Probability 2013 Ch02 Combinatorial Methods

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Probability

Chapter Outline

hapter two

ombinatorial Methods

2.1 Introduction

2.2

Counting

Principle2.3 Permutations

2.4 Combinations

2.5 Stirling’s Formula

Chien-Chao Tseng


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Probability

2.1 Introduction

Recall Theorem 1.3: If

– sample space is finite and

– sample points are all equally likely,

then P( A)= N ( A)/ N (S ) Some probability problems can be solved simply by

counting.

We study combinatorial

analysis

in chapter 2.

Combinatorial analysis deals with methods of counting.

– enable us to count systematically

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Probability

2.2 Counting Principle

Theorem 2.1 (Counting Principle)

If set E contains n elements and set F contains m elements

there are nm ways in which we can choose,

• First, an

element

of

E and

• then an element of F .

Theorem 2.2 (Generalized Counting Principle)

Let ,

, . . . ,

be sets with

,

, . . . ,

elements,

respectively.

– There are × × ×∙ ∙ ∙×ways in which we can • first, choose an element of ,

• then an

element

of

, • then an element of , • . . . ,

• and

finally

an

element

of

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Probability

Examples 2.1 and 2.2

Ex. 2.1 How many outcomes are there if we throw 5 dice?

Sol:

– Let , 1 5: set of all possible outcomes of th dice

– = {1, 2, 3, 4, 5, 6}

– Total outcomes of throwing 5 dice: 6 x 6 x 6 x 6 x 6 6

Ex.

2.2

In

tossing

4

fair

dice,

P(at least one 3 among these 4 dice)?

Sol:

– Let

A be

the

event

of

at

least

one

3

among

the

4

dice – Let be the event of no 3 in tossing 4 dice

– ( ) 5 x 5 x 5 x 5

– (

) (

)/ =

5

/6

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Examples 2.3 and 2.4

Ex. 2.3 Virginia wants to give her son, Brian, 14 different

baseball cards within a 7‐day period.

– If Virginia gives Brian cards no more than once a day, in

how many way can this be done? (7x7x …x7 = 7

) Ex. 2.4 Rose has invited n friends to her birthday party.

– All attend, and each one shakes hands with everyone else

at the party exactly once,

– What is the number of handshakes?

Sol 1:

– 1 Peoples, each shakes hands with other peoples – Each handshake counted twice

– ( 1)/2

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Number of Subsets of a Set

( A): Power set of A; set of all subsets of A.

Theorem 2.3 A set with n elements has 2 subsets. Proof :

– Let , , , ⋯ , be a set with n elements. Exists a one‐to‐one correspondence between subsets of A

and sequences of 0’s and 1’s of length n:

• To

a

subset

B

of A ,

associate

a

sequence ⋯ , – where = 0 if ∉ B, and = 1 if ∈ B. • Ex. n = 3, association for the subsets of A:

– ∅: 000, {}: 100, {}: 010, …, {, }: 011, …

By Generalized Counting Principle (Thm 2.2),

number of sequences of 0’s and 1’s of length n is

2 2 ⋯ 2 2.

Number of subsets of A is 2

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Probability

Tree Diagrams

Three diagrams systematically identify all possible cases of an

experiment

Ex. 2.8 Bill and John keep playing chess until

– one of them wins 2 games in a row or 3 games altogether. – In what percent of all possible cases does the game end

because Bill wins 3 games without winning 2 in a row?

Sol:

– 10 cases total

– Bill wins 3 games without

winning 2 in a row: 1 case

10% of cases – But probability 0.1

(not equiprobable)

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Example of Tree Diagram

Ex. 2.9 Mark has $4.

– He decides to bet $1 on the flip of a fair coin 4 times.

– What is the probability that (a) he breaks even; (b) he wins

money?

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Probability

Examples of Permutations (1/2)

Ex. 2.10 3 people, Brown, Smith, and Jones, must be

scheduled for job interviews.

– In how many different orders can this be done?

Sol: 3! Ex. 2.11 2 anthropology, 4 computer science, 3 statistics, 3

biology, and 5 music books are put on a bookshelf with a

random arrangement.

– P( A): Prob. that books of the same subject are together?

Sol:

– Total number of possible arrangements: 17!

– Anthropology books first: 2! X 4! X 3! X 3! X 5! ways

– Subjects can be ordered in 5! ways

!!!!!!

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Probability

Examples (2/2)

Ex. 2.12 If 5 boys and 5 girls sit in a row in a random order,

P(no two children of the same sex sit together)?

Sol:

– 10 persons to sit in a row: 10! ways – No two of the same sex sit together:

• Boys in positions 1, 3, 5, 7, 9, and girls in 2, 4, 6, 8, 10,

or vice versa.

– 5! × 5! possibilities for each case.

Desired probability is

!!!

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Distinguishable Permutations of Alike Objects

Theorem 2.4 (Distinguishable permutations of alike objects)

The number of distinguishable permutations of n objects of k

different types is

!! ! ⋯ !

– where

are alike ,

are alike , . . . ,

are alike , and

– n = + +∙ ∙ ∙+ ,

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Probability

Examples of Permutations with Alike Objects

Ex. 2.13 How many different 10‐letter codes can be made

using 3 a’s, 4 b’s, and 3 c’s?

Sol: 10!/(3!x4!x3!)

Ex. 2.14 In how many ways can we paint 11 offices so that

– 4 of them will be painted green, 3 yellow, 2 white, and the

remaining 2 pink?

Sol: 11!/(4!x3!x2!x2!)

Ex. 2.15 A fair coin is flipped 10 times.

– P(exactly

3

heads)? Sol: 10! 3! 7!⁄ 2⁄

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─ Set of all sequences of Hs and Ts of length 10: 2 elements ─ Exactly 3 Hs:

/3!=10!/(3! x 7!)

Sec.

2.4

Unordered

arrangement

(Combination)


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2.4 Combinations

Order in which elements are arranged is immaterial

※ Definition An unordered arrangement of r objects from a set

A containing n objects (r ≤ n) is called

– an r‐element

combination

of

A ,

or

– a combination of the elements of A taken r at a time.

Number of r ‐element combinations of n objects is given by

! ! ! ! Observation: + + + ⋯ + + + = 2

– Total no. of subsets of a set with n elements: 2

– No. of subsets of size r of a set of size n: Total no. of subsets: + + ⋯ + +

+

+

+ ⋯ +

+

+

= 2

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More Observations

,

,

For any 0 r n, and

(2.4)

̶ (2.4)

can

be

proved

algebraically

or

verified

combinatorially Proof: by combinatorial argument

Consider a set of n+1 objects {, , ⋯ , , }

• Number of r ‐

element

combinations

of

this

set:

– Separate combinations into two disjoint classes:

With : ( r‐1)‐element combinations of

, ⋯ , Without :

r‐element combinations of

, ⋯ ,

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Probability

Examples of Combinations (1/3)

Ex. 2.16 In how many ways can 2 math and 3 biology books be

selected from 8 math and 6 biology books?

Sol:

Ex. 2.17 45 instructors were selected randomly to ask

whether they are happy with their teaching loads.

– Responses of 32 were negative. – If Drs. Smith, Brown, and Jones were

among those questioned.

– P(all three gave negative responses)?

Sol:

– No. of possible groups with 32 negative responses:

– 3 (negative) selected and other 29 from remaining 42:

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Examples of Combinations (2/3) Ex. 2.18 In a small town, 11 of 25 school teachers are against

abortion, 8 are for abortion, and the rest are indifferent.

– A random sample of 5 schoolteachers is selected.

(a) P(all 5 are for abortion)?

(b) P(all 5 have the same opinion)?

Ex. 2.19 In Maryland’s lottery, player pick 6 integers between

1 and 49, order of selection being irrelevant.

– P(grand prize)?, P(2nd prize)?, P(3rd prize)?

Sol:

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Examples of Combinations (3/3) Ex. 2.20 7 cards are drawn from 52 without replacement.

– P(at least one of the cards is a king)?

Sol:

– No kings

– Desired probability: 1 No kings 1

Ex. 2.21 5 cards are drawn from 52. – P(full house)

Sol:

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Binomial Expansion (1/2)

Theorem 2.5 (Binomial Expansion)

– For any integer n ≥ 0 ,

Proof:

– Observations:

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iinn

i

n y x

i

n y x

0

)(


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Binomial Expansion (2/2)

– Expansion of

– Results in only terms of the form , 0 ≤ i ≤ n

– appears

times

Because

emerges only whenever – x’s of n−i of the n factors of ( x + y) multiplied by

– y’s of the remaining i factors of ( x + y).

Hence

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Examples (Binomial Expansion) (1/4)

Ex. 2.25 What is the coefficient of x2y3 in the expansion of

2 3?

Sol:

– Let 2 and 3

2 3=

– In the expansion of

• Coefficient of : • = 2 ∙ 3

In the expansion of 2 3, coefficient of : 2

∙ 3

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Example (Binomial Expansion) (2/4)

Ex. 2.26 Evaluate the sum

Sol: Two approaches

By set theory, total no. of subsets of a set of n elements: 2 – A set of n elements has

, 0 ≤ i ≤ n, subsets with i

elements.

Total no. of subsets of a set of n elements:

By binomial expansion, = ∑ – Let x = y = 1,

11=2= ∑ 11 =∑

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Ex. 2.27 Evaluate the sum

Sol:

So

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Ex. 2.28 Prove that

Sol:

– Let , , ⋯ , and , , ⋯ , be two disjoint sets, each with n elements .

Number of subsets with n elements of A∪B: . – Any subset with n elements of A∪B is the union of

• a subset with i elements of A and

• a subset with n − i elements of B, for some 0 ≤ i ≤ n.

For each i:

subsets,

Number of subsets with n elements of A∪B:∑

∑

∑

we have the identity

∑

.

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Multinomial Expansion

Ex. 2.30 Distribute n distinguishable balls into k distinguishable cells so that

• n1 balls are distributed into the first cell,

• n2 balls

into

the

second cell,

…,

• nk balls into the k th cell, where n1 + n2 +…+ nk =n.

– How many possible ways?

Sol:

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!!...!

!

21 k nnn

n

Sol:


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Theorem 2.6 Multinomial Expansion

Theorem 2.6 (Multinomial Expansion) In the expansion of

– The coefficient of the term

» where

• is

Therefore,

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k

k

nk

nn

nnnn k

x x xnnn

n

21

21

21... 21 !...!!

!

nk x x x )( 21

nk x x x )( 21

k n

k

nn x x x 21

21

nnnn k ...21

!...!!

!

21 k nnn

n


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2.5 Stirling’s Formula

Stirling’s formula can be used to estimate ! Theorem 2.7 (Stirling’s Formula)

Stirling’s formula usually gives excellent approximations in

numerical computations Note:

– !/ 2 becomes 1 at ∞,

– But, it is still close to 1, even for very small values of n.

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Probability 2013 Ch02 Combinatorial Methods