axiom of probability - uj.edu.sa sample space and events sample space: it is a set whose elements...
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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