chapter 3 probabilities..pdf

7/25/2019 Chapter 3 Probabilities..pdf

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Chapter 3

Probability

Larson/Farber 4th ed 1


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Chapter Outline

3.1 Basic Concepts of Probability

3.2 Conditional Probability and the Multiplication

Rule

3.3 The Addition Rule

3.4 Additional Topics in Probability and Counting



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Section 3.1

Basic Concepts of Probability



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Section 3.1 Objectives

Identify the sample space of a probability experiment

Identify simple events

Use the Fundamental Counting Principle

Distinguish among classical probability, empirical

probability, and subjective probability

Determine the probability of the complement of an

event Use a tree diagram and the Fundamental Counting

Principle to find probabilities



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Probability Experiments

Probability experiment An action, or trial, through which specific results (counts,

measurements, or responses) are obtained.

Sample Space

The set of all possible outcomes of a probability

experiment.

Event

Consists of one or more outcomes and is a subset of thesample space.



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Probability Experiments

Probability experiment: Roll a die

Sample space: {1, 2, 3, 4, 5, 6}

Event: {Die is even}={2, 4, 6}



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Example: Identifying the Sample Space

A probability experiment consists of tossing a coin and

then rolling a six-sided die. Describe the sample space.


Solution:There are two possible outcomes when tossing a coin:

a head (H) or a tail (T). For each of these, there are six

possible outcomes when rolling a die: 1, 2, 3, 4, 5, or

6. One way to list outcomes for actions occurring in a

sequence is to use a tree diagram.


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Solution: Identifying the Sample Space


Tree diagram:

H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6

The sample space has 12 outcomes:

{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}


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Simple Events

Simple event

An event that consists of a single outcome.

e.g. Tossing heads and rolling a 3 {H3}

An event that consists of more than one outcome is

not a simple event.

e.g. Tossing heads and rolling an even number{H2, H4, H6}



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Example: Identifying Simple Events

Determine whether the event is simple or not.

You roll a six-sided die. Event B is rolling at least a 4.


Solution:

Not simple (event B has three outcomes: rolling a 4, a 5,

or a 6)


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Fundamental Counting Principle


If one event can occur inmways and a second event

can occur in nways, the number of ways the two

events can occur in sequence is m* n. Can be extended for any number of events occurring

in sequence.



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Example: Fundamental Counting

Principle

You are purchasing a new car. The possible

manufacturers listed.

Manufacturer: Ford, GM, Honda

How many different ways can you select onemanufacturer? Use a tree diagram to check your

result.



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Principle


manufacturers, car sizes are listed.


Car size: compact, midsize

How many different ways can you select one

manufacturer, one car size? Use a tree diagram to

check your result.



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Principle


manufacturers, car sizes, and colors are listed.


Car size: compact, midsize

Color: white (W), red (R), black (B), green (G)

How many different ways can you select one

manufacturer, one car size, and one color? Use a treediagram to check your result.



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Solution: Fundamental Counting

Principle

There are three choices of manufacturers, two car sizes,

and four colors.

Using the Fundamental Counting Principle:

3 2 4 = 24 ways



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Types of Probability

Classical (theoretical) Probability

Each outcome in a sample space is equally likely.


Number of outcomes in event E( )Number of outcomes in sample space

P E


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Example: Finding Classical Probabilities

1. Event A: rolling a 3

2. Event B: rolling a 73. Event C: rolling a number less than 5


Solution:Sample space: {1, 2, 3, 4, 5, 6}

You roll a six-sided die. Find the probability of eachevent.


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Solution: Finding Classical Probabilities

1. Event A: rolling a 3 Event A = {3}


1( 3) 0.167

6P rolling a

2. Event B: rolling a 7 Event B= { } (7 is not inthe sample space)0

( 7) 06

P rolling a

3. Event C: rolling a number less than 5Event C = {1, 2, 3, 4}

4( 5) 0.667

6P rolling a number less than


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Range of Probabilities Rule

Range of probabilities rule

The probability of an eventEis between 0 and 1,

inclusive.

0 P(E) 1


[ ]0 0.5 1

Impossible Unlikely

Even

chance Likely Certain


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Complementary Events

Complement of event E

The set of all outcomes in a sample space that are not

included in eventE.

DenotedE (Eprime)

P(E) +P(E) = 1

P(E) = 1P(E)

P(E

) = 1

P(E)


E

E


21/80Copyright 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The bar graph shows the cell phone provider

for students in a class. One of these students

is chosen at random. Find the probabilitytheir provider is not AT&T.

A. 0.3

B. 0.6

C. 0.125D. 0.4

Slide 3- 21


22/80Copyright 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

The bar graph shows the cell phone provider

for students in a class. One of these students

is chosen at random. Find the probabilitytheir provider is not AT&T.

A. 0.3

B. 0.6

C. 0.125D. 0.4

Slide 3- 22


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Example: Probability of the Complement

of an Event

You survey a sample of 1000 employees at a company

and record the age of each. Find the probability of

randomly choosing an employee who is not between 25

and 34 years old.


Employee ages Frequency,f

15 to 24 54

25 to 34 366

35 to 44 233

45 to 54 180

55 to 64 125

65 and over 42

f= 1000


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Solution: Probability of the Complement

of an Event

Use empirical probability to

findP(age 25 to 34)


Employee ages Frequency,f

15 to 24 54

25 to 34 366

35 to 44 233

45 to 54 180

55 to 64 125

65 and over 42

f= 1000

366( 25 34) 0.366

1000

fP age to

n

Use the complement rule

366

( 25 34) 1 1000

6340.634

1000

P age is not to


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Example: Probability Using a Tree

Diagram

A probability experiment consists of tossing a coin and

spinning the spinner shown. The spinner is equally

likely to land on each number. Use a tree diagram to

find the probability of tossing a tail and spinning an oddnumber.



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Solution: Probability Using a Tree

Diagram

Tree Diagram:


H T

1 2 3 4 5 76 8 1 2 3 4 5 76 8

H1 H2 H3 H4 H5 H6 H7 H8 T1 T2 T3 T4 T5 T6 T7 T8

P(tossing a tail and spinning an odd number) =4 1

0.2516 4


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Example: Probability Using the


Your college identification number consists of 8 digits.

Each digit can be 0 through 9 and each digit can be

repeated. What is the probability of getting your college

identification number when randomly generating eightdigits?



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Solution: Probability Using the


Each digit can be repeated

There are 10 choices for each of the 8 digits

Using the Fundamental Counting Principle, there are

10 10 10 10 10 10 10 10

= 108 = 100,000,000 possible identification numbers

Only one of those numbers corresponds to your ID

number


1

100,000,000P(your ID number) =


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Section 3.1 Summary

Identified the sample space of a probabilityexperiment

Identified simple events

Used the Fundamental Counting Principle Distinguished among classical probability, empirical

probability, and subjective probability

Determined the probability of the complement of an

event

Used a tree diagram and the Fundamental Counting

Principle to find probabilities



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Section 3.2

Conditional Probability and the

Multiplication Rule



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Determine conditional probabilities

Distinguish between independent and dependent

events

Use the Multiplication Rule to find the probability oftwo events occurring in sequence

Use the Multiplication Rule to find conditional

probabilities



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Conditional Probability

Conditional Probability

The probability of an event occurring, given that

another event has already occurred

DenotedP(B |A) (read probability ofB, givenA)



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Example: Finding Conditional

Probabilities

Two cards are selected in sequence from a standard

deck. Find the probability that the second card is a

queen, given that the first card is a king. (Assume that

the king is not replaced.)


Solution:

Because the first card is a king and is not replaced, the

remaining deck has 51 cards, 4 of which are queens.

4( | ) (2 |1 ) 0.078

51

nd st P B A P card is a Queen card is a King


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Example: Finding Conditional

Probabilities

The table shows the results of a study in which

researchers examined a childs IQ and the presence of a

specific gene in the child. Find the probability that a

child has a high IQ, given that the child has the gene.


Gene

Present

Gene not

present Total

High IQ 33 19 52

Normal IQ 39 11 50

Total 72 30 102


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Solution: Finding Conditional

Probabilities

There are 72 children who have the gene. So, the

sample space consists of these 72 children.


33( | ) ( | ) 0.458

72P B A P high IQ gene present

Of these, 33 have a high IQ.

Gene

Present

Gene not

present TotalHigh IQ 33 19 52

Normal IQ 39 11 50

Total 72 30 102


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Independent and Dependent Events

Independent events

The occurrence of one of the events does not affect

the probability of the occurrence of the other event

P(B |A) = P(B) or P(A |B) =P(A)

Events that are not independent are dependent



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Example: Independent and Dependent

Events

1. Selecting a king from a standard deck (A), not

replacing it, and then selecting a queen from the deck

(B).


4( | ) (2 |1 )

51

nd st P B A P card is a Queen card is a King

4( ) ( ) 52P B P Queen

Dependent (the occurrence ofA changes the probability

of the occurrence ofB)

Solution:

Decide whether the events are independent or dependent.


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Example: Independent and Dependent

Events

Decide whether the events are independent or dependent.

2. Tossing a coin and getting a head (A), and then

rolling a six-sided die and obtaining a 6 (B).


1( | ) ( 6 | )

6P B A P rolling a head on coin

1( ) ( 6)

6P B P rolling a

Independent (the occurrence ofA does not change the

probability of the occurrence ofB)

Solution:


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The Multiplication Rule

Multiplication rule for the probability ofA and B

The probability that two eventsA andB will occur in

sequence is

P(A and B) = P(A) P(B| A)

For independent events the rule can be simplified to

P(A and B) = P(A) P(B)

Can be extended for any number of independentevents



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Example: Using the Multiplication Rule

Two cards are selected, without replacing the first card,

from a standard deck. Find the probability of selecting a

king and then selecting a queen.


Solution:Because the first card is not replaced, the events are

dependent.( ) ( ) ( | )

4 4

52 51

160.006

2652

P K and Q P K P Q K


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A coin is tossed and a die is rolled. Find the probability

of getting a head and then rolling a 6.


Solution:

The outcome of the coin does not affect the probabilityof rolling a 6 on the die. These two events are

independent.( 6) ( ) (6)

1 1

2 6

10.083

12

P H and P H P


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The probability that a particular knee surgery is

successful is 0.85. Find the probability that three knee

surgeries are successful.


Solution:

The probability that each knee surgery is successful is

0.85. The chance for success for one surgery is

independent of the chances for the other surgeries.

P(3 surgeries are successful) = (0.85)(0.85)(0.85)

0.614


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Find the probability that none of the three kneesurgeries is successful.


Solution:Because the probability of success for one surgery is

0.85. The probability of failure for one surgery is

10.85 = 0.15

P(none of the 3 surgeries is successful) = (0.15)(0.15)(0.15)

0.003


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Find the probability that at least one of the three kneesurgeries is successful.


Solution:

At least one means one or more. The complement to

the event at least one successful is the event none are

successful. Using the complement rule

P(at least 1 is successful) = 1P(none are successful)

10.003

= 0.997


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Section 3.3

Addition Rule



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Mutually Exclusive Events

Mutually exclusive

Two eventsA andB cannot occur at the same time


A

BA B

A andB are mutually

exclusive

A andB are not mutually

exclusive


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Example: Mutually Exclusive Events

Decide if the events are mutually exclusive.

EventA: Roll a 3 on a die.

EventB: Roll a 4 on a die.


Solution:

Mutually exclusive (The first event has one outcome, a

3. The second event also has one outcome, a 4. These

outcomes cannot occur at the same time.)


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Example: Mutually Exclusive Events

Decide if the events are mutually exclusive.

EventA: Randomly select a male student.

EventB: Randomly select a nursing major.


Solution:

Not mutually exclusive (The student can be a male

nursing major.)


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The Addition Rule

Addition rule for the probability of A or B

The probability that eventsA orB will occur is

P(A or B) = P(A) + P(B)P(A and B)

For mutually exclusive eventsA andB, the rule canbe simplified to

P(A or B) = P(A) + P(B)

Can be extended to any number of mutuallyexclusive events



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Example: Using the Addition Rule

You select a card from a standard deck. Find theprobability that the card is a 4 or an ace.


Solution:

The events are mutually exclusive (if the card is a 4, itcannot be an ace)

(4 ) (4) ( )

4 452 52

80.154

52

P or ace P P ace

4

4

44 A

A

A

A

44 other cards

Deck of 52 Cards


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You roll a die. Find the probability of rolling a numberless than 3 or rolling an odd number.


Solution:

The events are not mutually exclusive (1 is anoutcome of both events)

Odd

5

3 1

2

4 6

Less thanthree

Roll a Die


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Solution: Using the Addition Rule


( 3 )

( 3) ( ) ( 3 )2 3 1 4

0.6676 6 6 6

P less than or odd

P less than P odd P less than and odd

Odd

5

3 1

2

4 6

Less than

three

Roll a Die


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A blood bank catalogs the types of blood given by

donors during the last five days. A donor is selected at

random. Find the probability the donor has type O or

type A blood.


Type O Type A Type B Type AB Total

Rh-Positive 156 139 37 12 344

Rh-Negative 28 25 8 4 65Total 184 164 45 16 409


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The events are mutually exclusive (a donor cannot havetype O blood and type A blood)


Rh-Positive 156 139 37 12 344

Rh-Negative 28 25 8 4 65

Total 184 164 45 16 409

( ) ( ) ( )

184 164409 409

3480.851

409

P type O or type A P type O P type A


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Find the probability the donor has type B or is Rh-

negative.



Rh-Positive 156 139 37 12 344


Total 184 164 45 16 409


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Rh-Positive 156 139 37 12 344


Total 184 164 45 16 409

Solution:

The events are not mutually exclusive (a donor can have

type B blood and be Rh-negative)


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Rh-Positive 156 139 37 12 344


Total 184 164 45 16 409

( )

( ) ( ) ( )

45 65 8 102 0.249409 409 409 409

P type B or Rh neg

P type B P Rh neg P type B and Rh neg


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Section 3.3 Summary

Determined if two events are mutually exclusive

Used the Addition Rule to find the probability of two

events



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Section 3.4

Additional Topics in Probability and

Counting



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Determine the number of ways a group of objects can

be arranged in order

Determine the number of ways to choose several

objects from a group without regard to order Use the counting principles to find probabilities



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Permutations

Permutation

An ordered arrangement of objects

The number of different permutations of n distinct

objects is n! (n

factorial) n! = n(n1)(n2)(n3) 32 1

0! = 1

Examples:

6! = 654321 = 720

4! = 4321 = 24



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Example: Permutation of n Objects

The objective of a 9 x 9 Sudoku number

puzzle is to fill the grid so that each

row, each column, and each 3 x 3 grid

contain the digits 1 to 9. How manydifferent ways can the first row of a

blank 9 x 9 Sudoku grid be filled?


Solution:The number of permutations is

9!= 987654321 = 362,880 ways


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Permutations

Permutation of nobjects taken rat a time

The number of different permutations of n distinct

objects taken rat a time


!( )!

n r

nP

n r

where r n


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Example: Finding nPr

Find the number of ways of forming three-digit codes inwhich no digit is repeated.


Solution:

You need to select 3 digits from a group of 10 n = 10, r= 3

10 3

10! 10!

(10 3)! 7!

10 9 8 7 6 5 4 3 2 1

7 6 5 4 3 2 1

720 ways

P


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Example: Finding nPr

Forty-three race cars started the 2007 Daytona 500.How many ways can the cars finish first, second, and

third?


Solution:

You need to select 3 cars from a group of 43

n = 43, r= 3

43 3

43! 43!

(43 3)! 40!43 42 41

74, 046 ways

P


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Distinguishable Permutations

Distinguishable Permutations

The number of distinguishable permutations of n

objects where n1 are of one type, n2 are of another

type, and so on


1 2 3

!

! ! ! !k

n

n n n n

where n1 + n2 + n3 +

+ nk= n


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Example: Distinguishable Permutations

A building contractor is planning to develop asubdivision that consists of 6 one-story houses, 4 two-

story houses, and 2 split-level houses. In how many

distinguishable ways can the houses be arranged?


Solution:

There are 12 houses in the subdivision

n = 12, n1 = 6, n2 = 4, n3 = 2

12!6! 4! 2!

13, 860 distinguishable ways


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Combinations

Combination of nobjects taken rat a time A selection of robjects from a group of n objects

without regard to order


!( )! !

n r

nC

n r r


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Example: Combinations

A states department of transportation plans to develop anew section of interstate highway and receives 16 bids

for the project. The state plans to hire four of the

bidding companies. How many different combinations

of four companies can be selected from the 16 biddingcompanies?


Solution:

You need to select 4 companies from a group of 16

n = 16, r= 4

Order is not important


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Solution: Combinations


16 4

16!

(16 4)!4!

16!

12!4!16 15 14 13 12!

12! 4 3 2 1

1820 different combinations

C


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Example: Finding Probabilities

A student advisory board consists of 17 members. Three

members serve as the boards chair, secretary, and

webmaster. Each member is equally likely to serve any

of the positions. What is the probability of selecting at

random the three members that hold each position?



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Solution: Finding Probabilities

There is only one favorable outcome

There are

ways the three positions can be filled


17 3

17!

(17 3)!

17!17 16 15 4080

14!

P

1( 3 ) 0.0002

4080P selecting the members


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You have 11 letters consisting of one M, four Is, four

Ss, and two Ps. If the letters are randomly arranged in

order, what is the probability that the arrangement spells

the wordMississippi?



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There is only one favorable outcome

There are

distinguishable permutations of the given letters


11!34, 650

1! 4! 4! 2!

1( ) 0.00002934650

P Mississippi

11 letters with 1,4,4, and 2

like letters


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A food manufacturer is analyzing a sample of 400 corn

kernels for the presence of a toxin. In this sample, three

kernels have dangerously high levels of the toxin. If

four kernels are randomly selected from the sample,

what is the probability that exactly one kernel contains a

dangerously high level of the toxin?



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The possible number of ways of choosing one toxickernel out of three toxic kernels is

3C1 = 3

The possible number of ways of choosing three

nontoxic kernels from 397 nontoxic kernels is

397C3 = 10,349,790

Using the Multiplication Rule, the number of ways of

choosing one toxic kernel and three nontoxic kernelsis

3C1 397C3 = 3 10,349,790 3 = 31,049,370



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The number of possible ways of choosing 4 kernels

from 400 kernels is

400C4 = 1,050,739,900

The probability of selecting exactly 1 toxic kernel is


3 1 397 3

400 4

(1 )

31,049,370 0.02961,050,739,900

C CP toxic kernel

C


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Section 3.4 Summary

Determined the number of ways a group of objects

can be arranged in order

Determined the number of ways to choose several

objects from a group without regard to order Used the counting principles to find probabilities

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