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# Elements of probability. (Lecture 3)

## 1. Lecture #_3

Lecture #_3Elements of probability

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## 2.

Definition 1A simple event is an outcome of an experiment that

can not be decomposed into a simpler outcome

Example 1. Tossing 1 fair coin

1 , 2

1 head (H)

2 tail (T)

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## 3.

Example 2. Tossing 3 fair coins at once1 , 2 , 3 , 4 , 5 , 6 , 7 , 8

1 HHH 5 TTT

2 TTH 6 HHT

3 THT 7 HTH

4 HTT 8 THH

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## 4.

A random variable X( ) is a function of , e.g., X( ) isthe number of heads in one trial, hence,

X( 1 ) 3, X( 6 ) X( 7 ) X( 8 ) 2,

X( 2 ) X( 3 ) X( 4 ) 1, X( 5 ) 0

Definition 2

An event is a collection of one or more simple events

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## 5.

Example 3.A 2 Heads in one toss

A ω6,ω7 ,ω8

B More than one H in a trial

B ω1,ω6,ω7 ,ω8

C Exactly 2 tails (T)

C ω2,ω3,ω4

D At least one H or one T

D ω1,ω2,ω3,ω4,ω5,ω6,ω7 ,ω8

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## 6.

Definition 3The sample space of an experiment is the collection of

all possible simple events

Denote it by Ω

{ 1 , 2 }

For example 1:

For example 2 : { 1 , 2 , ..., 8 }

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## 7.

Venn diagram1

2 3

4

B

5

6 7

A

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Definition 4.

The probability p of a simple event is a number that

measures the likelihood that the event will occur

when the experiment is performed.

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## 8.

I.Classical definition of probabilities. If one has n

equiprobable simple events 1 , ..., n , then the

probability

of

p( A)

A i1 , i 2 , ..., im

containing m simple events is

m

P( A )

n

Evidently, that P( ) 1.

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## 9.

II. We may take p being equal to the relative frequencyf /n of a simple ith event if n is very large.

i

III. We may select p based on a priori knowledge of a

situation under study.

The case III is more frequently occurred. We guess or

formulate a hypothesis H concerning p and, using

statistics, test this hypothesis based on realizations

( x1 , ..., x n ) of a sample X n { X1 , ..., X n }.

Definition 5

The probability P( A) of an event A is calculated

by summing the probabilities of the simple events

belonging to A .

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## 10. Steps for calculating probabilities

1. List the simple events2. Assign probabilities to simple events

3. Determine the number of simple events

containing in the event

4. Sum the simple event probabilities to

obtain the event probability

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## 11.

For example 2p(ω i ) 1/8, i 1,8

p(A) p(2H) p{ω 6 ,ω 7 ,ω 8 }

1

8

1

8

1

8

3

8

p(B) P(more than 1H)

p{ω1 ,ω 6 ,ω 7 ,ω 8 }

1

8

1

8

1

8

1

8

1

2

p(D) p(Ω) 1 (certain event)

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## 12.

Compound eventsDefinition 6

The union of A & B is the event that occurs if

either A or B or both occur ( A B )

A

B

i , i 1, n ,

belong A or B or to both of them

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## 13.

Definition 7The intersection of A & B is the event that occurs if

both A and B occur ( A B )

i , belong to both A and B

A - the complement event to A

i of A do not belong to A!

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## 14.

Example 3A { 6 , 7 , 8 }, B { 1 , 6 , 7 , 8 }

6 , 7 , 8

A B { 6 , 7 , 8 }

A B { 1 , 6 , 7 , 8 }

1

A

B

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## 15.

Let a set { 1 , ..., n } be given, then all possibleevents (combinations of i th ) (empty set,

corresponding to the improbable event) is algebra S of

events

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## 16.

If two events A & B are mutually exclusive(incompatible), then

P( A B) P( A) P(B).

Suppose we have a target. A shooter produced a shot.

A to hit the target

B fail to hit

A&B are incompatible

P( A) P(B) 1

A B is the certain

event

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## 17.

Axioms of probabilityFor every (a set of all possible simple events) and S

(algebra of events) we postulate

1) Axiom 1. For every S P(A)

2) Axiom 2. P( ) 1

-is certain event

3) Axiom 3. If S and B S are incompatible, then

P( A B) P( A) P(B).

The following two rules are consequences of axioms:

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## 18.

Additive ruleP( A B ) P( A ) P(B ) P( A B )

Multiplicative rule

A B

A

B

P( A B ) P ( A | B )P ( B )

P( A B ) P ( B | A )P ( A )

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## 19.

The conditional probabilityr

m

k

A

B

A contains m elements of

B contains k elements of

A B contains r elements of

{ 1 ,..., n }, i th are equiprobable

r

P( A B )

n

k

P(B)

n

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## 20.

Let the event B occurred. What is the prob. for A tooccur?

r r / n P( A B )

P( A | B )

k k /n

P( B )

P( A B )

P( A | B )

P( B )

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## 21.

IndependenceEvents A and B are independent if the assumption

that B has occurred does not alter the probability that

A occurs, or if

P (A | B) = P (A)

or

P (B | A) = P (B)

Otherwise, A&B are dependent

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## 22.

Example 2A 6 , 7 , 8 two H in one tos sin g

B 1 , 6 , 7 , 8 than 1 H in one tossing

3

4

8

P(A)

P(B)

P( ) 1

8

8

8

A B 6 , 7 , 8 A

3

P( A B ) P( A )

8

3 4 3

P( A B ) P( A | B )P( B )

4 8 8

3 1 3

3

P( A ) P( B ) P( A B )

8 2 16

8

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## 23.

Hence, A and B are dependentP( A B ) 3 / 8 3

3

P( A | B )

P ( A ) !!

P( B )

1/ 2 4

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For independent events

P(A B) = P(A) P(B)

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## 24.

Random variable X( (discrete)X( is the number of H for example 2. From the

symmetry of a coin we may suppose that the

probability p to obtain a head in one trial is

p 1 / 2.(q 1 / 2 1 p) . Then

3 k 3 k

P ( X k ) p q

,

k

k 0,1,2,3.

2

3 2

3 1 1

1 1 3

P ( A ) P ( X 2 )

3

4 2 8

2 2 2

P(B ) P( X 1) P( X 2) P( X 3)

3

0

3

3 1 1

3 1 1

8 3 2 2

8 8 2

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## 25.

Events (X=2) & (X=3) are mutually exclusive, henceP ((X=2) (X=3)) = P (X=2)+P (X=3).

We may consider P(X=k) as both known and known

within the unknown parameter p.

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## 26.

Population moments (discrete case)Let P (X = k) be the p. d. of X, k {set of discrete

numbers}= .

P( X k ) 1

k K

Definition 9

The initial population moment of P (X = k) of order m

is

m k m P( X k ) E( X m )

k K

EX m - expectation of Xm or

expected value of Xm

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## 27.

Definition 10The central population moment of P (X = k) of order

m is

m (k EX)m P( X k ) E( X EX)m

k K

Definition 11 (p.210)

The mean or expected value of a discrete r. v. X is

EX kP( X k )

k K

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## 28.

Definition 12The variance of a discrete random variable X is

2 E( X ) 2 (k ) 2 P( X k )

k K

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## 29.

Example Binomial distributionP (X k )

n

k

pk (1 p)n k , k=0,n

K 0,1, ...,n

n

EX k

k 0

n

k

pk (1 p)n k

n

kn!

pk (1 p)n k

k 1 k !(n k )!

n

n(n 1)!

p pk 1qn k

k 1 (k 1)!(n k )!

k 1

n

np

n 1

k 1

pk 1qn k

k 1 r k r 1

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## 30.

p qnp,

r 0

n 1

n 1

r (n 1) r

n 1

(p q)

1

r p q

r 0

n 1

np

n 1

r

r (n 1) r

Thus,

EX np

By the same lines

2

E(X

)

npq

S tan dard deviation

npq (1a)

2

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