Axiom of Probability

Lecturer : Asrar Alyafie

Contents

2.1 Introduction

2.2 Sample Space and Events

2.3 Axioms of Probability

2.4 Some Simple Propositions

2.5 Sample Spaces Having Equally Likely Outcomes

2.6 Probability as A Continuous Set Function

2.7 Probability as A Measure of Belief

2.2 Sample Space and Events

Sample Space:

It is a set whose elements represent all possible outcomes of an experiment. It is usually denoted by S.

Some examples:

1. The sex of a new born child, then the sample space is:

S = {g, b}

2. Flipping two coins, then the sample space is:

S = {(H,H), (H, T), (T,H), (T, T)}

3- Tossing a dice once, then the sample is:

S = {1,2,3,4,5,6}

4- Tossing two dice, then the sample space consists of the 36 points (𝑛 𝑆 = 62)

S = {(i, j): i, j = 1, 2, 3, 4, 5, 6}

Event :It is a subset of a sample space. It is usually denoted by E.

An example:

1. Flipping two coins, then E is the event that a head

appears on the first coin.

E = {(H,H), (H, T)}

There are some important operations on Events:

Complement of event:

For any event E, we define the new event 𝑬𝒄, referred to

as the complement of E, to consist of all outcomes in the

sample space S that are not in E. That is, 𝐸𝑐 will occur if

and only if E does not occur.

Union:

For any two events E and F of a sample space S, we

define the new event E ∪ F, called the union of E and

F, to consist of all outcomes that are either in E or in

F or in both E and F.

For example:

For the sex of a new born child, E = {g} and F = {b},

then: E ∪ F = {g, b}

For example: Flipping two coins, E = {(H,H), (H, T)} and F = {(T,H)},

then: E ∪ F = {(H,H), (H, T), (T,H)}

Intersection:

Two events E and F, we may also define the new event

EF, called the intersection of E and F, to consist of all

outcomes that are both in E and in F. That is, the event

EF (sometimes written E ∩ F) will occur only if both E

and F occur.

For example :

E = {(H,H), (H, T), (T,H)} is the event that at least 1 head

occurs & F = {(H, T), (T,H), (T, T)} is the event that at

least 1 tail occurs.

then E ∩ F = EF = {(H, T), (T,H)}.

Subset:

Two events E and F, if all of the outcomes in E are also

in F, then we say that E is a subset of F, and write E ⊂ F .

Thus, if E ⊂ F, then the occurrence of E implies the

occurrence of F.

If E ⊂ F and F ⊂ E , we say that E and F are

equal and write E = F.

Commutative laws E ∪ F = F ∪ E EF = FE

Associative laws (E∪F) ∪ G = E ∪ (F ∪G) (EF)G = E(FG)

Distributive laws (E∪F) G = EG ∪ FG EF ∪G = (E∪G)(F ∪G)

2.3 Axioms of Probability

Axiom1: 0 ≤ 𝑃(𝐸) ≤ 1

Axiom 2: P(S) = 1

Axiom 3:

For mutually exclusively events A, B, and C in S:

1. P(A∪B) = P(A) + P(B)

2. P(A∪B∪C) = P(A) + P(B) +P(C)

In general, for mutually exclusively events 𝐸1, 𝐸2,…, 𝐸𝑛defined on S.

𝑃 ራ

𝑖=1

𝑛

𝐸𝑖 =

𝑖=1

𝑛

𝑃 𝐸𝑖

EXAMPLE 3a

If our experiment consists of tossing a coin and if we assume that

a head is as likely to appear as a tail, then we would have

P({H}) = ½

P({T}) = ½

EXAMPLE 3b

If a die is rolled and we suppose that all six sides are equally

likely to appear, then we would have P({1}) = P({2}) = P({3}) =

P({4}) = P({5}) = P({6}) = 1/6. From Axiom 3, it would thus

follow that the probability of rolling an even number would

equal

P({2, 4, 6}) = P({2}) + P({4}) + P({6})

= 1

6+

1

6+

1

6

= 3

6=

1

2

2.4 Some Simple Propositions

Proposition 4.1.

P(E c) = 1 − P(E)

Proposition 4.2.

If E ⊂ F, then P(E) ≤ P(F).

Proposition 4.3.

P(E ∪ F) = P(E) + P(F) − P(EF)

EXAMPLE: Let A and B defined on the sample space S

such that: P(A)=0.3 , P(B)=0.25, P(A∩B)=0.07.

Find P( A ∪ B)?

P( A∪ B) = P(A) + P(B) - P(A∩B)

= 0.3 + 0.25 – 0.07 = 0.48

2.5 Sample Spaces Having Equally

Likely Outcomes

EXAMPLE 5a

If two dice are rolled, what is the probability that the sum of the

upturned faces will equal 7?

Solution.

E={ (1, 6), (6, 1), (2,5), (5, 2), (3, 4), (4, 3) }

P(E)= 6/36 =1/6

EXAMPLE 5c

A committee of 5 is to be selected from a group of 6men and

9 women. If the selection is made randomly, what is the

probability that the committee consists of 3 men and 2

women?

Solution.

𝟔𝑪𝟑 ×𝟗𝑪𝟐

𝟏𝟓𝑪𝟓=

𝟔𝟑

𝟗𝟐

𝟏𝟓𝟓

=𝟐𝟒𝟎

𝟏𝟎𝟎𝟏

EXAMPLE 5b

If 3 balls are “randomly drawn” from a bowl

containing 6 white and 5 black balls, what is the

probability that one of the balls is white and the other

two black?

Solution.

𝟔𝑪𝟏 ×𝟓𝑪𝟐

𝟏𝟏𝑪𝟑=

𝟔𝟏

𝟓𝟐

𝟏𝟏𝟑

=𝟒

𝟏𝟏

P(A B)=P(A) . P(B) Independent Events

P(A B)=P(A) . P(B|A) dependent Events

P(A or B)=P(A) + P(B) Mutually Exclusive

P(AUB)= P(A) + P(B) – P(A∩B) Not mutually exclusive

The END

Thanks !