• Random Experiment : an experiment in which outcome varies in an unpredictable fashion when the experiment is repeated under the same condition
Specified by:
1. An experimental procedure
2. One or more measurements or observations tch-prob 1

EXAMPLE 2.1 :
Experiment E
1
: Select a ball form an urn containing balls numbered 1 to 50.
Note the number of the ball.
Experiment E
2
: 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 E
3
: Toss a coin three times and note the sequence of heads/ tails.
Experiment E
4
: Toss a coin three times and note the number of heads.
Experiment E
6
: A block of information is transmitted repeatedly over a noisy channel until an error-free block arrives at the receiver.
Experiment E
7
: Pick a number at random between zero and one.
Experiment E
12
: Pick two numbers at random between zero and one.
Experiment E
13
: Pick a number X at random between zero and one, then pick a number Y at random between zero and X .
tch-prob 2

• Compare E3 and E4: same procedure, different observations
• A random experiment with multiple measurements or observations: E2, E3, E12, E13
• sequential experiment consists of multiple subexperiments: E3, E4, E6, E12, E13
• dependent subexperiments: E13 tch-prob 3

•
Outcome , sample point:

– one and only one per experiment
– mutually exclusive, cannot occur simultaneously
•
Sample space : set of all possible outcomes
• Example 2.2: sample spaces
, , , , , , S
1 2 3 6 7 12 13
• number of outcomes:
– Finite
– countably infinite S6
– unaccountably infinite S7




Discrete sample space
Continuous
Sample space tch-prob 4

EXAMPLE 2.2 :
The sample spaces corresponding to the experiments in Example
2.1 are given below using set notation :
S
1
S
2



1, 2,......, 50

        w
 
S
3
 
, , , , , , ,
S
4
S
6

 

0,1, 2, 3

1, 2, 3,

1
  
S
7
  x
S
12

: 0 x
  
: 0
S
13

  
: 0
 
1 and y x 1

0 1

 tch-prob 5

Event : a subset of S, a collection of outcomes that satisfy certain conditions
Certain event: S null event:
 elementary event: event of a single outcome
Example 2.3: ,
2 4
,
7
A2: The ball is white and even-numbered
A4: The number of heads equals the number of tails
A7: The number selected is nonnegative tch-prob 6

Set Operations
1.
Union A B
2.
Intersection A B
3.
Complement
A c  
A
B equal A=B
- Commutative properties of set operations
A B

B A A B

B A
- Associative properties of set operations
A ( B C )

( A B ) C A ( B C )

( A B ) C
- Distributive properties of set operations
A ( B C )

( A B ) ( A C )
- De Morgan’s Rules ( A B ) c  c
A B c
( ) c  c
A B A B c
Venn diagrams: Fig. 2.2
tch-prob 7

Example 2.5: Configuration of a three-component system a. series all three
A
1
I A
2
I A
3
C1 C2 b. parallel at least one of three
A A A
1 2 3 c. two-out-of-three
( A
1
A
2
A
3
) ( c
A
1
A
2
A
3
) ...
n k
 n
1
A k

A A
1 2 k

1 k

1
A k

A A
1 2

A k k

1
A k
.....
.....
An
An
A k tch-prob
C1
C2
C3
C3
8

2.2 The Axioms of Probability
A probability law for the experiment E is a rule that assigns to each event a number P(A ), called the probability of A , that satisfies the following axioms:
Axiom II: P(S)=1
A i
A j total=1
P A B
      i
 j
P



 k


1
A k



   k

1
P A
 k



A set of consistency rules that any valid probability assignment must satisfy.
tch-prob 9

Corollary 1.
P A
 c 
 pf:
A A c 
1

P S

P A A
 c 


P A 


P A
 c 

Corollary 2.
P A 
 

1 pf: from Cor.1,


P A
 c 


1
Corollary 3. P  
 
0 pf: Let A=S , in Cor.1.
then
P


 k

1
A k


 k

1
P A
 k  for n

2 tch-prob 10

P A B




P A 


P B 


P A B


Corollary 5. pf: P A B




P A B
 c 


P B

A c 


P A B





P A B
 c 

 P A B

P B

A c 


P A B





A B
Corollary 6.
P



 k n

1
A k



  n j

1

P A j


 
P A j
A k

 ....
  n

1
P A
 1
....
An


P A 


P B 
 pf.


P A 


P A
 c
B



P A 
 tch-prob 11

Axioms + corollaries
 provide rules for computing the probability of certain events in terms of other events. However, we still need initial probability assignment
1.
Discrete Sample Spaces find the prob.of elementary events; all distinct elementary events are mutually exclusive,
Example 2.6., 2.7
2. Continuous Sample Spaces assign prob. to intervals of the real line or rectangular regions in the plane
Example 2.11
tch-prob 12

Discrete Sample Spaces
First, suppose that the sample space is finite, and is given by
S

 a
1
, a
2 a n

 
P

' '
,
1 2
, , a
' m


P 
 


P 
 

(by corollary ?)
 
P

 

(2.9)
If S is countably infinite, then Axiom Ⅲ ’ implies that the probability of an event such as
D

 b
1
'
, b
2
' 
 is given by
  
P

 
 
P

 
 
(2.10) tch-prob 13

If the sample space has n elements, S

 a
1
,  a n

, a probability assignment of particular interest is the case of equally likely outcomes.
The probability of the elementary events is
P
 
1

P
 
2
  
P
  n

1 n
(2.11)
The probability of any event that consists of k outcomes, say
B

 a
1
'
,  a k
'

, is
P
 

P
 
1
  
P
  k
'  k n

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.
(remember the classical definition?) tch-prob 14

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 .
tch-prob 15

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
 

P


P
 

P
 

P
 

P
 

5
10
P
 

P
 

P
 

P
 

3
10 .
P
 

P
 

P
 

P
 

P
 

P
 

5
10
.
From Corollary 5.
P

A

B
     
A

B


5
10

3
10

2
10

6
10
.
tch-prob 16

where we have used the fact that A

B

P

A

B

From Corollary 6,
P

A

B

C
       
A

B

  
C
  
C
   
C


5
10

3
10

5
10

2
10

2
10

1
10

1
10

9
10 tch-prob 17

EXAMPLE 2.7
Suppose that a coin is tossed three times. If we observe the
S sequence of heads and tails, then there are eight possible outcomes


HHH , HHT , HTH , THH , TTH , THT , HTT , TTT

. If we assume that the outcomes of S
3 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

P


HHT ,

HHT
 
HTH

P

,

THH
HTH

 


P
 
THH
 

3
8
S
 
0,1, 2,3

* If we count the number of heads in three tosses, then
P

" 2 heads in 3 tosses "


P
 

1
4 tch-prob 18

Continuous Sample Spaces
EXAMPLE 2.11
Consider Experiment E
12
, 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
A

 x

B

 y

0 .
5

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
P
 

1
2
P
 

1
2
P
 

1
2 tch-prob 19

FIGURE 2.8
a Two-dimensional sample space and three events.
y y
S
0 1
(a) Sample space x y

1
2 x

1
2 y
0 1
(b) Event
 x
1
2
0 1
1
(c) Event
 y
2 x x
 y
0 1
(d) Event
 x
 y
 x tch-prob 20

2.3 Computing probabilities using counting methods
In many experiments with finite sample space, the outcomes can be assumed to be equiprobable.
1.
Sampling with Replacement and with Ordering choose k objects from a set A that has n distinct objects, with replacement and note the order. The experiment produces an ordered k -tuple
 x x x
1, 2,..., k

The number of distinct ordered k-tuples =
  n k
Pr[2 balls with same number] =
5
25
5
2 
25 tch-prob 21

2.
Sampling without Replacement and with ordering choose k objects from a set A that has n distinct objects, without replacement.
number of distinct ordered k-tuples = n (n-1)…(n-k+1)
Ex.2.13 5 balls, select 2
Pr [1st number >2nd number ]= 10/20 tch-prob 22

Permutations of n Distinct Objects
Consider the case when k=n number of permutations of n objects = n !
For large n , stirling’s formula is useful n !

2
 n n
  n
Ex. 2.16 12 balls randomly placed into 12 cells, where more than 1 ball is allowed to occupy a cell.
12!
Pr [ all cells are occupied ]= ~ 5.37 10

5 tch-prob 23

3.
Sampling without Replacement and without ordering
~ ” combination of size k ” = C k n each combination has k ! possible orders
C k
C k k n

!

 n k n n



 k


!
 n k 1 n k 1



 n !
 
 n
 
!
!
k
“binomial coefficient”, read “ n choose k ”
Ex. 2.18 The number of distinctive permutation of k white balls and n-k black balls
 n !

!
 tch-prob 24

Ex. 2.19 A batch of 50 items contains 10 defective items.
Select 10 items at random, Pr [ 5 out of 10 defective ]=?


10 40
 

5




50
5
10




Exercise prob. 47: Multinomial
Ex.2.20 Toss a die 12 times. Pr [ each number appears twice ]= ?
12!
2!2!2!2!2!2!
12
6

{1,1,2,2,3,3, …….6,6} permutation tch-prob 25

4. Sampling with Replacement and without ordering


 n
  k  


 
 n k k
  n

1

 Pick n k balls from an urn of balls. ( k may be greater than n)
Suppose k=5, n=4 object 1 2 3 4
XX X XX
The result can be summarized as XX // X / XX
= k: x
3!5!

8
 
3 n-1: /
=
 k n 1 !

!

1 !


Ex. Place k balls into n cells.
tch-prob 26

2.4 Conditional Probability
We are interested in knowing whether A and B are related, i.e. , if B occurs, does it alter the likelihood of A occurs.
Define the conditional probability as
B
A







P B 


 for P[B]>0
- renormalize B ] tch-prob 27

Example 2.21. A ball is selected from an urn containing 2 black balls (1,2) and 2 white balls (3,4).
A: Black ball selected ½
B: even-numbered ball ½
C: number > 2 ½



P A

I B
P B 


  0.25
0.5

0.5

P A 




P A

I C



  0
0.5
P A 

Not related
Related tch-prob 28



I



P B P A B



 is useful in finding prob. in sequential experiments
Ex.2.22 An urn contains 2 black, 3 white balls.
Two balls are selected without replacement.
Sequence of colors is noted.
Sol. 1. b b two subexperiments b w w b w w
B
2
1/4
1/10
[
1
I B
2
]

[
1
] [ / B
2 1
]
B
1
2/5
3/4
W
2
3/10
3/5
W
1
B
2
2/4
3/10
Tree diagram
W
2
2/4
3/10 tch-prob 29

Let be mutually exclusive, whose union =
2 n
S , partition of S .



I

A I ( B
1
U B
2

A B
1

U

 
A B
2


...

B n
)
A B n 
P A 

 

I
1


 

I
2




I n


B1 B2
A
“ Theorem on total probability ”
P A 


P B P A B
1 1



P B P A B
2 2


P A B P B n
S
Bn n

Ex. 2.24 In prev. example (2.22) find P[W
2
] =Pr [second ball is white]
P W
2


P W B P B
2 1 1


P W W P W
2 1
 
1

    
4 5 2 5 5 tch-prob 30

Suppose event A occurs, what is the prob. of event Bj ?
Bayes’ Rule
P B
 j
A



P A

I B
P A 
 j

 
P A B
 j


P B
 k n

1
P A B k
 j


P B k


Before experiment: P [ Bj ]= a priori probability
After experiment: A occurs
P [ Bj/A ]= a posteriori probability tch-prob 31

Ex.2.26. Binary symmetric channel
Suppose
P 
A
0
 
P 
A
1
  1
2
1-p
A i
0 p 1
P B
1


P B A P A
1 0 0


P B A A
1 1 1
 e
2
P A B
0 1


P A B
1 1



P B A P A
1 0
P B
 
 
1

0
 e


  e / 2
1/ 2
 e ,
1-e e e
1-e
1 1
2 2
B j
0
1 tch-prob 32

2.5 Independence of Events
If knowledge of the occurrence of an event B does not alter the probability of some other event A , then A is independent of B .



P A B

P B 


 
Two events A and B are independent if


I



P A P B 
 tch-prob 33

Ex. 2.28. A ball is selected from an urn of 4 balls
{1,b},{2,b},{3,w},{4,w}
A: black ball is selected
B: even-numbered ball
C: number >2
P[A]=P[B]=P[C]=0.5


P A C




 4

0
P A P B 

P A B
P A C






P A
P A




If two events have non zero prob., and are mutually exclusive, then they cannot be independent.
0

P  I 

P A P B

0 tch-prob 34

Three events A, B , and C are independent if
Ex. 2.30.
P B

I D


P A

I B



P A P B 

P A

I C 


P A P C 

P B

I C



P B P C 
   
P A

I B I C



P A P B P C 

4
P B P D 

B
P B

I F

 4
P D

I F


P B

I D I F
4


P
P
P B P D P F 

D
F
F
Common application of the independence concept is in making the assumption that the event of separate experiments are independent.
Ex.2.31. A fair coin is tossed three times. P[HHH]=1/8, ….
tch-prob 35

2.6. Sequential Experiments
• Sequence of independent experiments
A k the k , ,

, A
P A
1
I A
2
I

I
• Bernoulli trial:
A n 

P A P A
1 2
 ...
P A n

Perform an experiment once, note whether an event A occurs.
success, if A occurs prob= p failure, otherwise.
prob= 1-p
Pn(k) = prob. of k successes in n independent repetitions of a Bernoulli trial = ?
tch-prob 36

Ex.2.34. toss a coin 3 times. P [ H ]= p
P k 0

1 p
 3
P k 2

3 p 2
 

 
3 p
 
2 
3
 p 3
Binomial probability law p n
   n
  k p k
  k=0,1,…,n tch-prob 37

Binomial theorem
 a b
 n k n

0 n
Let a=b=1 ,

2 n k n

0 n
 
Let a

,
  p , 1
 k n


0 k p k

1
 p
  k n


0 p n
 
 
10
 majority decoding is used. What is prob. of error?

2
  
    
 
  3  

6 tch-prob 38

Multinomial Probability Law
P B j
 p j
Bj ’s: partition of S ; mutually exclusive p
1
 
2 p
M

1 n independent repetitions of experiment B
1
B
2 ....
k j number of times Bj occurs.
[( , ,...,
1 2 k
M
)]
 n !
k k
2
!...
k
M
!
p p
1 2 k
2
...
p
M kM
....
Ex.2.38. dart 3 areas p

0.2,0.3,0.5
throw dart 9 times, P[3 on each area]=?
P

3,3,3
 
9!
3!3!3!
      3
B
M
S tch-prob 39

Geometric Probability Law
Repeat independent Bernoulli trials until the occurrence of the first success.
        m

1 p , m

1,2,...
m

1 times m



1
   p m



1

1
 p
 m

1  p


1 p


1
P[ more than k trials are required before a success occurs ]

[ ] p

 m k 1

1
 p
 m

1  p
 k
P[The initial k trials are failures] tch-prob 40

Sequence of Dependent Experiments
Ex.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 : 1 ball #1, 2 balls #0
Urn 1 : 5 balls #1, 1 ball #0
Initially, flip a coin. Thereafter, the urn used in the subexperiment corresponds to the number selected in the previous subexperiment.
t h
0
0
1
0
1
1
0
1
1
0
0
1
0
1 ....
Urn 1
Urn 2
S
0
S
1
S
2
Trellis diagram tch-prob
S
3
41

P S
0
S
1



1
S
0
   conditional prob.
P S
0
S
1
S
2



2
S
0 1 0
S
1
P S
0
S
1
,...
P S
2
S
0 1 1
S
0
  
0
P S
2 1

1
S
0
P S
S n


 n

1
P S n

1
S n

2
Markov property

...

0
P S
....
Ex.2.42 calculate with
P  0011 

P    1 0   0 0   
P
 
P


1
3
5
6
P
P

2
3

1
6
P
2
P [1] tch-prob 42

Problem 95 on Page 83
95 Consider a well-shuffled deck of cards consisting of 52 distinct cards, of which four are aces and four are kings.
a. Pr[Obtaining an ace in the first draw]=
4

1
52 13 b. Draw a card from the deck and look at it.
Pr[Obtaining an ace in the second draw]=
A
3
51
A
4
51
If does not know the previous outcome
1
1 12 4
13 51 13 51
13 tch-prob 43

c. Suppose we draw 7 cards from the deck.
Pr[7 cards include 3 aces]=

4

48
 3


52
7
4




Pr[7 cards include 2 kings]=

4

48
 2


52
7
5




Pr[7 cards includes 3 aces and 2 kings]=

4

4

44

3


2 
52
7



2

 d. Suppose the entire deck of cards is distributed equally among four players. Pr[each player gets an ace]=
4 48 3 36 2 24 13
4!
52 39 26 13
4!
tch-prob 44

Birthday problem.
p =No two people in a group of n people will have a common birthday.
p


1
365


1

2
365
 
 
1
 n

1
365

 p p

1

2
0.01
n

23 n

56
Lotto Problem .
Six numbers and one special number are picked from a pool of 42 numbers. …
1.
What is the probability of winning the grand prize?
2.
What is the probability of winning 2 nd prize?
3.
What is the probability that number 37 is drawn next time? tch-prob 45