Probability

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‫العظيم‬ ‫هللا‬ ‫صدق‬
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Institute of Aviation Engineering and Technology
BAS 201 – Mathematics (3-A)

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Instructor: Dr. Khaled Ramadan Mohamed
E-mail: ramadank637@gmail.com
Lecture time: Saturday 1st slot & 2nd slot and 3rd slot

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Assessment tool Marks Percentage
Assignments, Class participation,
attendance, quizzes, and Midterm
40 40%
Final exam 60 60%
Total 100 100%
System Assessment

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Students will be familiar with the following by the end of the
course:
1. Basic Concepts of Probability Theory
2. Random Variables
3. Probability Distribution of Random Variables
4. Population and Sample

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Students will be familiar with the following by the end of
lecture one:
1. Introduction
2. Algebra of Events (Boolean algebra)
3. Axioms of Probability
4. Independence
5. Conditional Probability
6. Theorem of Total Probability

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1- Introduction
1. The origin of probability theory lies in physical
observations associated with games of chance.
2. It was found that if an "unbiased" coin is tossed
independently n times, the ratio of the number of heads to
the total number of tosses, is very likely to be very close to
1/2.
3. This is so-called "classical definition" of probability.
4. we cannot use this idea as the basis of mathematical theory
of probability.

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1- Introduction
Question?
What are the difference between { }, ( ), [ ]
( ) is a point on the plain
[ ] is an interval on x-axis
{ } is a set of infinite points

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1- Introduction
Example 1.1 Page 4
Let 𝐴 be the event that we have an odd number when rolling a
die. Find 𝑃(𝐴).
Solution
Define the sample space, which represents the set of all
possible outcomes as Ω = {1,2,3,4,5,6}
Define the event 𝐴, which represents a set of outcomes as
𝐴 = {1,3,5}
𝑃 𝐴 =
3
6
=
1
2

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1- Introduction
Example 1.2 Page 4
Let 𝐵 be the event that we have a number ≥5 when rolling a
die. Find 𝑃(𝐵).
Solution
𝐵 = {5, 6}
𝑃 𝐵 =
2
6
=
1
3

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1- Introduction
Example 1.3 Page 5
Let 𝐶 be the event that we have a sum of 7 when rolling a pair
of die. Find 𝑃(𝐶).
Solution
𝐶 = { 1,6 , 2,5 , 3,4 , 6,1 , 5,2 , (4,3)}
𝑃 𝐶 =
6
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=
1
6

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2- Algebra of Events (Boolean algebra)
➢ Now, we are interested in introducing some operations by
which new events are from old ones.
➢ These operations are "or," "and," and "not.”
Union (∪) Intersection (∩) Complement (𝐴𝑐)

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➢ Let 𝐴 and 𝐵 be events in the same sample space Ω.
➢ Define the intersection of 𝐴 and 𝐵 (denoted by 𝐴 ∩ 𝐵)
➢ Intersection as the set consisting of those points belonging to
𝐴 and 𝐵.
𝐴 ∩ 𝐵
Ω

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Example
Consider an unbiased die that tossed, let 𝐴 be the event that we
have an odd number, and 𝐵 be the event that we have a number
greater than or equal to 4. Find 𝐴 ∩ 𝐵, and 𝑃(𝐴 ∩ 𝐵)
Solution
Ω = {1,2,3,4,5,6}
𝐴 = {1,3,5}
𝐵 = {4, 5, 6}
𝐴 ∩ 𝐵 = {5}
𝑃(𝐴 ∩ 𝐵) =
1
6

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➢ Define the union of 𝐴 and 𝐵 (denoted by 𝐴 ∪ 𝐵)
➢ Union is the set consisting of those points belonging to
either 𝐴 or 𝐵 or both.
𝐴 ∪ 𝐵
Ω
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃(𝐴 ∩ 𝐵)

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Example
greater than or equal to 4. Find 𝐴 ∪ 𝐵, and 𝑃(𝐴 ∪ 𝐵)
Solution
Ω = {1,2,3,4,5,6}
𝐴 = {1,3,5}
𝐵 = {4, 5, 6}
𝐴 ∩ 𝐵 = {5}
→
𝑃(𝐴 ∩ 𝐵) = 1/6
𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵 − 𝑃 𝐴 ∩ 𝐵 =
3
6
+
3
6
−
1
6
=
5
6
𝑃(𝐴) = 3/6
→ 𝑃(𝐵) = 3/6
→

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➢ If the events 𝐴 and 𝐵 are mutually exclusive or disjoint
events.
➢ 𝐴 ∩ 𝐵 = ∅ → 𝑃 𝐴 ∩ 𝐵 =zero
➢ 𝐴 ∪ 𝐵 = 𝐴 + 𝐵 → 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵

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➢ Define the complement of 𝐴 (denoted by 𝐴𝑐
)
➢ Complement the set consisting of those points which do not
belong to 𝐴.

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Example
greater than or equal to 4. Find 𝑃 𝐴𝑐
, and 𝑃(𝐵𝑐
)
Solution
Ω = {1,2,3,4,5,6}
𝐴 = {1,3,5} & 𝐴𝑐
= {2,4,6}
𝐵 = {4, 5, 6} & 𝐵𝑐
= {1,2,3}
𝑃 𝐴𝑐
= 3/6
𝑃 𝐵𝑐
= 3/6

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➢ Define the difference of 𝐴 and 𝐵 (denoted by 𝐴 𝐵)
➢ Difference is the set consisting of those points belonging to
𝐴 and do not belong to 𝐵.
𝐴 𝐵
Ω
)
𝑃 𝐴 𝐵 = 𝑃 𝐴 − 𝑃(𝐴 ∩ 𝐵

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Example
greater than or equal to 4. Find 𝑃 𝐴 𝐵
Solution
Ω = {1,2,3,4,5,6}
𝐴 = {1,3,5}
𝐵 = {4, 5, 6}
𝐴 𝐵 = {1,3}
𝑃 𝐴 𝐵 = 2/6

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Some important properties:
1. 𝐴 ∪ 𝐵 = 𝐵 ∪ 𝐴 2. 𝐴 ∪ 𝐵 ∪ 𝐶 = (𝐴 ∪ 𝐵) ∪ 𝐶
3. 𝐴 ∩ 𝐵 = 𝐵 ∩ 𝐴 4. 𝐴 ∩ 𝐵 ∩ 𝐶 = (𝐴 ∩ 𝐵) ∩ 𝐶
5. 𝐴 ∪ 𝐴 = 𝐴 6. 𝐴 ∪ 𝐴𝑐
= Ω
7. 𝐴 ∩ 𝐴 = 𝐴 8. 𝐴 ∩ 𝐴𝑐
= ∅
9. 𝐴 ∩ Ω = 𝐴 10. 𝐴 ∪ ∅ = 𝐴
11. 𝐴 ∪ Ω = Ω 12. 𝐴 ∩ ∅ = ∅

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Example 1.4 page 8
An experiment involves choosing an integer N between 0 and 9
(the sample space consists of the integers from 0 to 9, inclusive).
Let 𝐴 = 𝑁 ≤ 5 , 𝐵 = 3 ≤ 𝑁 ≤ 7 , 𝐶 = {𝑁 is even and 𝑁 > 0}.
List the points that belong to the following events.
i. 𝐴 ∩ 𝐵 ∩ 𝐶
ii. 𝐴 ∪ 𝐵 ∩ 𝐶𝑐
iii. 𝐴 ∪ 𝐵 ∩ 𝐶𝑐
iv. (𝐴 ∩ 𝐵) ∩ 𝐴 ∪ 𝐶 𝑐

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Solution
𝐴 = 0, 1, 2, 3, 4, 5 , 𝐵 = 3, 4, 5, 6, 7 , 𝐶 = {2, 4, 6, 8}
i. 𝐴 ∩ 𝐵 ∩ 𝐶 = 0, 1, 2, 3, 4, 5 ∩ 3, 4, 5, 6, 7 ∩ 2, 4, 6, 8 = {4}
ii. 𝐴 ∪ 𝐵 ∩ 𝐶𝑐
= 0, 1, 2, 3, 4, 5 ∪ 3, 4, 5, 6, 7 ∩ 0, 1, 3, 5, 7, 9 = 0, 1, 2, 3, 4, 5 ∪ 3, 5, 7
= 0, 1, 2, 3, 4, 5, 7
iii. 𝐴 ∪ 𝐵 ∩ 𝐶𝑐
= 0, 1, 2, 3, 4, 5 ∪ 3, 4, 5, 6, 7 ∩ 0, 1, 3, 5, 7, 9 = 0, 1, 2, 3, 4, 5, 6, 7
∩ 0, 1, 3, 5, 7, 9 = {0, 1, 3, 5, 7}
iv. 𝐴 ∩ 𝐵 ∩ 𝐴 ∪ 𝐶 𝑐
= 0, 1, 2, 3, 4, 5 ∩ 3, 4, 5, 6, 7 ∩ 0, 1, 2, 3, 4, 5 ∪ 3, 4, 5, 6, 7 𝑐
𝐴 ∩ 𝐵 ∩ 𝐴 ∪ 𝐶 𝑐
= 3, 4, 5 ∩ 7, 9 = ∅

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3- Axioms of Probability
Let "𝑃" the a real-valued function defined as a class of events
of the sample space 𝑆. The probability that an event 𝐴 ⊂ Ω,
where ⊂ means a subset. This event occurs 𝑃(𝐴), if it satisfies
the following axioms:
Axiom 1: 0 ≤ 𝑃(𝐴) ≤ 1, for all events 𝐴 ⊂ Ω
Axiom 2: 𝑃 𝑆 = 1
Axiom 3: If 𝐴, 𝐵 are mutually exclusive events (i. e. 𝐴 ∩ 𝐵 = ∅),
then 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃(𝐵)

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Axiom 4: If 𝐴1, 𝐴2 , … . . , 𝐴𝑛 are a sequence of mutual
exclusive events (i. e. 𝐴𝑖 ∩ 𝐴𝑗 = ∅, for 𝑖 ≠ 𝑗), then 𝑃(
)
𝐴1 ∪ 𝐴2
∪ ⋯ . .∪ 𝐴𝑛 = 𝑃 𝐴1 + 𝑃(𝐴2) + ⋯ + 𝑃(𝐴𝑛)
Axiom 5: If 𝐴1, 𝐴2 , … . . , 𝐴𝑛 are a sequence of mutual
exclusive events (i. e. 𝐴𝑖 ∩ 𝐴𝑗 = ∅, for 𝑖 ≠ 𝑗), then 𝑃(
)
𝐴1 ∪ 𝐴2
∪ ⋯ . .∪ 𝐴𝑛 = 𝑃 𝐴1 + 𝑃(𝐴2) + ⋯ + 𝑃(𝐴𝑛)

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Basic theorems
1. 𝑃 ∅ = 0, ∅ is the impossible event
2. 𝑃 𝐴𝑐
= 1 − 𝑃(𝐴)
3. If 𝐴, 𝐵 are two events, such that 𝐴 ⊂ 𝐵, then 𝑃(𝐴) ≤ 𝑃(𝐵)
4. If 𝐴, 𝐵 are any two events, then 𝑃 𝐴 𝐵 = 𝑃 𝐴 − 𝑃(𝐴 ∩ 𝐵)
5. 𝐴 ∪ 𝐴𝑐
= Ω, then 𝑃 𝐴 + 𝑃 𝐴𝑐
= 1
6. If 𝐴, 𝐵 are disjoint events, then 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵
7. If 𝐴, 𝐵 are any two events, then 𝑃 𝐴 ∪ 𝐵 = 𝑃 𝐴 + 𝑃 𝐵
− 𝑃 𝐴 ∩ 𝐵

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Basic theorems
8. If 𝐴, 𝐵, 𝐶 are three independent events, then 𝑃 𝐴 ∩ 𝐵 ∩ 𝐶
= 𝑃 𝐴 𝑃 𝐵 𝑃(𝐶)
9. If 𝐴, 𝐵, 𝐶 are three not independent events, then
𝑃 𝐴 ∩ 𝐵 ∩ 𝐶 = 𝑃 𝐴 𝑃 𝐵|𝐴 𝑃(𝐶|𝐴 ∩ 𝐵)
10. 0 ≤ 𝑃𝑖 ≤ 1
11. σ𝑖=1
𝑛
𝑃𝑖 = 1

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Example 1.5 page 11
i. No heads appear
ii. At least one head appears
iii. Exactly two tails appear
iv. Two successive heads appear
Three fair coins are tossed, find the probability that:

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Solution
Ω={HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
i. A={no heads appear}={TTT} → P(A)=1/8
ii. B={At least one head appear}={ HHH, HHT, HTH, HTT, THH, THT, TTH }
→ P(B)=7/8
i. C={Exactly two tails appear}={HTT, THT, TTH} → P(C)=3/8
ii. D={Two successive heads appear}={HHH, HHT, THH} → P(D)=3/8

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Up to Page 11
End of lecture one.

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