Discrete StructuresChapter 4

Counting and ProbabilityNurul Amelina Nasharuddin

Multimedia Department

Outline

• Rules of Sum and Product• Permutations• Combinations: The Binomial Theorem• Combinations with Repetition: Distribution• Probability

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Combinations with Repetition

Example:

How many ways are there to select 4 pieces of fruits from a bowl containing apples, oranges, and pears if the order does not matter, only the type of fruit matters, and there are at least 4 pieces of each type of fruit in the bowl

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Answer• Some of the results:

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Possible Selection Representation

A A A O X X X | X |

A A A A X X X X | |

A A O P X X | X | X

P P P P | | X X X X

Answer

The number of ways to select 4 pieces of fruit = The number of ways to arrange 4 X’s and 2 |’s, which is given by

= 6! / 4!(6-4)! = C(6,4) = 15 ways.

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4

134

Combinations with Repetition

• In general, when we wish to select, with repetition, r of n distinct elements, we are considering all arrangements of r X’s and n-1 |’s and that their number is

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r

nr

nr

nr 1

)!1(!

)!1(

Combinations with Repetition

• An r-combination of a set of n elements is an unordered selection of r elements from the set, with repetition is:

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)!1(!

)!1(1),1(

nr

nr

r

nrrnrC

Example (1)

A person throwing a party wants to set out 15 assorted cans of drinks for his guests. He shops at a store that sells five different types of soft drinks. How many different selections of 15 cans can he make?

(Here n = 5, r = 15)

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Example (1)

• 4 |’s (to separate the categories of soft drinks)• 15 X’s (to represent the cans selected)

= 19! / 15!(19-15)!

= C(19,15)

= 3876 ways.

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15

1515

Example (2)

A donut shop offers 20 kinds of donuts. Assuming

that there are at least a dozen of each kind when we

enter the shop, we can select a dozen donuts in

(Here n = 20, r = 12).

= C(31, 12) = 141,120,525 ways.

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12

12012

Example (3)

A restaurant offers 4 kinds of food. In how many ways can we choose six of the food?

 

C(6 + 4 - 1, 6) = C(9, 6)

= C(9, 3)

= 9! = 84 ways.

3! 6!

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Which formula to use?

Order Relevant Order Does Not Relevant

Repetition is Allowed nk

Repetition is Not Allowed

P(n, k)

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Different ways of choosing k elements from n

k

nk 1

k

n

Counting and Probability

Discrete Probability

• The probability of an event is the likelihood that event will occur.

• “Probability 1” means that it must happen while “probability 0” means that it cannot happen

• Eg: The probability of… – “Manchester United defeat Liverpool this season” is 1– “Liverpool win the Premier League this season” is 0

• Events which may or may not occur are assigned a number between 0 and 1.

Discrete Probability

Consider the following problems:• What’s the probability of tossing a coin 3

times and getting all heads or all tails?

• What’s the probability that a list consisting of n distinct numbers will not be sorted?

Discrete Probability

• An experiment is a process that yields an outcome• A sample space is the set of all possible outcomes

of a random process• An event is an outcome or combination of

outcomes from an experiment• An event is a subset of a sample space• Examples of experiments: - Rolling a six-sided die - Tossing a coin

ExampleExperiment 1: Tossing a coin.• Sample space: S = {Head or Tail} or we could write: S =

{0, 1} where 0 represents a tail and 1 represents a head.

Experiment 2: Tossing a coin twice• S = {HH, TT, HT, TH} where some events:

– E1 = {Head},

– E2 = {Tail},

– E3 = {All heads}

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Definition of Probability• Suppose an event E can happen in r ways out of a total of n

possible equally likely ways.• Then the probability of occurrence of the event (called its

success) is denoted by

• The probability of non-occurrence of the event (called its failure) is denoted by

• Thus,

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n

rEP )(

n

r

n

rnEP

1)(

1)()( EPEP

Definition of Probabilityusing Sample Spaces

• If S is a finite sample space in which all outcomes are equally likely and E is an event in S, then the probability of E, P(E), is

• where

N(E) is the number of outcomes in E

N(S) is the total number of outcomes in S

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||

||

)(

)()(

S

E

SN

ENEP

Example (1)

What’s the probability of tossing a coin 3 times and getting all heads or all tails?Can consider set of ways of tossing coin 3 times: Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}Next, consider set of ways of tossing all heads or all tails: Event, E = {HHH, TTT}

Assuming all outcomes equally likely to occur P(E) = 2/8 = 0.25

Example (2)

• Five microprocessors are randomly selected from a lot of 1000 microprocessors among which 20 are defective. Find the probability of obtaining no defective microprocessors.

There are C(1000,5) ways to select 5 micros.There are C(980,5) ways to select 5 good micros.The prob. of obtaining no defective micros isC(980,5)/C(1000,5) = 0.904

Probability of Combinations of Events

• Theorem: Let E1 and E2 be events in the sample space S. Then

P(E1 E2) = P(E1) + P(E2) – P(E1 E2)

• Eg: What is the probability that a positive integer selected at random from the set of positive integers not greater than 100 is divisible by either 2 or 5

E1: Event that the integer selected is divisible by 2

E2: Event that the integer selected is divisible by 5

P(E1 E2) = 50/100 + 20/100 – 10/100 = 3/5

Exercise

a) If any seven digits could be used to form a telephone number, how many seven-digits telephone numbers would not have repeated digits?

b) How many seven-digit telephone numbers would have at least one repeated digit?

c) What is the probability that a randomly chosen seven-digit telephone number would have at least one repeated digit?

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Answer

a) 10 x 9 x 8 x 7 x 6 x 5 x 4 = 604800

b) [no of PN with at least one digit] = [total no of PN] – [no of PN with no repeated digit] = 107 – 604800 = 9395200

c) 9395200 / 107 = 0.93952

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Counting Elements of Sets

The Principle of Inclusion/Exclusion Rule for Two or Three Sets

If A, B, and C are finite sets, thenN(AB) = N(A) + N(B) – N(A B)

and N(ABC) = N(A) + N(B) + N(C) – N(AB) – N(AC) – N(BC) +

N(ABC)

Example (1)

• In a class of 50 college freshmen, 30 are studying BASIC, 25 studying PASCAL, and 10 are studying both.

How many freshmen are not studying either computer language?

A: set of freshmen study BASIC B: set of freshmen study PASCAL N(AB) = N(A)+N(B)-N(AB) = 30 + 25 – 10 = 45 Not studying either: 50 – 45 =5

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20

10

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Example (2)

• A professor takes a survey to determine how many students know certain computer languages. The finding is that out of a total of 50 students in the class,

30 know Java;18 know C++;26 know SQL;

9 know both Java and C++;16 know both Java and SQL;

8 know both C++ and SQL;47 know at least one of the 3 languages.

Example (2)

a. How many students know none of the three languages?

b. How many students know all three languages?c. How many students know Java and C++ but not

SQL? How many students know Java but neither C++ nor SQL

Answer:a. 50 – 47 = 3b. ?c. ?

Example (2)

• J = the set of students who know Java

• C = the set of students who know C++

• S = the set of students who know SQL

• Use Inclusion/Exclusion rule.