br7ch04 - Web4students

Understandable Statistics

Seventh Edition

By Brase and Brase

Prepared by: Lynn Smith

Gloucester County College

Chapter Four

Elementary Probability Theory

Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .

1

Probability

•

Probability is a numerical measurement of likelihood of an event.

•

The probability of any event is a number between zero and one.

•

Events with probability close to one are more likely to occur.

•

If an event has probability equal to one, the event is certain to occur.

2 Copyright (C) 2002 Houghton Mifflin Company. All rights reserved .

Probability Notation

If A represents an event,

P(A) represents the probability of A.


3

Three methods to find probabilities:

•

Intuition

•

Relative frequency

•

Equally likely outcomes


4

Intuition method based upon our level of confidence in the result

Example: I am 95% sure that I will attend the party.


5

Probability as Relative Frequency

Probability of an event = the fraction of the time that the event occurred in the past = f n where f = frequency of an event n = sample size


6

Example of Probability as Relative Frequency

If you note that 57 of the last 100 applicants for a job have been female, the probability that the next applicant is female would be:

57

100


Law of Large Numbers

In the long run, as the sample size increases and increases, the relative frequencies of outcomes get closer and closer to the theoretical (or actual) probability value.


Equally likely outcomes

No one result is expected to occur more frequently than any other.


9

Probability of an event when outcomes are equally likely = number of outcomes favorable to event total number of outcomes


10

Example of Equally Likely

Outcome Method

When rolling a die, the probability of getting a number less than three =


2



6

1

3

11

Statistical Experiment activity that results in a definite outcome


12

Sample Space set of all possible outcomes of an experiment


13

Sample Space for the rolling of an ordinary die:

1, 2, 3, 4, 5, 6


14

For the experiment of rolling an ordinary die:

•

P(even number) =

3 = 1

6 2

•

P(result less than four) = 3 = 1

6 2

•

P(not getting a two) =

5

6


15

Complement of Event A the event not A


16

Probability of a Complement

P(not A) = 1 – P(A)


17

Probability of a Complement

If the probability that it will snow today is 30%,

P(It will not snow) = 1 – P(snow) =

1 – 0.30 = 0.70


18

Probabilities of an Event and its Complement

•

Denote the probability of an event by the letter p.

•

Denote the probability of the complement of the event by the letter q.

• p + q must equal 1

• q = 1 - p


Probability

Related to Statistics

•

Probability makes statements about what will occur when samples are drawn from a known population.

•

Statistics describes how samples are to be obtained and how inferences are to be made about unknown populations.


Independent Events

The occurrence (or non-occurrence) of one event does not change the probability that the other event will occur.


21

If events A and B are independent,

P(A and B) = P(A)

•

P(B)


22

Conditional Probability

•

If events are dependent, the occurrence of one event changes the probability of the other.

• The notation P(A|B) is read “the probability of A, given B.”

•

P(A, given B) equals the probability that event A occurs, assuming that B has already occurred.


For Dependent Events:

•

P(A and B) = P(A)

•

P(B, given A)

•

P(A and B) = P(B)

•

P(A, given B)


24

The Multiplication Rules:

•

For independent events:

P(A and B) = P(A)

•

P(B)

•

For dependent events:

P(A and B) = P(A)

•

P(B, given A)

P(A and B) = P(B)

•

P(A, given B)


25

For independent events:

P(A and B) = P(A)

•

P(B)

When choosing two cards from two separate decks of cards, find the probability of getting two fives.

P(two fives) =

P(5 from first deck and 5 from second) =


1

13



1

13



1

169

26

For dependent events:

P(A and B) = P(A)

•

P(B, given A)

When choosing two cards from a deck without replacement, find the probability of getting two fives.

P(two fives) =

P(5 on first draw and 5 on second) =

4

52



3

51



12

2652



1

221


27

“And” versus “or”

•

And means both events occur together.

•

Or means that at least one of the events occur.


28

For any events A and B,

P(A or B) =

P(A) + P(B) – P(A and B)


29

When choosing a card from an ordinary deck, the probability of getting a five or a red card:

P(5 ) + P(red) – P(5 and red) =

4

52



26

52



2

52



28

52



7

13


30

When choosing a card from an ordinary deck, the probability of getting a five or a six:

P(5 ) + P(6) – P(5 and 6) =

4

52



4

52



0

52



8

52



2

13


31

For any mutually exclusive events A and B,

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


32

When rolling an ordinary die:

P(4 or 6) =

1

6



1

6



2



6

1

3


33

Survey results:

Education: Males Females Row totals

College

Graduates

54 62 116

Not College

Graduates

31 40 71

Column totals

85 102

P(male and college grad) = ?

187(Grand total)


Survey results:


College

Graduates

54 62 116

Not College

Graduates

31 40 71

Column totals

85 102 187(Grand total)

54

P(male and college grad) =

187


35

Survey results:


College

Graduates

54 62 116

Not College

Graduates

31 40 71

Column totals

85 102

P(male or college grad) = ?

187(Grand total)


Survey results:


College

Graduates

54 62 116

Not College

Graduates

31 40 71

Column totals

85 102 187(Grand total)

P(male or college grad) =

147

187


37

Survey results:


College

Graduates

54 62 116

Not College

Graduates

31 40 71

Column totals

85 102 187(Grand total)

P(male, given college grad) = ?


Survey results:


College

Graduates

54 62 116

Not College

Graduates

31 40 71

Column totals

85 102 187(Grand total)

P(male, given college grad) =

54

116


39

Counting Techniques

•

Tree Diagram

•

Multiplication Rule

•

Permutations

•

Combinations


40

Tree Diagram a method of listing outcomes of an experiment consisting of a series of activities


41

Tree diagram for the experiment of tossing two coins start

H

T

H

T

H

T


Find the number of paths without constructing the tree diagram:

Experiment of rolling two dice, one after the other and observing any of the six possible outcomes each time .

Number of paths = 6 x 6 = 36


Multiplication of Choices

If there are

n possible outcomes for event E

1 and m possible outcomes for event E

2

, then there are

n x m or nm possible outcomes for the series of events E

1 followed by E

2

.


Area Code Example

Until a few years ago a three-digit area code was designed as follows.

The first could be any digit from 2 through 9.

The second digit could be only a 0 or 1.

The last could be any digit.

How many different such area codes were possible?

8



2



10 = 160


Ordered Arrangements

In how many different ways could four items be arranged in order from first to last?

4  3  2  1 = 24


46

Factorial Notation

• n! is read "n factorial"

• n! is applied only when n is a whole number.

• n! is a product of n with each positive counting number less than n


47

Calculating Factorials

5! = 5 • 4 • 3 • 2 • 1 =

120

3! = 3 • 2 • 1 = 6


48


Definitions

1! = 1

0! = 1

49

Complete the Factorials:

4! = 24

10! = 3,628,800

6! = 720

15! = 1.3077 x 10 12


50

Permutations

A permutation is an arrangement in a particular order of a group of items.

There are to be no repetitions of items within a permutation.)


51

Listing Permutations

How many different permutations of the letters a, b, c are possible?

Solution: There are six different permutations: abc, acb, bac, bca, cab, cba.


52

Listing Permutations

How many different two-letter permutations of the letters a, b, c, d are possible?

Solution: There are twelve different permutations: ab, ac, ad, ba, ca, da, bc, bd, cb, db, cd, dc.


Permutation Formula

The number of ways to arrange in

order n distinct objects, taking them

r at a time, is:


P n , r

 n



!

 n r



!

54

Another notation for permutations: n

P r


55

Find P

7, 3

P

7 , 3



( 7

7 !



3 )!



7

4 !

!



5040

24



210


56

P

3, 3

Applying the Permutation

Formula

P

4, 2

P

30

6, 2 = __________

P

8, 3

P 210

15, 2 = _______


57

Application of Permutations

A teacher has chosen eight possible questions for an upcoming quiz. In how many different ways can five of these questions be chosen and arranged in order from #1 to #5?

Solution: P

8,5

=

8 !

3 !


= 8• 7 • 6 • 5 • 4 = 6720

58

Combinations

A combination is a grouping in no particular order of items.


59

Combination Formula

The number of combinations of n objects taken r at a time is:

C n , r



( n

 n !

r ) !

r !


60

Other notations for combinations: n

C r or





 n r 






61

Find C

9, 3

C

9 , 3



3 !

( 9

9 !



3 )!



9 !

3 !

6 !



362880

6 ( 720 )



84


62

Applying the Combination

Formula

C

5, 3

=

10

C

3, 3

=

C

6, 2

=

15

______


C

7, 3

=

C

15, 2

=

105

63

Application of Combinations

A teacher has chosen eight possible questions for an upcoming quiz. In how many different ways can five of these questions be chosen if order makes no difference?

8

Solution: C

8,5

=

5 !

3

!

!

= 56


64

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