Chapter 1 Probability and Statistics

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Introduction

Math 3410Statistics- chapter 1 introduction to probability

Julian Chan

Department of MathematicsWeber State University

2013
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Introduction

Objectives

Objectives:

What is probability and a statistic

Some visual tools (histograms)

Sample spaces and events

Counting

Permutations

Combinations
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Introduction

Why probability and statistics

1 Probability and statistics allows us to make inferences about apopulation. For instance if we throw a rock we can use Newtons laws ofmotion to determine exactly where the rock will land given that it waslaunched with inital velocity v0 and at a certain height h0.

2 What is not taken into account with Newtons law is any random events!

3 For instance the equipment might not work exactly the same each time,or you may have not calculated cartain constants correctly such asfriction or the initial velocity. It could also be that the constants are notknow and we wish to be able to determine them. Probability andstatistics allows us the determine an appropiate range for these values!

4 In addition given a certain radius we can calculate the probability that theobject lands close to the theoretical destination.
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Introduction

The starting point

Many of these topics are delt with in the second semester of this course(3420) however some of the topics are discussed in 3410.

The starting point of our studies will be probability theory.

The course will come back to the study of statistics once we have laid afoundation.

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Introduction

What is a probability

A quality control engineer at an integrated circuit manufacturing planttakes a samle of 100 RAM chips from the assembly line and finds that 10are defective. The company can tolerate 5% defective production in the

long run. The quality control engineer has to determine whether the longrun defective rate is within the tolerable range.

We now take the time to identify several key concepts to this process andto probability and statistics.

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Introduction

Foundations

To answer the question or to compute a prbability in general we define apopulation.

A population is the group of individuals for which the conclusions orinferences are to be drawn.

In this instance the population is all possible RAm chips that have or willbe produced.

A sample is a subset of population that is observed or used to makestatistical inferences.

In this instance the sample is the 100 ram chips taken from the assembly

line(, and here the sample size is n = 100).
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Introduction


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Introduction

Example

An electrical enineer is studying the peak demand at a power plant. It isobserved that on 80 of the 100 days randomly selected for study frompast records (of peak demand), the peak demand ocurred between 6 and

7 P.M.

We can use the LLN to compute

Introduction
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Introduction

Example

An electrical enineer is studying the peak demand at a power plant. It isobserved that on 80 of the 100 days randomly selected for study frompast records (of peak demand), the peak demand ocurred between 6 and

7 P.M.

We can use the LLN to computeP(demand between six and seven) = 80/100 = 80%.
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Introduction


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Histogram

This observational way of calculating probabilities allows us to give a

graphical way of summarizing probabilities which is called a histogram.Suppose that a car dealership offers 3 payment options to its customers.Option zero represents that fact that the customer did not use any oftheir options. We have that 20 customers did not use their options, 30customers took option 1, 40 customers took option 2, and 10 customerstook option 3. We can estimate the probability of option zero, one two,

and three as 20%, 30%, 40%, and 10%. This is summarized in thehistogram from which one can see that a histogram is equivalent togiving the probabilities.

Why a histogram?

It allows us to visualize the data!

It allows us to discuss or see the spread of the data which we will discusslater

It allows us to see the shape of the data which we will also discuss later.

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Histogram

Histogram

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Histogram

We now plot the CDF which is P(x k). In this instance k is the integer0, 1, 2, 3. The CDF has a largest value of 100%.

Obtain the values of the CDF by first noting thatP(x 0) = P(x = 0) = 20, then P(x 1) = P(x = 0) +P(x= 1) = 50,P(x2) = P(x= 0) + P(x = 1) + P(x= 2) = 90,P(x3) = P(x= 0) + P(x = 1) + P(x= 2) + P(x = 3) = 100.

Introduction
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CDF

Cummulative distribution

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Histogram

The following chart represents the total proportion of individuals in theUS who have a debt of up to the indicated amount.

How many individuals have a debt between 40k and 60k?How many individuals have a debt between 40k and 80k?

What does the histogram of this data set look like?

Introduction
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Debt

Is this s CDF or a Histogram?

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Histogram

To obtain the Histogram we successively subtract the previousprobabilities.

P(0 < x 20) = 15

P(20 < x 40) = P(0 < x 40) P(0 < x 20) = 35 15 = 20

P(40 < x 60) = P(0 < x 60) P(0 < x 40) = 53 35 = 18

P(60 < x 80) = P(0 < x 80) P(0 < x 60) = 65 53 = 12

P(80 < x 100) = P(0 < x 100) P(0 < x 80) = 85 65 = 20

P(100 < x< ) = P(0 < x < ) P(0 < x 100) = 100 85 = 15

This information allows us to make the following histogram!

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CDF

Cummulative distribution

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Organizing data

1 When data is collected from a survey or designed experiment, they mustbe organized into a manageable form. Data that is not organized isreferred to as raw data.

Introduction
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Organizing data

1 When data is collected from a survey or designed experiment, they mustbe organized into a manageable form. Data that is not organized isreferred to as raw data.

2 We often use a frequency distribution to organize data. A frequency

distribution lists each category of data and the number of occurrences foreach category of data.

3 The relative frequency distribution is the proportion of observationswithin a category and is found using the formula:

relative frequency = frequency/sum of all frequencies

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Example

1 To determine the proportion of colors in an MMs bag of candy. A bag ofMMs is randomly selected and the colors are recorded below.

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Example


2 brown, brown, yellow, red, red, red, brown, orange, blue, green, blue,brown, yellow, yellow, brown, red, red, brown, brown, brown, green, blue,

green, orange, orange, yellow, yellow, yellow, red, brown, red, brown,orange, green, red, brown, yellow, orange, red, green, yellow, yellow,brown, yellow, orange.

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Example


2 brown, brown, yellow, red, red, red, brown, orange, blue, green, blue,brown, yellow, yellow, brown, red, red, brown, brown, brown, green, blue,

green, orange, orange, yellow, yellow, yellow, red, brown, red, brown,orange, green, red, brown, yellow, orange, red, green, yellow, yellow,brown, yellow, orange.

3 To determine what colors appears most we construct a frequencydistribution.

Introduction
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Frequency distribution

Table: Frequency distribution

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Color Frequency

Brown 12

Yellow 10Red 9

Orange 6

Blue 3

Green 5

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Relative Frequency distribution

Table: Relative Frequency distribution

1

Color Relative Frequency

Brown 1245

Yellow 1045Red 9

45

Orange 645

Blue 345

Green 545

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Bar graphs

1 Now that we have the frequency and relative frequency distributions ofthe MMs we are ready to construct bar(histogram) graphs.

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Bar graphs

1 Now that we have the frequency and relative frequency distributions ofthe MMs we are ready to construct bar(histogram) graphs.

2 A bar graph (histogram) is constructed by labeling each category of data

on either the horizontal or vertical axis and the frequency or relativefrequency of the category on the other axis. Rectangles of equal widthare drawn for each category. The height of each rectangle represents thecategorys frequency or relative frequency.

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Bar graph

1

Figure: Bar graph

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Classes

1 Classes are categories into which data are grouped.

2 When a data set consists of a large number of different discrete data

values or when a data set consists of continuous data, we must createclasses by using intervals of numbers.

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Classes

1 It is imporant to determine the number of classes that you would like tohave.

2 The number of classes can affect the chape of the hisogram, and isrelated to the class width by the following formula

Class width = (upper class limit lower class limit)/number of classes

3 Where upper class limit is the largest data point and lower class limit isthe smallest.

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Hourly wage

1 We would like to determine if the hourly wage of Americans has increasedor decreased in the last year. To do this we take a random sample of 30individuals hourly wage (in dollars). The data is as follows:

2 8.2, 10.4, 11.3, 15.7, 21.2, 26.6, 28.8, 30.1, 14.4, 12, 9, 30.4, 30.7, 50,43.8, 60, 19.9, 33.3, 44.5, 21.2, 17.5, 19.9, 11.1, 11.9, 13.6, 17.7, 18.2,14.5, 24.7, 10.

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Histogram

1 The smallest data point is 8.2, and the largest is 60

2 We choose to divide our data into 5 classes thus each class width is givenby (60 8.2)/5 = 10.36.

3 Note that the number of classes can affect the shape of the histogramand best practices are discussed on page 197.

4 We first make the following frequency table:

Introduction

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Frequency distribution

Table: Frequency distribution

1

Wage Frequency

[8.2,18.56) 15

[18.56, 28.92) 7

[28.92, 39.28) 4

[39.28, 49.64) 2

[49.64, 60] 2

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Hi
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Histogram

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Figure: Histogram

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Foundations

Another was to calculate a probability of an event or outcome A is usingthe formula:

P(A) =number of ways event A can occurr

number of ways the experiment can proceed

One advantage of this formula is that is does not require an experimentor data to compute a probability.

Drawback it assumes that the possible out comes of an experiment areequally likely to occur (this is not a big deal since we can account for thisin later chapters such at 3).

Introduction

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Foundations

Before we can effectively use this method we must define a few things.

The set of all possible outcomes is called the Sample Space denoted by S.

A subset A ofS is called an event and individual elements ofS are called

sample points or experimental units.Consider two events A1 and A2 of a common sample space S. A1 and A2are mutually exclusive if and only ifA1 A2 = 0. Similarly eventsA1,A2,A3, . . . are mutually exclusive if and only ifAi Aj = 0.
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Introduction

Example


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Example

1 A home computer is tied to a mainfram computer via a telephonemodem. The home computer will dial repeatively until contact is made.Once contact has been made the dialing process will of course end. Let Cdenote contact and N denote no contact.

2

What is the sample space?3 S = C,NC,NNC,NNNC, . . .

4 Consider the events A1 = NNC,NNNC and A2 = C,NNNNC these twoevents are exclusive, but not impossible!

5 We can illistrate the sample space with the tree diagram.

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Sample


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Sample

1

Figure: Sample space

Introduction

AND GATE


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AND GATE

You can tink ofA B as an AND GATE in computer science.

Consider the sample S = 1, 2, 3, 4, 5, 6, A = 1, 2, 5, 6, and B= 2, 4, 6

You can think ofA B as the OR GATE

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AND GATE
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AND GATE

Table: AND GATE

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outcome A B AB

2,6 on on on1,5 on off off

4 off on off

3 off off off

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OR GATE
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OR GATE

Table: OR GATE

1

outcome A B AB

2,6 on on on1,5 on off on

4 off on on

3 off off off

Introduction

Complement


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Complement

1 The complement of an event A is denoted by AC or A

and is the set ofevents in S that is not in A.

2 You can think of this as the NOT GATE in computer science (ie it doesthe opposite).

3 Here AC = 3, 4 and BC = 1, 3, 5.

4 We can apply the complement to the AND GATE and the OR GATE toget the following.

Introduction

AND GATE complement
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AND GATE complement

Table: AND GATE

1

outcome A B AC BC (AB)C AC BC

2,6 on on off off off off

1,5 on off off on on on

4 off on on off on on

3 off off on on on on

Introduction

OR GATE complement
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OR GATE complement

Table: OR GATE

1

outcome A B AC BC (AB)C AC BC

2,6 on on off off off off

1,5 on off off on off off 4 off on on off off off

off

3 off off on on on on

Introduction

Demorgans laws and properties of sets
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Demorgan s laws and properties of sets

In addition to making a connection to computer science we have also

discovered DeMorgans Laws!

(DeMorgan) (A B)C = AC BC, and

(A B)C = AC BC.

Here are some other properties of sets.

(A B) C = (A C) (B C)(A B) C = (A C) (B C)

There are many other properties of sets and from time to time we willfind some of them very useful.

The relationship between boolean algebra and computer science is quite

intense.

All these relationships can be illistrated with a Venn diagram

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Venn diagram
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Venn diagram

1

Figure: DeMorgans law with a Venn diagram


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Introduction

Example


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p

Basic counting principle: consder an experiment E1 occurs in n1 ways, E2occurs in n2 ways, . . .Ek occurs in nk ways then the experiment E1 andE2 and . . . and Ek occurs in n1n2 nk ways.

If we let k= 2 this says that if we have two events or experiments then

both of them occur in n1 n2 ways.Consider the following example: A home owner doing some remodelingrequires the services of both a plummer and an electrican. If there are 9and 11 such contractors respectively then there are 9 11 = 99 ways inwhich the home owner can hire an electrican and a plummer!

Introduction

Example


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p

You decide to open a pizza shop where the customers get to make theirown pizza. They have to choose one type of dough, one cheese, one

meat, and one topping to the pizza! If there are 3 types of dough, 5cheeses, 6 meats, and 7 toppings to the pizza how many different pizzascan be made?

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Example


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p



The answer is (Im hungry now) to use the general counting principle:3 5 6 7 = 630.

What happens if you allow for an extra price to have 3 toppings and 2meats?

Consider the experiment of recording the birthday of 20 randomlyselected persons. Ignoring leap years and assuming one year is 365 dayscompute the probability that no two individuals share a birthday.

Introduction

Example


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The answer is (Im hungry now) to use the general counting principle:3 5 6 7 = 630.

What happens if you allow for an extra price to have 3 toppings and 2meats?

Consider the experiment of recording the birthday of 20 randomlyselected persons. Ignoring leap years and assuming one year is 365 dayscompute the probability that no two individuals share a birthday.

There are 365 days for which the first person can have a birthday, 364days in which the second person can have a birthday, . . . . Finally we getthat

P(A) =365 364 363 346

36520

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Example


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An arrangement ofn distinct objects is called a permutation. Thenumber of ways of ordering n distinct objects taken r at a time will bydesignated by the symbol Prn .

A permutation is an ordered subset.

A permutation of objects has a definite order to it!

We now introduce factorial notation.N! = N (N 1) (N 2) . . . 3 2 1

4! = 4 3 2 1 = 24

3! = 3 2 1 = 6

0! = 1 (definition) Think of the number of ways to permute zero objects

The birthday example was an example of a permutation.

Introduction

Example
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The names of 3 employees are to be randomly drawn WITHOUTREPLACEMENT from a bowl containing the names of 30 employees of a

small company. The first name gets first prize, the second second prize,and the third third prize. How many different winner combinations arethere if 30 employees enter?

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Example


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30 29 28 = 24360.

What role does making a first second and third have?

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Example


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30 29 28 = 24360.


Order (permutation)!Suppose that there are N boxes each of a different color and you mustplace selection of them in a room that can hold r boxes. In how manyways can you do this?

N (N 1) (N 2) (N r+ 1) = n!(nr)!

= prn. This give a way for a

formula for permuting r out ofN objects!Why was coloring important?

Introduction

Example


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The names of 3 employees are to be randomly drawn WITHOUTREPLACEMENT from a bowl containing the names of 30 employees of asmall company. The first name gets first prize, the second second prize,and the third third prize. How many different winner combinations arethere if 30 employees enter?

30 29 28 = 24360.


Order (permutation)!Suppose that there are N boxes each of a different color and you mustplace selection of them in a room that can hold r boxes. In how manyways can you do this?

N (N 1) (N 2) (N r+ 1) = n!(nr)!

= prn. This give a way for a

formula for permuting r out ofN objects!Why was coloring important?

It made it a permutation so order matters and it is not a combination (ifit was colorless it would be a combination).

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Example


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A student must take a math course, a physics course, a statistics, an artclass, an acting class and a language class in his first year. How manydifferent schedules will the student have if they choose to take 4 courses

this semester?

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Example
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A student must take a math course, a physics course, a statistics, an artclass, an acting class and a language class in his first year. How manydifferent schedules will the student have if they choose to take 4 courses

this semester?p46 = 6 5 4 3 = 360

Introduction

Example
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A password must be exactly 8 characters long and can consist of anyletter (upper or lower case) and digits. How many password combinationsare there?

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Example
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62 61 60 53 = p638 = a really big number.

What if the pass word was only 3 characters long?

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Example
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62 61 60 53 = p638 = a really big number.

What if the pass word was only 3 characters long?

The number of combinations would only be 62 61 60 = 226920 whichcould easily be hacked!!!

Introduction

Combinations
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A combination is a permutation in which order does not matter!

A combination is an unordered subset.

Consider the following example: How many ways can we select twocanadates out of 5 for a job position?

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Combinations
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A combination is a permutation in which order does not matter!

A combination is an unordered subset.

Consider the following example: How many ways can we select twocanadates out of 5 for a job position?

Here we can use the permutation formula P

2

5 = 5 4 but this gives anordering to the canadates (a first place and a second place which theproblem does not assert). In other words the ordering of the canadatesdoes not matter (in this instance) so we have to divide by the number ofdifferent orderings in this instance there are 2 possible orderings so theansewer is 5!

2!(52)!= 10.

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Example


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Let us now consider choosing r boxes out ofn, but now all the boxes areindistinguishable (they are not colored), and we would like to be able todetermine the number of different combinations of boxes.

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Example
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Let us now consider choosing r boxes out ofn, but now all the boxes areindistinguishable (they are not colored), and we would like to be able todetermine the number of different combinations of boxes.

Here there are r boxes chosen so we can again use the permutationformulat Prn, but since order does not matter it does not matter in whchway we place the r boxes we must divide by this number to get

n

r

=

n!

r!(n r)!

Introduction

Poker
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1

Figure: Example of a straight

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Poker
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1

Figure: Example of a full house

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Example
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The number of possible 5 card poker hands drawn from a deck of 52 is:

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Example
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525 = 2, 598, 960.

Compute the probability of being dealt a straight.
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Introduction

Example

Th b f ibl d k h d d f d k f i


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525 = 2, 598, 960.

Compute the probability of being dealt a straight.

A straight must start with a card that is 5 or higher which gives 10possibilities. Once the values have been decided upon there are 45 waysto decide on suit thus giving 10 45 possible ways to get a straight. Nowthe prbability of a straight is 10 45/

525 = .0039.

What is the probability of a full house?

First we piack the value of the 3 of a kind

131

= 13 then we assign suits

to those 3 cards wich can be done in

43

= 4 ways. Now pick the value

for the pair

121

= 12 then we assign suits to those 2 cards wich can be

done in 42 = 4 ways. Hence the probability of a full house is131

121

43

42

525

= .0014

Introduction

Example
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A batch consists ofn sound articles, k faulty ones. A sample ofm articlesare taken from random from the batch for inspection. It turned out thatthe first g articles from this sample are sound what is the probability thatthe next sample is sound is also?

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Example
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A batch consists ofn sound articles, k faulty ones. A sample ofm articlesare taken from random from the batch for inspection. It turned out thatthe first g articles from this sample are sound what is the probability thatthe next sample is sound is also?The total articles that remain is n + k g and the number of sound onesare n g therefore the probability is (n g)/(n + k g).

Introduction

Example
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A university warehouse has received a shipment of 25 printers, of which10 are lazer printer, and 15 are ink jet printers. If 6 of these 25 printersare to be checked by a particular technician what is the probability thatexactly 3 of the selected are lazer printers?

Introduction

Example
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A university warehouse has received a shipment of 25 printers, of which10 are lazer printer, and 15 are ink jet printers. If 6 of these 25 printersare to be checked by a particular technician what is the probability thatexactly 3 of the selected are lazer printers?

There are25

6

ways to proceeed with this experiment. The number ofways of getting exactly 3 lazer printers is the number of ways of choosing3 lazer jet and 3 inkjet which by the generalized counting principle is

153

103

hence the probability is

153

103

/

256

= .3083

Introduction

Permutation or combination(find that number)?
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A professor is to make 4 problems for students to practice permutationsand combinations. If there are to be 4 problems in how many ways cantwo problems be concerning combinations and 2 regarding permutations?A batch of 36 bolts and nuts containing 18 bolts and 18 nuts is divided atrandom into two equal parts. What is the probability that both parts willhave the same number of nuts and bolts

Four married couples are going to sit in 8 seats in a row at a concert.How many ways can they sit? In how many ways can they sit if eachcouple must sit next to eachother?Consider a lotter with 30 participants. There is a first second and thirdprize. Find the number of different ways we can find a winner for firstsecond and third?

Introduction

Solution
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Since we distinguish between permutation and combination this is acombination problem and the answer is

42

.

P=(189 )(

189 )

(3618)= .26

8! ways for them to sit (just a permutation). If they have to sit next to

eachother think of each couple as a unit. There are 4! ways to permutethe units amoung the 8 seats, and with in each unit we can rearrangehusband and wife getting 24 rearrangements so we get 4! 24.This is a permutaion since order matters and the number of ways inwhich this can occur is P30

3.

Introduction

Example

I h l 12 ffi li h i h fi i f
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In how many ways can we place 12 traffic lights into the configuration of3 green 5 reg and 4 yellow?

Introduction

Example

I h l 12 t ffi li ht i t th fi ti f
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The 3 green 5 reg and 4 yellow we think of blocks of indistinguishableobjects. We first choose the green giving us 12choose3, next we choosewith the remaining lights

95

, and finally

44

. Using the genral counting

principle or the permutation of indistinguishable objects formula weobtain

123

95

44In chemistry an element has several different states of the electron. We

are interested in determining the number of different configurations for anatom in a certain state. For this configuration there are 6 electrons andfour different states of interest.We want to determine the number of different ways in which an electronconfiguration has 2 elements in the first orbital, and 3 in the second andone in the third orbital.

Introduction

Example

I h l 12 t ffi li ht i t th fi ti f


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The 3 green 5 reg and 4 yellow we think of blocks of indistinguishableobjects. We first choose the green giving us 12choose3, next we choosewith the remaining lights

95

, and finally

44

. Using the genral counting

principle or the permutation of indistinguishable objects formula weobtain

123

95

44In chemistry an element has several different states of the electron. We

are interested in determining the number of different configurations for anatom in a certain state. For this configuration there are 6 electrons andfour different states of interest.We want to determine the number of different ways in which an electronconfiguration has 2 elements in the first orbital, and 3 in the second andone in the third orbital.We can use the permutations of indistinguishable objects or6choose2

43

11

Introduction

Homework
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Homework from chapter 1.

Introduction

Homework


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Homework from chapter 1.1,2,5,7,11,14,15,20,22,26.

Chapter 1 Probability and Statistics

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