p  ,
PHTH   PH PT PH   p 2 1  p ,
PTHH   PT PH PH   p 2 1  p ,
PTTH   PT PT PH   p 1  p  2 ,
PTHT   PT PH PT   p 1  p  2 ,
PHTT   PH PT PT   p 1  p  2 , and
PTTT   PT PT PT   1  p  3 ,
Let k be the number of heads in three trials, then
Pk  0  PTTT   1  p  ,
3
Pk  1  PTTH , THT , HTT   3 p1  p  ,
2
Pk  2  PHHT , HTH , THH   3 p1  p  , and
2
Pk  3  PHHH   p 3 .
THEOREM Let k be the number of successes in n independent
Bernoulli trials, then the probabilities of k are given by the binomial
probability law:
n k
nk
pn k     p 1  p 
for k  0,..., n,
k 
where pn k  is the probability of k successes in n trails,
n
n!
  
 k  k!n  k !
is the binomial coefficient.
(2.32)
(2.33)
We now prove the above theorem. Following Example 2.34 we
see that each of the sequences with k successes and n – k failures
nk
has the same probability, namely p k 1  p  . Let N n k  be the number
of distinct sequences that have k successes and n – k failures, then
pn k   Nn k  p k 1  p 
nk
.
(2.34)
The expression N n k  , is the number of ways of picking k positions
out o fn for the successes. It can be shown that
n
N n k     .
k 
(2.35)
EXAMPLE 2.35
Verify that Eq. (2.32) gives the probabilities found in Example 2.34.
In Example 2.34, let “toss results in heads” correspond to a
“success,” then
p3 0 
3! 0
3
3
p 1  p   1  p  ,
0!3!
3! 1
2
2
p3 1 
p 1  p   3 p1  p  ,
1!2!
3! 2
1
1
p3 2 
p 1  p   3 p 2 1  p  ,
2!1!
3! 3
p3 3 
p 1  p 0  p 3 ,
3!0!
Binomial theorem
 n  k nk
a  b    a b .
k 0  k 
n
n
(2.36)
If we let a = b = 1,
n
n n
n
2      N n k  ,
k 0  k  k 0
If we let a = p and b = 1 – p in Eq. (2.36), then
n
n k
nk
1     p 1  p    pn k  ,
k 0  k 
k 0
which confirms that the probabilities of the binomial probabilities sum
to 1.
n
pn k  1 
n  k  p p k 
k  11  p  n
(2.37)
EXAMPLE 2.36
Let k be the number of active (nonsilent) speakers in a group of eight
noninteracting (i.e., independent) speakers. Suppose that a speaker
is active with probability 1/3 . Find the probability that the number of
active speakers is greater than six.
For i = 1,…, 8, let Ai denote the event “ith speaker is active.” The
number of active speakers is then the number of successes in eight
Bernoulli trials with p = 1/3.
 8  1 
Pk  7  Pk  8    
 7  3 
7
8
 2   8  1 
     
 3   8  3 
 .00244  .00015  .00259
EXAMPLE 2.37 Error Correction Coding
A communication system transmits binary information over a channel
that introduces random bit errors with probability   10 3 . The
transmitter transmits each information bit three times, and a decoder
takes a majority vote of the received bits to decide on what the
transmitted bit was. Find the probability that the receiver will make an
incorrect decision.
 
 
2
3








P k  2    .001 .999   .001  310 6 .
3
 2
3
 3
The Multinomial Probability Law
Suppose that n independent repetitions of the experiment are
performed. Let kj be the number of times event Bj occurs, then the
vector (k1, k2,…, kM ) specifies the number of times each of the
events Bj occurs. The probability of the vector (k1, k2,…, kM ) satisfies
the multinomial probability law :
Pk1 , k 2 ,, k M  
n!
p1k1 p2k2  pMk M ,
k1!k 2 ! k M !
where k1  k2   kM  n .
(2.38)
EXAMPLE 2.38
A dart is thrown nine times at a target consisting of three areas. Each
throw has a probability of .2, .3, and.5 of landing in areas 1, 2, and 3,
respectively. Find the probability that the dart lands exactly three
times in each of the areas.
Let n = 9 and p1 = .2, p2 = .3, and p3 =.5 :
P3,3,3 
9!
.23 .33 .53  .04536
3!3!3!
EXAMPLE 2.39
Suppose we pick 10 telephone numbers at random from a telephone
book and note the last digit in each of the numbers. What is the
probability that we obtain each of the integers form 0 to 9 only once?
Let M = 10, n = 10, and pj = 1/10 if we assume that the 10
integers in the range 0 to 9 are equiprobale.


10! 10
.1  3.6 10 4 .
1!1!1!
The Geometric Probability Law
Consider a sequential experiment in which we repeat independent
Bernoulli trials until the occurrence of the first success. Let the
outcome of this experiment be m, the number of trials carried out until
the occurrence of the first success. The sample space for this
experiment is the set of positive integers. The probability, p(m), that m
trials are required is found by noting that this can only happen if the
first m -1 trials result in failures and the mth trial in success. The
probability of this event is


pm  P A1c A2c  Amc 1 Am  1  p 
m1
p
m  1,2,,
(2.39)
where Ai is the event “success in ith trial.” The probability assignment
specified by Eq. (2.39) is called the geometric probability law.


m 1
m 1
 pm  p q m1  p
1
 1,
1 q
where q = 1 – p,
The probability that more than K trials are required before a success
occurs has a simple form :
Pm  K   p

q
m 1
m  K 1
 pq

K
j
q

j 0
1
 pq
1 q
K
 qK .
(2.40)
EXAMPLE 2.40 Error Control by Retransmission
Computer A sends a message to computer B over an unreliable
telephone line. The message is encoded so that B can detect when
errors have been introduced into the message during transmission. If
B detects an error, it request A to retransmit it . If the probability of a
message transmission error is q = .1, what is the probability that a
message needs to be transmitted more than two times ?
Each transmission of a message is a Bernoulli trial with
probability of success p = 1 – q. The probability that more than two
transmissions are required is given by Eq. (2.40):
Pm  2  q 2  102 .
Sequences of Dependent Experiments
EXAMPLE 2.41
A sequential experiment involves repeatedly drawing a ball from one
of two urns, noting the number on the ball, and replacing the ball in
its urn. Urn 0 contains a ball with the number 1 and two balls with the
number 0. The urn from which the first draw is made is selected at
random by flipping a fair coin. Urn 0 is used if the outcome is heads
and urn 1 if the outcome is tails. Thereafter the urn used in a
subexperiment corresponds to the number on the ball selected in the
previous subexperiment.
FIGURE 2.17
Trellis diagram for a Markov chain.
0
0
0
0
0
0
h
1
1
1
0
t
1
2
1
2
0
0
0
1
1
1
1
1
1
1
2
3
4
(a) Each sequence of outcomes corresponds
to a path through this trellis diagram
0
2
3
1
3
1
6
1
0
1
2
3
0
1
0
1
3
1
6
1
3
1
6
5
6
2
3
5
6
1
5
6
1
(b) The probability of a sequence of outcomes is the
product of the probabilities along the associated path
To compute the probability of a particular sequence of outcomes,
say s0, s1, s2. Denote this probability by Ps0  s1 s2 
.
Let A  s2  and B  s0  s1, then since PA  B  PA | BPB
we have
Ps0  s1 s2   Ps2 | s0  s1Ps0  s1
 Ps2 | s0  s1Ps1| s0 Ps0 .
(2.41)
Now note that in the above urn example the probability
Psn | s0   sn1 depends only on sn1 since the most recent
outcome determines which subexperiment is performed :
Psn | s0   sn1  Psn | sn1.
(2.42)
Therefore for the sequence of interest we have that
Ps0  s1 s2   Ps2 | s1Ps1| s0 Ps0 .
(2.43)
Sequential experiments that satisfy Eq. (2.42) are called Markov
chains. For these experiments, the probability of a sequence s0, s1, …,
sn is given by
Ps0, s1 ,, sn   Psn | sn 1 Psn 1 | sn  2  Ps1 | s0 Ps0 
(2.44)
EXAMPLE 2.42
Find the probability of the sequence 0011 for the urn experiment
introduced in Example 2.41.
P0011  P1 | 1P1 | 0P0 | 0P0 ,
1
2
P1 | 0 
and
P0 | 0 
3
3
5
1
P1 | 1 
and
P0 | 1  ,
6
6
 5  1  2  1  5
P0011        .
 6  3  3  2  54
P0 
1
 P1
2