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Engineering Probability and
StatisticsStatistics
Probability: Introduction and Basic
Counting Principles
Introduction
The Role of Probability in Statistics
• When you toss a single coin, you will see
either a head (H) or a tail (T). If you toss the
coin repeatedly, you will generate an infinitely coin repeatedly, you will generate an infinitely
large number of Hs and Ts – the entire
population.
Introduction
The Role of Probability in Statistics
• Now suppose you are not sure whether the
coin is fair; that is, you are not sure whether
the makeup of the population is 50-50.the makeup of the population is 50-50.
• Statistics is deterministic, while Probability is
probabilistic.
Events and Sample Space
Experiment – the process which an observation (or measurement) data is obtained through either uncontrolled events in nature or controlled situations in a laboratory.
Sample Space – the set whose elements are all the possible outcomes of an experiment.
Sample Points – elements in a sample space
Events and Sample Space
Finite Sample Space – has a finite number of
outcomes
ex. Outcomes of a single coin tossed
S = {H, T}S = {H, T}
Infinite Sample Space – has an infinite number
of outcomes
ex. Waiting time at the bus stop
Events and Sample Space
Event – a subset of the sample space.
Simple Event – an event that contains one
sample point.sample point.
Null Space {} or Empty Set Ø – has no outcomes,
cannot occur.
Set Operations
A U B = the event that occurs if A occurs or B
occurs (or both)
A ∩ B = the event that occurs if A occurs and B
occursoccurs
A’ = complement of A; the event that A does not
occur
Set Operations
Two events A and B are said to be mutually
exclusive events if they are disjoint, i.e.,
A ∩ B = Ø or {}
Example 1
Consider tossing a die and recording the number
that comes up.
Sample Space: S = {1, 2, 3, 4, 5, 6}
LetLet
A = event that odd number occurs
B = event that even number occurs
C = event that a perfect square occurs
Example 1
A = {1, 3, 5} B= {2, 4, 6} C = {1, 4}
Then
B U C = {1, 2, 4, 6}B U C = {1, 2, 4, 6}
A ∩ C = {1}
C’ = {2, 3, 5, 6}
Note that A and B are mutually exclusive, i.e.,
A ∩ B = Ø
Example 2
Two coins are tossed simultaneously. Then the
possible outcomes are
S = {HH, HT, TH, TT}
Let
A = event that a head appears on any coin
B = event that both outcomes of the coins are
the same
Example 2
A = {HH, HT, TH} B = {HH, TT}
n(A) = 3 n(B) = 2
ThenThen
A ∩ B = {HH}
A U B = S
Venn Diagram
Example:
100 students took part in the survey asking about their favorite subjects. The following are the responses:40 chose Science40 chose Science
35 chose Math
30 chose English
20 chose both Science & Math
18 chose both Science & English
15 chose both Math & English
9 chose all 3 subjects
Solution
ENTIRE SAMPLE SPACE
ScM
9
11 9
S
Sc
E
9 69
9
11
6 39TOTAL = 100
More on Example Problem
• n(S) = 100
• n(Sc U M U E) = 61
• n(Sc ∩ M ∩ E) = 9• n(Sc ∩ M ∩ E) = 9
• n(M only) = 9
• n(Sc U M U E)’ = 39
• n(Sc ∩ M ∩ E)’ = 91
Basic Counting Techniques
Listing Method
Consider the experiment that a coin and a die are
tossed simultaneously. The sample space istossed simultaneously. The sample space is
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
Basic Counting Techniques
Tree Method
Basic Counting Techniques
Multiplication Rule
“If a certain experiment can be performed in n1
ways and corresponding to each of these ways
another experiment can be performed in nanother experiment can be performed in n2
ways, then the combined experiment can be
performed in n1 • n2 ways.”
Therefore
Scoin • Sdie = (2)(6) = 12 ways
Example 1
If repetition is not allowed, (a) how many three-digit numbers can be formed from the digits 1, 2, 4, 5, and 6? (b) How many of these are odd?
A three-digit number is composed of hundred’s (102), ten’s (101), and unit’s (100) digits. Therefore,
5 • 4 • 3 = 60 three-digit numbers
102 101 100
Example 1
If we restrict ourselves only to those odd
numbers, then there are
4 • 3 • 2 = 24 three-digit numbers4 • 3 • 2 = 24 three-digit numbers
102 101 100
Permutation of Distinct Objects
Factorial Method
The number of permutations of n distinct objects is n!
n! = n•(n – 1)•... •2•1n! = n•(n – 1)•... •2•1
Example:
The letters a, b, and c are to be arranged. The possible arrangements are
abc bac cab
acb bca cba
Example
The number of ways is computed to be
3! = 6 ways
Permutation of Distinct Objects
Permutation (nPr)
The number of permutations of n objects taken r
at a time is
! P
( )!
nn r
n r=
−
Example
Consider the case if only two of the three letters
from the previous example are to be arranged,
then the possible arrangements are
ab ac bc
ba ca cb
There are 6 ways for the arrangement.
Example
By computation,
3 2
3!
(3 2)!P =
−
3 2
(3 2)!
6 waysP
−=
Permutation of Distinct Objects
Circular Permutation
The number of permutations of n distinct
objects arranged in a circle is
(n – 1)!
Permutation of Distinct Objects
Example:
Suppose that the letters a, b, and c are to be
arranged in a circular way, then the possible
arrangements arearrangements are
There are only two ways.
a ab c
c b
Example
By computation,
(n – 1) = (3 – 1)!
= 2 ways= 2 ways
Example 2
If 4 Americans, 3 Chinese and 3 Africans are to
be seated in a round table, how many seating
arrangements are possible
a. regardless of nationality?a. regardless of nationality?
b. when people of the same nationality sit next
to each other?
Example 2
Solution
a. Regardless of the nationality, the number of arrangements of the 10 people in a circle is
(10 – 1)! = 9! = 362,880 ways
b. Considering the three groups
(3 – 1)! = 2! = 2 ways(3 – 1)! = 2! = 2 ways
Considering each nationality
Americans: 4! = 24 ways
Chinese: 3! = 6 ways
Africans: 3! = 6 ways
Altogether:
2! • 4! • 3! • 3! = 1,728 ways
Permutation with Repetition
The number of distinct permutations of n
distinct objects of which n1 are of the first
kind, n2 of the second kind, ..., nk of the kth
kind iskind is
Also applicable for partitioning or groupings of
all the n objects.
1 2 1 2
!, ,... ! ! ...k k
n n
n n n n n n
= • • •
Example 1
In how many ways can the letters of the word
indeterminate be arranged?
Solution:Solution:
Since there are 13 letters in the given word and
out of these, there are some letters with
repetition.
Solution
Let n1 = number of letter “i” = 2
n2 = number of letter “n” = 2
n3 = number of letter “d” = 1
n4 = number of letter “e” = 3
n5 = number of letter “t” = 2n5 = number of letter “t” = 2
n6 = number of letter “r” = 1
n7 = number of letter “m” = 1
n8 = number of letter “a” = 1
Then 13 13!129,729,600 ways
2, 2,1,3, 2,1,1,1 2!2!1!3!2!1!1!1!
= =
Example 2
In how many ways can 10 people be assigned in
groups of 1, 2, 3, and 4 members.
Solution:Solution:
10 10!12,600 ways
1,2,3,4 1!2!3!4!
= =
Combination
The combination of n objects taken at r at a
time, where order does not count, is
!n n !( , ) or or
!( )!n r
n nC n r C
r r n r
= −
Example
A school wants to buy 6 computers for itslaboratory from a local supplier. The supplierhas 10 computers in stock, 4 of which areforeign-made.
a. Find how many ways there are to buy 6computers from the supplier.
b. Find how many ways there are to buycomputers if the school prefers 4 local and 2foreign-made computers.
Solution
a. 10C6 = 210 ways
b. (6C4)(4C2) = 90 ways
Special Cases for Permutation
1. Clustering/Grouping
Example:
Six people are seated in a row. In how many
ways can they be arranged if two of themways can they be arranged if two of them
would want to be sitting next to each other?
Solution:
5!2! = 240 ways
Special Cases for Permutation
2. Complement of Clustering/Grouping
Example:
Consider the previous example of six people
seated in a row. In how many ways can theyseated in a row. In how many ways can they
be arranged if two people don’t want to sit
next to each other?
Solution:
6! – 5!2! = 480 ways
Additive Rule for Permutation
1. Inclusive Range (using phrases “at most” & “at least”)
Example:
Consider 10 books: 3 Algebra, 4 Trigonometry, 3 Physics,to be arranged in a book shelf where only 5 books canbe placed. How many waysbe placed. How many ways
a. can one arrange books on the shelf?
b. can one arrange 2 Algebra, 2 Trigonometry, and 1 Physics book?
c. can one arrange with at most 3 Trigonometry books?
d. can one arrange with at least 2 Algebra books?
Solution:
a. 10P5 = 30,240 ways
b. 3P2·4P2·3P1 = 216 ways
c. T = 3: 4P3·6P2 = 720 ways
Additive Rule for Permutation
c. T = 3: 4P3·6P2 = 720 ways
T = 2: 4P2·6P3 = 1440 ways
T = 1: 4P1·6P4 = 1440 ways
T = 0: 4P0·6P5 = 720 ways
Total 4320 ways
d. A = 2: 3P2·7P3 = 1260 ways
A = 3: 3P3·7P2 = 252 ways
Total 1512 ways
Additive Rule for Permutation
2. Exclusive Range (using phrases “greater/more than” & “less than”)
Example:
Consider 10 books: 3 Algebra, 4 Trigonometry, 3 Physics, to be arranged in a book shelf where only 5 books can
Additive Rule for Permutation
to be arranged in a book shelf where only 5 books can be placed. How many ways
a. can one arrange with less than 3 Trigonometry books?
b. can one arrange with greater than 2 Algebra books?
c. can one arrange without Physics books?
a. T = 2: 4P2·6P3 = 1440 ways
T = 1: 4P1·6P4 = 1440 ways
T = 0: 4P0·6P5 = 720 ways
Total 3600 ways
Additive Rule for Permutation
Total 3600 ways
b. A = 3: 3P3·7P2 = 252 ways
c. P = 0: 3P0·7P5 = 2520 ways
Additive Rule for Combination
1. Inclusive Range (using phrases “at most” & “at least”)
Example:
Consider 10 students: 4 boys and 6 girls to beselected to form a team of 5 quizzers. How manyselected to form a team of 5 quizzers. How manyways
a. can a team be formed?
b. can a team be formed with 2 boys and 3 girls?
c. can a team be formed with at most 3 girls?
d. can a team be formed with at least 3 boys?
Additive Rule for Combination
Solution:
a. 10C5 = 252 ways
b. 4C2·6C3 = 120 ways
c. G = 3: 6C3·4C2 = 120 waysc. G = 3: 6C3·4C2 = 120 ways
G = 2: 6C2·4C3 = 60 ways
G = 1: 6C1·4C4 = 6 ways
G = 0: 6C0·4C5 = Ø ways (since there are only 4 boys)
Total 186 ways
Additive Rule for Combination
d. B = 3: 4C3·6C2 = 60 ways
B = 4: 4C4·6C1 = 6 ways
Total 66 ways
Additive Rule for Combination
2. Exclusive Range (using phrases “greater/more than” & “less than”)
Example:
Consider 10 students: 4 boys and 6 girls to beselected to form a team of 5 quizzers. How many
Consider 10 students: 4 boys and 6 girls to beselected to form a team of 5 quizzers. How manyways
a. can a team be formed with more than 2 boys?
b. can a team be formed with less than 5 girls?
c. can an all-boys team be formed?
Additive Rule for Combination
Solution:
a. B = 3: 4C3·6C2 = 60 ways
B = 4: 4C4·6C1 = 6 ways
Total 66 ways
b. G = 4: 6C4·4C1 = 60 waysb. G = 4: 6C4·4C1 = 60 ways
G = 3: 6C3·4C2 = 120 ways
G = 2: 6C2·4C3 = 60 ways
G = 1: 6C1·4C4 = 6 ways
G = 0: 6C0·4C5 = Ø ways
Total 246 ways
Additive Rule for Combination
c. Since there are only four boys, then the
number of ways that an all boys team can be
formed is
6C0·4C5 = Ø ways6C0·4C5 = Ø ways