counting principles.pdf

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