semester 2010 (june

Department of Computer Engineering, Faculty of Engineering,Kasetsart University, THAILAND

1st semester 2010 (June – September)

Assoc. Prof. Anan Phonphoem, Ph.D.

Department of Computer Engineering, Faculty of Engineering,

Kasetsart University, THAILAND

Anan Phonphoem, Dept.of Computer Engineering, Kasetsart Universty 3

P[☺]

is a function that maps event

in the sample space to real number

From experiment: Roll a dice

Outcomes:

number = 1,2,3,4,5,6

Sample space:

S = {1,2,3,…,6}

Event examples:

E1 = {number < 3} = {1,2}

E2 = {number is odd} = {1,3,5}

P[E1] = 2/6 = 1/3

P[E2] = 3/6 = 1/2

16 June 2010

Axiom 1: For any event A, P[A] 0

Axiom 2: P[S] = 1

Axiom 3: For events A1, A2,…, An of mutual exclusive

events

P[A1A2…An] = P[A1]+P[A2]+…+P[An]

4Anan Phonphoem, Dept.of Computer Engineering, Kasetsart Universty16 June 2010

• If we know P[A] before an experiment

• P[A] 1

• Advanced knowledge almost certainly occur

• P[A] 0

• Advanced knowledge almost certainly not occur

• P[A] ½

• Advanced knowledge maybe occur

• P[A] is a priori probability of A

5Anan Phonphoem, Dept.of Computer Engineering, Kasetsart Universty16 June 2010

• In practice, it maybe impossible to find the

precise outcome of an experiment

• However, if we know that Event B has

occurred

• Probability of A when B occurs can be described

(the outcome of Event A is in set B)

• Still don’t know P[A]

Anan Phonphoem, Dept.of Computer Engineering, Kasetsart Universty 616 June 2010

• Notation: P[A|B]

• “Probability of A given B”

• The condition probability of the event A given the occurrence of the event B

• Definition:


P[A|B] =P[AB]

P[B]

• P[B] > 0

B

AA B

16 June 2010


S

A

B

P[A|B] =P[AB]

P[B]

P[A|S] =P[AS]

P[S]

P[A]

1=

= P[A]

S

A

B

A

A B

B

A B

S

A

BB

A B

16 June 2010

http://www.fifa.com/worldcup/teams/index.html

What is the probability that

South Korea will win the world

cup ?

What is the probability that

South Korea will win the world

cup, given that all teams in

Europe are withdrawn ?

Assume: all teams are equally likely to win the game

1

32

1

19

Let B1, B2,…,Bn be mutually exclusive events whose union equals sample space S partition of S


B1 B2B3

B4 Bn…A

i = 1

n

P[A Bi]P[A] =

For any event A

A = AS = A(B1 B2…Bn)

P[A] = P[AB1] + P[AB2] +…+ P[ABn]

Theorem:

16 June 2010

• Let B1, B2,…,Bn be mutually exclusive events whose union equals sample space S

• P[Bi] > 0


B1 B2B3

B4 Bn…A

P[A] = P[AB1] + P[AB2] +…

P[A] = P[A|B1]P[B1] + P[A|B2]P[B2] +…

i = 1

n

P[A Bi]P[A] = Theorem:

i = 1

n

P[A|Bi]P[Bi]P[A] = Theorem:

16 June 2010


Definition: Event A and B are independent iff

P[AB] = P[A]P[B]

P[A|B] =P[AB]

P[B]

=P[A]P[B]

P[B]

P[B|A] = P[B]

P[A|B] = P[A]

16 June 2010

P[A] = 0.3

P[A|B] = 0.3


No matter event B occurs or not,

event A is not affected

16 June 2010

16

Independent Disjoint

P[AB] 0 P[AB] = 0

P[AB] = P[A]P[B] P[AB] = P[A]+P[B]

BA

BA

Note: Independent = Disjoint iff P[A]=0 or P[B]=0

Anan Phonphoem, Dept.of Computer Engineering, Kasetsart Universty16 June 2010

• The sequence is equally likely

• R1 = The first light was red

• R2 = The second light was red

• G2 = The second light was green

• Are R2 and G2 independent?

• Are R1 and R2 independent?


1 2 3

• 3 traffic lights, observe a sequence of lights

16 June 2010

• Sample Space:

• S = {rrr, rrg, rgr, rgg, grr, grg, ggr, ggg}

• Are R2 and G2 independent?


rrr

rrg

grr

grg

rgr

rgg

ggr

ggg

• P[R2]= P[{rrr, rrg, grr, grg}] = 4/8 = ½

• P[G2]= P[{rgr, rgg, ggr, ggg}] = ½

• P[R2G2] = 0

• P[R2]P[G2] = (½)* (½) = ¼

R2 and G2 are not independent

R2 and G2 are disjoint

rrr

rrg

grr

grg

R2

rgr

rgg

ggr

ggg

G2

16 June 2010

• P[R1]= P[{rrr, rrg ,rgr, rgg}] = ½

• P[R2]= P[{rrr, rrg ,grr, grg}] = ½

• P[R1R2] = P[{rrr ,rrg}] = 2/8 = 1/4

• P[R1]P[R2] = (½) * (½) = ¼

R1 and R2 are independent

R1 and R2 are not disjoint


rrr

rrg

grr

grg

rgr

rgg

ggr

ggg

rrr

rrg

grr

grg

R2

R1

rrr

rrg

rgr

rgg

• Are R1 and R2 independent?

16 June 2010

20

Definition: Event A1,A2 and A3 are independent iff

1) A1 and A2 are independent



4) P[A1A2A3] = P[A1] P[A2] P[A3]

Definition: Event A and B are independent iff

P[AB] = P[A] P[B]Is it sufficient ?

P[ABC] = P[A] P[B] P[C]

Anan Phonphoem, Dept.of Computer Engineering, Kasetsart Universty

NO !

16 June 2010

Note:

• Given: P[A] = P[B] = P[C] = 1/5• Given: P[AB] = P[AC] = P[BC] = P[ABC] = 1/25• Independence in pairs (number 1-3) may not

independent

• Only number 4) is insufficient to guarantee the independence

• Ex.: One of the event is Null


• Assume that the event of separate

experiments are independent

• Example:

• Assume that outcome of a coin toss is independent

of the outcomes of all prior and all subsequent coin

tosses

• P[H] = P[T] = ½

• P[HTH] = P[H] P[T] P[H] = 1/2*1/2*1/2 = 1/8


• Experiment: in sequence

subexperiments subexperiments

• Each subexp. may depend on the previous one

• Represented by a Tree Diagram

• Model Conditional Prob. Sequential Experiment


Leaf1

Outcome (node)

Outcomes of

the complete

Experiment (Leaf)

Branch

Prob. value

16 June 2010

• Timing coordination of 2 traffic lights

• P[the second light is the same color as the first] = 0.7

• Assume 1st light is equally likely to be green or red

• Find P[The second light is green] ?

• Find P[wait for at least one light] ?


26

• P[G1] = P[R1] = 0.5

0.5

0.5

0.7

0.3

0.3

0.7

G1

R1

G2

G2

R2

R2

G1G2 = 0.35

G1R2 = 0.15

R1G2 = 0.15

R1R2 = 0.35

• P[G2G1] = P[G2|G1]P[G1] = (0.7)(0.5) = 0.35


27

P[The second light is green] ?

P[G2] = P[G2G1] + P[G2R1] = 0.35 + 0.15 = 0.5

P[G2] = P[G2|G1]P[G1] + P[G2|R1]P[R1]

0.5

0.5

0.7

0.3

0.3

0.7

G1

R1

G2

G2

R2

R2

G1G2 = 0.35

G1R2 = 0.15

R1G2 = 0.15

R1R2 = 0.35


28

P[wait for at least one light] ?

W = {G1R2 R1G2 R1R2}

P[W] = P[G1R2] + P[R1G2] + P[R1R2]

= 0.15 + 0.15 + 0.35 = 0.65

0.5

0.5

0.7

0.3

0.3

0.7

G1

R1

G2

G2

R2

R2

G1G2 = 0.35

G1R2 = 0.15

R1G2 = 0.15

R1R2 = 0.35


16 June 2010 Anan Phonphoem, Dept.of Computer Engineering, Kasetsart Universty 29

0.4W1

L1

Korea Republic

Greece Argentina Nigeria

D1

W2

L2

D2

W2

L2

D2

W2

L2

D2

W3

L3

D3

W3

L3

D3

W3

L3

D3

0.1

0.5

0.3

0.1

0.6

0.6

0.1

0.3

0.6

0.1

0.30.6

0.1

0.3

0.3

0.1

0.6

0.3

0.1

0.6

W1W2W3 = 0.4 * 0.3 * 0.6 = 0.072

W1W2D3 = 0.4 * 0.3 * 0.1 = 0.012

W1W2L3 = 0.4 * 0.3 * 0.3 = 0.036

31

If experiment A has n possible outcomes,

and experiment B has k possible outcomes,

Then there are nk possible outcomes

when you perform both experiments


32

1st draw: select 1 out of 52 52 outcomes

2nddraw: select 1 out of 51 (one card has been drawn)

51 outcomes

3rddraw: select 1 out of 50 50 outcomes

Total outcomes = (52)(51)(50)

Example:

Shuffle a deck and select 3 cards in order.

How many outcomes?



(n-k)!(n-k)!

(n)k =n!

(n-k)!

n(n-1)(n-2)…(n-k+1) (n-k)(n-k-1)…(1)

(n-k)!=

(n)k = n(n-1)(n-2)…(n-k+1)

= n(n-1)(n-2)…(n-k+1)

Theorem:

The number of k-permutations, (n)k ,

(ordered sequence) of n distinguishable objects is

34

Example: You are allowed to choose only three from five items (in ordered sequence).

How many possible outcomes?

(n)k =n!

(n-k)!

(5)3 =5!

(5-3)!

5 x 4 x 3 x 2!

2!= = 60

A:

B:

C:

DifferentAnan Phonphoem, Dept.of Computer Engineering, Kasetsart Universty16 June 2010

35

Theorem: Given n distinguishable objects,

There are nk ways to choose with replacement

a sample of k objects


36

Example: You are allowed to choose only three from five items (with replacement).

How many possible outcomes?

nk = 5 x 5 x 5 = 125

A:

B:

C:


37

Theorem:

The number of ways to choose k objects

out of n distinguishable objects is

( )n

k=

(n)k

k!n!

k!(n-k)!=


38

2 subexperiments: ( ) then (k)k

( ). (k)k = (n)k

nknk

53( ) = 10

No order

(3)3 = 6

Order

1 2 3

(5)3 = 60

Order


39

• Perform repeated trials

• p = a success probability

• (1-p) = a failure probability

• Each trial is independent

• Sk,n = the event that k successes in n trials

( )n

kP[Sk,n] = pk(1-p)n-k


• 3 trials with 2 successes

• 000 001 010 011 100 101 110 111

• How many way to choose 2 out of 3


• What is the probability of success for each way ?• p2

* (1-p)

( )3

2P[S2,3] = p2(1-p)3-2

( )nk= = = 3( )3

2

• Example: In the first round of a programming contest,

probability that a program will pass the test is 0.8 .

• From 10 candidates, what is the probability that x candidates

will pass?


10

x( )P[Ax,10] = (0.8)x(1-0.8)10-x

P[A8,10] = (45)(0.1678)(0.04) = 0.3

Solution:

A = {program pass the test}, P[A] = 0.8

Testing a program is an independent trial

And what is P[x = 8]?

42

Let probability that a computer works = p

Series: P[A] = P[A1A2] = p2

Parallel: P[B] = ?

Series Parallel


P[B] = 1 – P[Bc]

= 1 – P[B1cB2

c]

= 1 – (1 – p)2

• Probability meaning

• Sample space, Event, Outcome

• Set Theory

• Probability measurement

• Conditional Probability

• Independence

• Sequential experiments tree diagram

• Counting Methods

• Independent Trials


44

• Homework Problems posted in the class

Website

• Due date on Friday, June 25 – In class


semester 2010 (june

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