Probability and

Statistics

Dr. Sondos Fadl

Dept. Information Technology,
Faculty of Computers and Information
Menoufia University

Lecture 02
CHAPTER OUTLINE

•DEFINITIONS
•SAMPLE SPACES AND EVENTS
• Random Experiments
• Sample Spaces
• Events
• Counting Techniques
 Counting Techniques

• In many of the examples in last lecture, it is easy to

determine the number of outcomes in each event.
• In more complicated examples, determining the
outcomes that comprise the sample space (or an event)
becomes more difficult.
• Instead, counts of the numbers of outcomes in the
sample space and various events are used to analyze the
random experiments. These methods are referred to as
counting techniques. Some simple rules can be used to
simplify the calculations.

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EXAMPLE

An automobile manufacturer provides vehicles equipped

with selected options. Each vehicle is ordered
➢ With or without an automatic transmission
➢ With or without air conditioning
➢ With one of three choices of a stereo system
➢ With one of four exterior colors
If the sample space consists of the set of all possible
vehicle types, what is the number of outcomes in the sample
space?
S={…}
First Rule: Multiplication

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The size of the sample space equals the number of
branches in the last level of the tree, and this quantity
equals 2 × 2 × 3 × 4 = 48.
Example1:
How many sample points are there in the sample space
when a pair of dice is thrown once?

Solution: The first die can land face-up in any one of 𝑛1

= 6 ways. For each of these 6 ways, the second die can
also land face-up in 𝑛2 = 6 ways.

Therefore, the pair of dice can land in

𝑛1∙𝑛2 = (6) ∙(6) = 36 possible ways.

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Example 2:
In the design of a casing for a gear housing, we can use four
different types of fasteners, three different bolt lengths, and
three different bolt locations.

How many different designs are possible?

Solution: From the

multiplication rule, 4×3×3=36
different designs are possible.

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Example 3:
In how many different ways can a true-false test consisting
of 10 questions be answered?

Solution: Each of the 10 questions can be chosen in two

ways, because each question is either true or false.
Therefore, the product rule shows there are:

2 × 2 × ⋯ × 2 = 210 = 1024 ways to answer the test.

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Example 4:
How many different gray codes created form bit strings of
length three are there?

Solution: Each of the seven bits can be chosen in two

ways, because each bit is either 0 or 1.
Therefore, the product rule shows there are a total of 23
= 8 different bit strings of length three.

2× 2 × 2 = 8 11
Example 5:
How many bit strings of length 5, but always end with
1's?

Solution:
b1 b2 b3 b4 b5
Count 2 2 2 2 1

There are 2 ∙ 2 ∙ 2 ∙ 2 ∙ 1 = 16 bit strings

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Example 6:
How many different Passwords are available if each
password contains a sequence of three letters followed by
four digits.

Solution:

There are 26 choices for each of the three letters and 10

choices for each of the three digits.

26·26·26·10·10·10·10 = 175,760,000 possible passwords

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 Second Rule: Permutations
Another useful calculation finds the number of ordered
sequences of the elements of a set. Consider a set of
elements, such as 𝑆 = {𝑎, 𝑏, 𝑐}.

A permutation of the elements is an ordered sequence

of the elements. For example, 𝑎𝑏𝑐, 𝑎𝑐𝑏, 𝑏𝑎𝑐, 𝑏𝑐𝑎, 𝑐𝑎𝑏,
and 𝑐𝑏𝑎 are all of the permutations of the elements of 𝑆.

3×2×1= 6

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 1. linear permutations

Ex. The number of permutations of the four letters 𝑎, 𝑏, 𝑐,

and 𝑑 will be 4! = 24.

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 ‫ﻛﺎم واﺣد وھﺣطﮭم ﻓﻲ ﻛﺗم ﻣﻛﺎن ﺧﻠﺻت‬

2. Permutations of Subsets
In some situations, we are interested in the number of
arrangements of only some of the elements of a set.

The number of permutations of subsets of 𝒓 elements

selected from a set of 𝒏 different elements is

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Example 1:
Consider a set of elements, such as 𝑆 = {𝑎, 𝑏, 𝑐, 𝑑, 𝑒}.

What is the number of different of subsets of 3 elements

selected from 𝑆 is?

Solution: 𝒓 = 𝟑, 𝒏 = 𝟓

𝑛! 5! 5×4×3×2×1
𝑃𝑛 = = = = 60
𝑛−𝑟 ! 2 ! 2×1

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Example 2:
A printed circuit board has eight different locations in
which a component can be placed. If four different
components are to be placed on the board, how many
different designs are possible?

Solution: 𝒓 = 4, 𝒏 = 8

8!
𝑃48 =8×7×6×5=
4!
= 1680 different designs are possible

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3. Permutations of Similar Objects:

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 ‫ﺑﻘﺳم ﻋﻰ ﻋدد ﻣرات اﻟﺗﻛرار‬

Example 1:
In a Statistics class, the teacher needs to have 20 students
standing in a row. Among these 20 students, there are 12
boy, and 8 girl. How many different ways can they be
arranged in a row if only their class level will be
distinguished?
Solution: 𝒏 = 𝟐𝟎 , 𝒏𝟏 = 𝟏𝟐, 𝒏𝟐 = 𝟖

𝑛! 20!
= = = 125,970
𝑛1! 𝑛2! 12! 8!

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Example 2:
A part is labelled by printing with four thick lines, three
medium lines, and two thin lines. If each ordering of the
nine lines represents a different label, how many different
labels can be generated by using this scheme?

Solution: the number of possible part labels is

𝟗!
= 𝟏𝟐𝟔𝟎
𝟒! 𝟑! 𝟐!

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 Third Rule: Combinations

Another counting problem of interest is the number of

subsets of 𝑟 elements that can be selected from a set of 𝑛
elements. Here, order is not important. These are called
combinations.

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Combinations
The number of combinations, subsets of 𝑟 elements that
𝑛
can be selected from a set of 𝑛 elements, is denoted as
𝑟
or 𝐶𝑟𝑛 and

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Example 2:
A printed circuit board has eight different locations in
which a component can be placed. If five identical
components are to be placed on the board, how many
different designs are possible?

Solution: 𝒓 = 5, 𝒏 = 8

8 8!
𝐶4 = = 56
5! 3!

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Example1:
How many possible selections of 3 balls from a box
contains 10 colored balls?

Solution: 𝒏 = 𝟏𝟎 , 𝒓 = 𝟑

𝑛 = 𝑛! 10!
= = 120
𝑟 𝑟! 𝑛 − 𝑟 ! 3! 7!

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