probability - anne gloag's math page...

Chapter

Probability

3

Section 3.1 Basic Concepts of Probability

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

Section 3.1 Objectives

Probability experiment

An action, or trial, through which specific results (counts, measurements, or responses) are obtained.

Outcome

The result of a single trial in a probability experiment.

Sample Space

The set of all possible outcomes of a probability experiment.

Event

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

Probability Experiments

Probability experiment: Roll a die

Outcome: {3}

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

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

Probability Experiments

A probability experiment consists of tossing a coin and then rolling a six-sided die. Describe the sample space.

Example: Identifying 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.

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}

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}

Simple Events

Fundamental Counting Principle

If one event can occur in m ways and a second event can occur in n ways, 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.

Fundamental Counting Principle

You are purchasing a new car. The possible manufacturers, car sizes, and colors are listed.

Manufacturer: Ford, GM, Honda

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 tree diagram to check your result.

Example: 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

Solution: Fundamental Counting Principle

In a particular college class, there are male and female students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts (a) through (j) below. (Note that you can't find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.)

Let F be the event that a student is female.

Let M be the event that a student is male.

Let S be the event that a student has short hair.

Let L be the event that a student has long hair.

Exercise

a. The probability that a student does not have long hair. b. The probability that a student is male or has short hair. c. The probability that a student is a female and has long hair. d. The probability that a student is male, given that the student has long hair. e. The probability that a student has long hair, given that the student is male. f. Of all the female students, the probability that a student has short hair. g. Of all students with long hair, the probability that a student is female. h. The probability that a student is female or has long hair. i. The probability that a randomly selected student is a male student with short hair. j. The probability that a student is female.

a. P(L')=P(S) b. P(M or S) c. P(F and L) d. P(M|L) e. P(L|M) f. P(S|F) g. P(F|L) h. P(F or L) i. P(M and S) j. P(F)

Solutions

Classical (theoretical) Probability

Each outcome in a sample space is equally likely.

Types of Probability

Number of outcomes in event E( )

Number of outcomes in sample spaceP E

1. Event A: rolling a 3

2. Event B: rolling a 7

3. Event C: rolling a number less than 5

Example: Finding Classical Probabilities

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

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

event.

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

Solution: Finding Classical Probabilities

1( 3) 0.167

6P rolling a

2. Event B: rolling a 7 Event B= { } (7 is not in

the sample space) 0( 7) 0

6P rolling a

3. Event C: rolling a number less than 5

Event C = {1, 2, 3, 4} 4

( 5) 0.6676

P rolling a number less than

Empirical (statistical) Probability

Based on observations obtained from probability experiments.

Relative frequency of an event.

Types of Probability

Frequency of event E( )

Total frequency

fP E

n

1. A company is conducting a telephone survey of randomly selected individuals to get their overall impressions of the past decade (2000s). So far, 1504 people have been surveyed. What is the probability that the next person surveyed has a positive overall impression of the 2000s? (Source: Princeton Survey Research Associates International)

Example: Finding Empirical Probabilities

Response Number of times, f

Positive 406

Negative 752

Neither 316

Don’t know 30

Σf = 1504

Solution: Finding Empirical Probabilities

Response Number of times, f

Positive 406

Negative 752

Neither 316

Don’t know 30

Σf = 320

event frequency

406( ) 0.270

1504

fP positive

n

Law of Large Numbers

As an experiment is repeated over and over, the empirical probability of an event approaches the theoretical (actual) probability of the event.

http://www.statcrunch.com/app/index.php?dataid=205337

Law of Large Numbers

Range of probabilities rule

The probability of an event E is between 0 and 1, inclusive.

0 ≤ P(E) ≤ 1

Range of Probabilities Rule

[ ] 0 0.5 1

Impossible Unlikely Even

chance Likely Certain

Complement of event E The set of all outcomes in a sample space that are not included in

event E. Denoted E ′ (E prime) P(E) + P(E ′) = 1 P(E) = 1 – P(E ′)

P(E ′) = 1 – P(E)

Complementary Events

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.

Example: Probability of the Complement of an Event

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

Σf = 1000

Use empirical probability to find P(age 25 to 34)

Solution: Probability of the Complement of an Event

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

Σf = 1000

P(age 25 to 34)

f

n

366

1000 0.366

• Use the complement rule

P(age is not 25 to 34) 1366

1000

634

1000 0.634

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 odd number.

Example: Probability Using a Tree Diagram

Tree Diagram:

Solution: Probability Using a Tree Diagram

H T

1 2 3 4 5 7 6 8 1 2 3 4 5 7 6 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

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 eight digits?

Example: Probability Using the Fundamental Counting Principle

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

Solution: Probability Using the Fundamental Counting Principle

1

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

Section 3.2

Conditional Probability and the Multiplication Rule

Determine conditional probabilities

Distinguish between independent and dependent events

Use the Multiplication Rule to find the probability of two events occurring in sequence

Use the Multiplication Rule to find conditional probabilities


Conditional Probability

The probability of an event occurring, given that another event has already occurred

Denoted P(B | A) (read “probability of B, given A”)

Conditional Probability

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.)

Example: Finding Conditional Probabilities

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 ) 0.07851

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

The table shows the results of a study in which researchers examined a child’s 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.

Example: Finding Conditional Probabilities

Gene

Present

Gene not

present

Total

High IQ 33 19 52

Normal IQ 39 11 50

Total 72 30 102

There are 72 children who have the gene. So, the sample space consists of these 72 children.

Solution: Finding Conditional Probabilities

P(B | A) P(high IQ | gene present )

33

72 0.458

Of these, 33 have a high IQ.

Gene

Present

Gene not

present

Total

High IQ 33 19 52

Normal IQ 39 11 50

Total 72 30 102

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

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).

Example: Independent and Dependent Events

4( | ) (2 |1 )

51

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

4( ) ( )

52P B P Queen

Dependent (the occurrence of A changes the probability of the occurrence of B)

Solution:

Decide whether the events are independent or dependent.

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).

Example: Independent and Dependent Events

1( | ) ( 6 | )

6P B A P rolling a head on coin

1( ) ( 6)

6P B P rolling a

Independent (the occurrence of A does not change the probability of the occurrence of B)

Solution:

Multiplication rule for the probability of A and B

The probability that two events A and B 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 independent events

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.

Example: Using the Multiplication Rule

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

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 probability of 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

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

Find the probability that none of the three knee surgeries is successful.


Solution:

Because the probability of success for one surgery is

0.85. The probability of failure for one surgery is

1 – 0.85 = 0.15

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

≈ 0.003

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


Solution:

“At least one” means one or more. The complement to

the event “at least one is successful” is the event “none

are successful.” Using the complement rule

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

≈ 1 – 0.003

= 0.997

Two events are independent if the following are true:

P(A|B) = P(A)

P(B|A) = P(B)

P(A AND B) = P(A) ⋅ P(B)

Two events A and B are independent if the knowledge that one occurred does not affect the chance the other occurs.

For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent.

More on Independent Events

Sampling may be done with replacement or without replacement. With replacement: If each member of a population is replaced

after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.

Without replacement: When sampling is done without replacement, then each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.

If it is not known whether A and B are independent or dependent, assume they are dependent until you can show otherwise

With and without replacement

Section 3.3

Addition Rule

Determine if two events are mutually exclusive

Use the Addition Rule to find the probability of two events


Decide if the events are mutually exclusive.

Event A: Roll a 3 on a die.

Event B: Roll a 4 on a die.

Example: Mutually Exclusive Events

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.)

Decide if the events are mutually exclusive.

Event A: Randomly select a male student.

Event B: Randomly select a nursing major.

Example: Mutually Exclusive Events

Solution:

Not mutually exclusive (The student can be a male

nursing major.)

Addition rule for the probability of A or B

The probability that events A or B will occur is

P(A or B) = P(A) + P(B) – P(A and B)

For mutually exclusive events A and B, the rule can be simplified to

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

Can be extended to any number of mutually exclusive events

The Addition Rule

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

Example: Using the Addition Rule

Solution: The events are mutually exclusive (if the card is a 4, it cannot be an ace)

(4 ) (4) ( )

4 4

52 52

80.154

52

P or ace P P ace

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


Solution: The events are not mutually exclusive (1 is an outcome of both events)

Solution: Using the Addition Rule

( 3 )

( 3) ( ) ( 3 )

2 3 1 40.667

6 6 6 6

P less than or odd

P less than P odd P less than and odd

The frequency distribution shows the volume of sales (in dollars) and the number of months in which a sales representative reached each sales level during the past three years. If this sales pattern continues, what is the probability that the sales representative will sell between $75,000 and $124,999 next month?


Sales volume ($) Months

0–24,999 3

25,000–49,999 5

50,000–74,999 6

75,000–99,999 7

100,000–124,999 9

125,000–149,999 2

150,000–174,999 3

175,000–199,999 1

A = monthly sales between $75,000 and $99,999

B = monthly sales between $100,000 and $124,999

A and B are mutually exclusive


Sales volume ($) Months

0–24,999 3

25,000–49,999 5

50,000–74,999 6

75,000–99,999 7

100,000–124,999 9

125,000–149,999 2

150,000–174,999 3

175,000–199,999 1

( ) ( ) ( )

7 9

36 36

160.444

36

P A or B P A P B

A blood bank catalogs the types of blood, including positive or negative Rh-factor, given by donors during the last five days. A donor is selected at random. Find the probability that 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 65

Total 184 164 45 16 409


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


Rh-Positive 156 139 37 12 344


Total 184 164 45 16 409

P(type O or type A) P(type O ) P(type A)

184

409

164

409

348

409 0.851

Find the probability that the donor has type B blood or is Rh-negative.


Solution:

The events are not mutually exclusive (a donor can have

type B blood and be Rh-negative)


Rh-Positive 156 139 37 12 344


Total 184 164 45 16 409



Rh-Positive 156 139 37 12 344


Total 184 164 45 16 409

P(type B or Rh neg)

P(type B) P(Rh neg) P(type B and Rh neg)

45

409

65

409

8

409

102

409 0.249

A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.

Tree Diagrams

In an urn, there are 11 balls. Three balls are red (R) and 8 balls are blue (B). Draw two balls, one at a time, with replacement. "With replacement" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.

Larson/Farber 5th edition 62

EXAMPLE

1. List the 24 BR outcomes: B1R1, B1R2, B1R3, ...

2. Using the tree diagram, calculate P(RR).

3. Using the tree diagram, calculate P(RB OR BR).

4. Using the tree diagram, calculate P(R on 1st draw AND B on 2nd draw).

5. Using the tree diagram, calculate P(R on 2nd draw given B on 1st draw).

6. Using the tree diagram, calculate P(BB).

7. Using the tree diagram, calculate P(B on the 2nd draw given R on the first draw).

Larson/Farber 5th edition 64

Calculate

1. B1R1; B1R2; B1R3; B2R1; B2R2; B2R3; B3R1; B3R2; B3R3; B4R1; B4R2; B4R3; B5R1; B5R2; B5R3; B6R1; B6R2; B6R3; B7R1; B7R2; B7R3; B8R1; B8R2; B8R3 2. P(RR)=(3/11)⋅(3/11)=9/121 3. P(RB OR BR)=(3/11)⋅(8/11) + (8/11)⋅(3/11) = 48/121 4. P(R on 1st draw AND B on 2nd draw)=P(RB)=(3/11)⋅(8/11)=24/121 5. P(R on 2nd draw given B on 1st draw)=P(R on 2nd | B on 1st)=24/88=3/11 6. P(BB) = 64/121 7. P(B on 2nd draw | R on 1st draw) = 8/11 There are 9 + 24 outcomes that have R on the first draw (9 RR and 24 RB).

Answers

An urn has 3 red marbles and 8 blue marbles in it. Draw two marbles, one at a time, this time without replacement from the urn. "Without replacement" means that you do not put the first ball back before you select the second ball. Below is a tree diagram. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, (3/11)⋅(2/10)=6/110.

Example

1. P(RR) =

2. P(RB OR BR)=

3. P(R on 2d | B on 1st) =

4. P(R on 1st and B on 2nd) =

5. P(BB) =

6. P(B on 2nd | R on 1st) =

Calculate

1. P(RR)=(3/11)⋅(2/10)=6/110

2. P(RB or BR)=(3/11)⋅(8/10) + (8/11)(3/10) =48/110

3. P(R on 2d | B on 1st) = 310

4. P(R on 1st and B on 2nd) = P(RB) = (3/11)(8/10) = 24/110

5. P(BB) = (8/11) ⋅ (7/10)

6. The probability is 24/30

Answers

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