251solnH1 2/13/08 (Open this document in 'Page Layout' view!) 1
H. Introduction to Probability
1. Experiments and Probability
Text problems 4.1, 4.2.
2. The Venn Diagram and the Addition Rule.
Downing & Clark, pg. 96 (pg 85 in 3 rd
ed) Basics 1, Application 2, 13. H1 (H0A), Text problems 4.3, 4.4, 4.10, 4.11, 4.8!, 4.9!.[4.3,
4.4, 4.8*, 4.9*]., (4.3,4.4, 4.8*, 4.9*)., H4, H5 (H1, H2).
3. Conditional and Joint Probability, Bayes’ Rule.
Text 4.16a-c, 4.18 [4.14a-c, 4.16] (4.13a-c, 4.15). H2, H3 (H0B, H0C), D&C pg 113 (pg 103 in 3 rd ed) 14 (Note error in text - 5/6 of the people in the city support Jones, 5/9 of the people in the country support Jones), 15, 16. H6 (H3).
4. Statistical Independence.
Text 4.16d, 4.22, 4.21!, 4.24, 4.30, 4.31, 4.33 [4.14d*, 4.19*, 4.20*, 4.22*, 4.28, 4.29, 4.31, 4.68*] (4.18*, 4.19*, 4.21*, 4.26, 4.27,
4.29). H8, H9 (H5, H6). pg. 97( pg. 85 in 3 rd ed) Applications 4, 5, 8, 9, 10, 11, 44. H7(H4).
5. Review.
Sections 1 and 2 are in this document.
----------------------------------------------------------------------------------------------------
Exercise 4.1: Two coins are tossed. a) Give an example of a simple event. b) Give an example of a joint event. c) What is the complement of a head on the first toss?
Solution: According to the Instructor’s Solutions Manual
(a)
(b)
Simple events include tossing a head or tossing a tail.
Joint events include tossing two heads (HH), a head followed by a tail (HT), a tail followed by a head (TH), and two tails (TT).
(c) Tossing a tail on the first toss
Exercise 4.2: An urn contains 12 red balls and 8 white balls. One ball is to be selected. a) Give an example of a simple event. b) What is the complement of a red ball?
Solution: According to the Instructor’s Solutions Manual
(a)
(b)
Simple events include selecting a red ball or selecting a white ball.
Selecting a white ball
Downing and Clark, pg. 85, Basics 1: When do you add the probabilities of 2 events?
Solution: The Addition Rule reads exclusive (disjoint), P

A

B


0
P
, so

A
P


A
B

     
B
    
A

B

, but if A and B are mutually
. So you add probabilities to find the probability that either one of two mutually exclusive events will occur.
Downing and Clark, pg. 85, Application 2: Wethersfield has an area of 14 sq. mi., the world has an area of 200000000 sq. mi.. what is the probability that a randomly thrown meteorite will hit Wethersfield?
14
Solution:
200000000

.
00000007
Downing and Clark, pg. 85, Application 13: Suppose that your chance of a job offer at your first-choice firm is 40%, your chance of a job offer at your second-choice firm is 40% and your chance of a job offer at both is 16%, what is your probability of an offer from either firm? ( Note: a better way to state this problem is to assume that you only applied to 2 firms and ask what is the chance that you get a job.)
Solution: Let J 1 be a job offer from the first-choice firm and J 2 be a job offer from the second-choice firm. The problem says
P

J 1

J 2
 
P
P
 

     
.
40
J 1
,

P
J
 
2



.
40
.
40

and
.
40

P
.
16

J 1

J

.
64 .
2


.
16 . By the Addition Rule

251solnH1 2/13/08 (Open this document in 'Page Layout' view!)
PROBLEM H1 (Old H0A): A pair of dice is tossed once. x
Diagram for dice problems. y
1
2
3
4
5
6
1
2

3

4

5

6

7

2
3

4

5

6

7

8

3
4

5

6

7

8

9

4
5

6

7

8

9

10

5
6

7

8

9

10

11

6
7

8

9

10

11

12

Define Event A as rolling an even number (i.e. the sum of both faces is even)
Define Event B as rolling a 5.
Define Event C as a 1 on at least one of the dice.
Find the following probabilities – say what points are in the event and add their probabilities: a. P
 
, P
 
, P
 
, P
 
, P
 
, P
 
Solution: Copy and mark diagram for each part of this problem. There is 1 way to roll a two, 3 ways to roll a 4, 5 ways to roll a 6, 5 ways to roll a 8, 3 ways to roll a 10 and one way to roll a 12. 1 + 3 + 5 +5 + 3 + 1
= 18. Each of these points has a probability of
1
36
. So P
 
 18 
36
1
2
.
There are 4 ways to roll a 5. P

There are 11 points with a 1 in them.
4
36

P
 
1
9
 11
.
.
36
There are 18 ways to roll an odd number. P
 18  1 
1

P
 
36 2
P
 


36

4

 8 
1

P
 
.
36 9
There are 36 – 11 = 25 points without a 1 in them. b. P

A

B

, P

A

C

, P

B

C

, P

A

B

, P

P
C
 

B


1
,

P
 

1
P

A

B

,
 11
P

36
A


C
25

,
.
36
P

B

C

, P

A

B

Solution: Copy and mark diagram for each part of this problem.
There is no way to roll both a 5 and an even number at the same time. P

A

B


0 .
The intersection of A and C is all points that add to an even number and have a 1 in them. These are (1,
1), (1, 3), (1, 5), (3, 1) and (5, 1). P

A

C

 5
36
. This is the joint probability of A and C .
There are only two points (1, 4) and (4, 1) that can include a 1 and add to 5. P

B

C

 2
36
 1
18
.
No points in A are in B so P

A

B


P
 
 1
2
.
There is 1 way to roll a two, 3 ways to roll a 4, 4 ways to roll a 5, 5 ways to roll a 6, 5 ways to roll a 8, 3 ways to roll a 10 and one way to roll a 12. 1 + 3 + 4 + 5 +5 + 3 + 1 = 22. Each of these points has a probability of
1
36
. So P

A

B

 22
36
.
In addition to the 18 points in A , there are 6 other points with 1 in them. So P

A

C

 24 
36
2
3
.
There are 4 ways to roll a 5. But only 2 of them, (3, 2) and (2, 3) are not in C P

B

C

 13
36
.
There are 18 points that add to an even number and 14 more that add to an odd number that is not a 5.
P

A

B



18

14

36
 32
36
 8
9
.
11 of the points with 1 in them do not add to 5. P

C

B

 9
36
 1
4
.
2

251solnH1 2/13/08 (Open this document in 'Page Layout' view!) c. Show that the addition rule works for each of the four unions in b).
Solution:
P

A

B
     
A

B
P

A

C
     
A

C


 18
36

 18
36
P

B

C
     
B

C


4
36
 11
36

4
36
 11
36

0

 2
36
22
36
5
36

,
24
36
 13
36
,
,
P

A

B


P
 

P
  
A

B

 18
36
 32
36
 18
36
 32
36 d. Which of the following are mutually exclusive? A and B ? A and C ? B and C ? A and B ?
Solution: Since 5 is not an even number A and B are mutually exclusive. A and C cannot be mutually exclusive because there are some even numbers that are sums of 1 and another number. B and C are not mutually exclusive because there are two points that include a 1 and total 5. exclusive because numbers that are not 5 include many even numbers.
A and B are not mutually e. Which of the following are collectively exhaustive? A , B , and C ? B , B , and C ?
Solution: A , B , and C are not collectively exhaustive because, for example, none of these include the point (6,5). B , B , and C are collectively exhaustive unless you can think of a point that is not in one of the three events.
Exercise 4.3: The old version of this problem read as follows. Take the following contingency table
A
A
B


10
20
B
20
40


and find the probability of a) A , b) B , c) not A , d) A and B , e) A and not B , f) not A and not B , g) A or B , h) A or not B , i) not A or not B . Only a), c), d) and h) appear in this problem in the 10 th edition.
A
Solution: Take the given contingency table
A
B

10


 20
B
20
40



and add the rows and columns to get
3
A
A
B


10
 20
30
B
20
40



60
30
60
. To get the probabilities, divide by 90.
A
A
90
B

.
1111


 .
2222
.
3333
B
.
2222
.
4444



.
6667
.
3333
.
6667
1 .
0000
According to the Instructor’s Solutions Manual (edited)
(a) P
 

30/90 = 1/3 = 0.3333 (g) P

A

B


30
30

30
60
–
10
20

50
70

5

0 .
5 5
(b)
(c)
P
 

30/90 = 1/3 = 0.3333
P
 

60/90 = 2/3 = 0.6667 (h) P

A

B


(d)
(e)
90

90
–
90

90

9
7

0 .
7 7
P

A

B


(f) P

A

B


90 90 90 90 9
P

A

B


10/90 = 1/9 = 0.1111
20/90 = 2/9 = 0.2222 (i) P

A

B
40/90 = 4/9 = 0.4444 Note that



A
60
90


B

60
90
–
40
90

80
90

8
9

0 .
8 8
is the complement of

A

B

.

251solnH1 2/13/08 (Open this document in 'Page Layout' view!) 4
A
Exercise 4.4: Take the given contingency table
A
B

10


 25
B
30
35



.
Find a) P
 
b) P

A and B
 c) P

A and B
 d) P

A or B

A
Solution: Complete the table.
A total
B


10
 25
35
B
30
35



65 total
40
. To make a joint probability table, divide through by
60
100
A
100.
A total
B

.
10


 .
25
.
35
B
.
30
.
35



.
65 total
.
40
.
60
. You should be aware that this table is
1 .
00
A
A


 P
P

A
B

B


P
A

 
B P
P

B
A

B

 
A

P
 
B



 total
P
P total
The text answer book has: (a) P
1 .
00
 

60/100 = 3/5 = 0.6 (b) P

A

B


10/100 = 1/10 = 0.1
(c) P

A

B


35/100 = 7/20 = 0.35 (d)
A
60
100
B

65
100
–
35
100
But wouldn’t it be easier to just read the joint probability table and say: (a)
(c) P

A

B


.
35 (d) P

A

B


P

A

P
B

A

B


     
A

B



90
100
P
.
60
 


9
10

.
65


.
60
0 .
9
(b)
.
35

P

A

B

.
90 ?

.
10
Exercise 4.10,4.11 [4.8 in 9 th ] (Not in 8 th edition): As the result of a study at a semiconductor manufacturing facility dies were classified as to whether they had particles on them and to whether the wafer produced was good or bad. The table reads:
Quality
Condition of Die
No Particles Particles
Good
Bad
Total
320
80
400
14
36
50
Total
334
116
450
(a) Give an example of a simple event. (b) Give an example of a joint event. (c) What is the complement of a good wafer? (d) Why is ‘good wafer’ and a die ‘with particles’ a joint event? (e) If a wafer is selected at random, what is the probability that it was produced from a die with no particles? (f) If a wafer is selected at random, what is the probability that the wafer is bad? (g) If a wafer is selected at random, what is the probability that the wafer is bad and it was produced from a die with no particles? (h) If a wafer is selected at random, what is the probability that the wafer is good and it was produced from a die with no particles? (i) If a wafer is selected at random, what is the probability that the wafer is good or it was produced from a die with no particles? (j) If a wafer is selected at random, what is the probability that the wafer is bad or it was produced from a die with particles? (k) Explain the difference between the results in
(h) and (i).

251solnH1 2/13/08 (Open this document in 'Page Layout' view!) 5
Solution: Define the following events:
Y
N
“Yes, there are particles.”
“There are no particles.”
G
B
“The wafer is good.”
“The wafer is bad.”
If we use the events above, we get the table below. If we divide by 450, we get the second table.
G
B
N



320
80
400
Y
14
36



50
334
116
450
G
B
N



.
7111
 .
1778
.
8889
Y
.
0311
.
0800



.
1111 1
.
.
.
7422
2578
0000
According to the Instructor’s Solutions Manual (Edited heavily)
(a)
(b)
(c)
Give an example of a simple event - G , “A wafer is good.”
Give an example of a joint event - on the die.”
G

N , “A wafer is good and no particle was found
What is the complement of a good wafer? – , “bad wafer.”
(d) Why is ‘good wafer’ and a die ‘with particles’ a joint event? - A joint event involves occurance of 2 or more simple events like. A wafer can be a “good wafer” and have been produced by a die “with particles”.
(e) If a wafer is selected at random, what is the probability that it was produced from a die with no particles? - P
 

.
8889 or P (no particles)

400
450

0.8889
(f) If a wafer is selected at random, what is the probability that the wafer is bad? -
P
 

.
2578 or P
 bad
 
116
450

0.2578
(g) If a wafer is selected at random, what is the probability that the wafer is bad and it was produced from a die with no particles? - P

B

N


.
1778 or
P
 bad and no particles
 
80
450

0.1778
(h) If a wafer is selected at random, what is the probability that the wafer is good and it was produced from a die with no particles? - P

G

N


.
7111 or
P
 good and no particles
 
320
450

0.7111
(i)
(j)
If a wafer is selected at random, what is the probability that the wafer is good or it was produced from a die with no particles? -
P

G

N
     
G

N


.
7422

.
8889

.
7111

.
9200 or
P
 good or no particles
 
334

400

320
450 450 450

0.92
If a wafer is selected at random, what is the probability that the wafer is bad or it was produced from a die with particles? - P

B

Y
     
B

Y


251solnH1 2/13/08 (Open this document in 'Page Layout' view!) 6

.
2578

.
1111

.
0800

.
2889 or
P
 bad or particles
 
116

50

36
450 450 450

0.2889
.
(k) Explain the difference between the results in (h) and (i). - The probability of “good or no particles” includes the probability of “good and no particles”, the probability of “good and particles” and the probability of “bad and no particles”.
Exercise [4.9 in 9 th ] (Not in 8 th or 10 th edition): The table reads:
Book Airline Tickets on the
Internet?
Research Airline Tickets on the Internet? Yes No
Total
Yes
No
Total
88
20
108
124
168
292
212
188
400
Define the following events:
B “Books airline tickets on the Internet.”
B
R
“Does not book airline tickets on the Internet”
“Researches airline prices on the Internet.”
R “Does not research airline prices on the Internet.”
If we use the events above, we get the table below. If we divide by 400, we get the second table.
R
R
B



88
20
108
B
124
168
292



212
188
400
R
R
B

.
22


 .
05
.
27
B
.
.
31

42


.
73
.
.
53
47
1 .
00
According to the Instructor’s Solutions Manual (Edited heavily)
(a)
(b)
Give an example of a simple event - , “Researches ticket prices on the Internet.”
Give an example of a joint event - R

B , “Researches ticket prices on the Internet and book tickets on the Internet.”
(c)
(d)
What is the complement of “researches airline prices on the Internet”? - R ,
“Does not research ticket prices on the Internet.”
Why is R

B , “researches airline prices on the Internet” and “books airline tickets on the
Internet” a joint event? An employee can research airline ticket prices on the Internet and also book airline ticket prices on the Internet.
If a corporate travel manager is selected at random, what is the probability that he or she:
(e) Researches airline prices on the Internet? - P
 

P (Researches ticket prices on the
(f)
Internet) = 212/400 = 0.53
Books airline tickets on the Internet? - P
 

P (books tickets on the Internet) = 108/400
(g)
= 0.27
Researches airline prices on the Internet and books airline tickets on the Internet ? -
P

R

B


P (researches ticket prices on the Internet and books tickets on the Internet)
= 88/400 = 0.22.
(h) Does not research airline prices on the Internet and does not book airline tickets on the
Internet ? - P

R

B


P (does not research ticket prices on the Internet and does not book tickets on the Internet) = 168/400 = 0.42
.

251solnH1 2/13/08 (Open this document in 'Page Layout' view!) 7
(i) Researches airline prices on the Internet or books airline tickets on the Internet ? -
P
P

R
B

B

P


R
P (researches ticket prices on the Internet or books tickets on the Internet) =
P B

R


108/400 + 212/400 – 88/400 = (20+88+124)/400

.
27

.
53

.
22 = 0.58
(k)
(j) Does not research airline prices on the Internet or book airline tickets on the Internet ? -
P (does not research ticket prices on the internet or book ticket on the Internet) = 188/400
+ 292/400 – 168/400

P
    
R

B


P

R

B
 
R

B
 
R

B


=
(20+168+124)/400

.
05

.
42

.
31 = 0.78
Explain the difference between the results in (g) and (i) - The probability of

R

B

,
“researches ticket prices on the Internet or book tickets on the Internet” includes the probability of

R

B

, “researches ticket prices on the Internet and books tickets on the Internet”, the probability of

R

B

, “researches ticket prices on the Internet but does not book tickets on the Internet”, and the probability of

R

B

, “does not research ticket prices on the Internet but books tickets on the Internet”.
4.8
Exercise 4.8 and 4.9 [Not in 8 th or 9 th ]: 2000 community members were sampled with the following results:
Drives to Work
Yes
No
Total
Homeowner
824
176
1000
Renter
681
319
1000
4.9
(a)
(b)
(c)
(d)
Example of a simple event: “Is a homeowner.”
Example of a joint event: “Is a homeowner who drives to work.”
What is the complement of “Drives to Work”? “Does not drive to work.”
Why is “Drives to Work” and “Is a homeowner” a joint event? A joint event involves occurance of 2 or more simple events. A person can drive to work and is also a homeowner.
Before we go any further, make the table above into a joint probability table by dividing by 2000.
(a)
(b)
(c)
Use H for homeowner, R for renter and D for “drives to work.”
D
D
H



.
4120
 .
0880
.
5000
R
.
3405
.
1595



.
5000
.
7525
.
2475
1 .
0000
P (drives to work) = 1505/2000

P = 0.7525
P (drives to work and is a homeowner) = 824/2000
P (drives to work or is a homeowner)

P

D

H


P
P

D
D


H
P H

= 0.4120

D

H

Total
1505
495
2000
(d)
= 1505/2000 + 1000/2000 – 824/2000 = (176+824+681)/2000 = 0.8405
Explain the differences between the answers in (b) and (d). The probability of “drives to work or is a homeowner” includes the probability of “drives to work and is a homeowner”, the probability of “drives to work but is not a homeowner”, and the probability of “does not drive to work but is a homeowner”.

251solnH1 2/13/08 (Open this document in 'Page Layout' view!) 8
PROBLEM H4 (Old H1) : . Find a formula for
Solution: Remember:
P

A

B

C

P A

P
P

A
B


B
P
     
C

A

B

P A
A
A


B
C

B C D

.
and the extended Addition Rule
 
B

C
 
A

B

C

.
We can see an alternate pattern of adding and subtracting longer and longer intersections. If we guess that it continues, the next item would be:
P


A
P


A
B


B
C


C
D

         
P

P
A

A
B


P
D
B


P
 
A
C


C

D

P A
B


B
C
 

D
A


C
A
 

B
A


C
D

 
D

P B

C
 
B

D
 
C

D

If anyone really wants it, I will put a proof of this in this space.
But note that it might be easier to compute P

A

B

C

D


P




A

B

C

D





1

P

A

B

C

D

(especially if the events are independent)
PROBLEM H5 (Old H2): A firm's employees are 15% African-American, 10% Hispanic, and 3%
Asian. If 3% are both African-American and Hispanic, what percent of the employees are minority?
Solution: Using probabilities from the problem statement, let us define the following events: is African-American, P
 

.
15 ; H , Employee is Hispanic, P
 

.
10 ; A
B , Employee
, Employee is Asian,
P
 

.
03 . In addition we are told that P

B

H


.
03 . If we use the extended Addition Rule and assume that all the joint probabilities that are not given are zero, we get
P

B

H

A
       
B

H
 
B

A
 
H

A
 
B

H

.
15

.
10

.
03

.
03

0

0

0

.
25 .

A

Parts not copied ©2003 Roger Even Bove
th
Exercise 4.8: yes no
B
B yes
T



60
15
75 no
T
60
65



125
120
80
200
This is a contingency table representing the results of a survey of 200 business people to see if they had bank credit cards or travel/entertainment cards. In the rows B represents having a bank credit card and, in the columns, T represents having a travel/entertainment card.
The text asks you to a) name a simple event, b) name a joint event, c) explain what B means and why it is the complement of , d)explain why B

T is a joint event, e) find P
 
, f) find P
 
, g) find
P

B

T

, h) find P

B

T

, i) find P

B

T

and j) find
P
 
Solution: According to the Instructor’s Solutions Manual (edited)
(a) Since simple events have only one criterion specified, an example could be any one of the following:
(1) Having a bank credit card, ,
(2) Not having a bank credit card, B ,
(3) Having a travel/entertainment credit card, ,
(4) Not having a travel/entertainment credit card, T .
(b) Since joint events specify two criteria simultaneously, an example could be any one of the following:
(1) Having a bank credit card and not having a travel/entertainment credit card, B

T ,
(2) Not having a bank credit card and not having a travel/entertainment credit card,
(3) Having a bank credit card and having a travel/entertainment credit card, B

T ,
B

T ,

251solnH1 2/13/08 (Open this document in 'Page Layout' view!) 9
(4) Not having a bank credit card and having a travel/entertainment credit card B

T .
(c) B , “Not having a bank credit card” is the complement of having a bank credit card, since it involves all events other than having a bank credit card.
(d) B

T , Having a bank credit card and having a travel/entertainment credit card is a joint event
because two criteria are specified simultaneously.
(e)
 

P (has a bank credit card) = 120/200 = 3/5 = 0.6
(f) P
P B
 

(g) P

T
B
P (has a travel/entertainment credit card) = 75/200 = 3/8 = 0.375

T


P (has a bank credit card and a travel/entertainment credit card) = 60/200 = 3/10
= 0.3
(h) P

B

T


P (does not have a bank credit card and does not have a travel/entertainment
credit card) = 65/200 = 13/40 = 0.325
(i) P

B

T


P (has a bank credit card or has a travel/entertainment credit card)
(j)
=
P

120
200
B


T

75
200

–
60
200

135
200

27
40

0 .
675
P (does not have a bank credit card or has a travel/entertainment credit card)
80
=
200

75
200
–
15
200

140
200

7
10

0 .
7 yes no
Exercise [4.9 in 9 th ] : Again, we have a contingency table. yes S no S
T



1197
127
1324
T
33
143



176
1230
270
. 1500 people were
1500
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(a) surveyed about whether an internet retailer got a product to them in time for the holidays and whether they were satisfied with the experience. T represents ‘Product received in time’ and S represents ‘Satisfied with the experience.’ a) Name the simple events. b) Name the joint events. c)
What does the complement of
Find and explain P

S

T

S signify. d) Find and explain
. g) Find and explain
 
P
 
. e) Find and explain
P S

T . h) Find and explain P

S

T

P
 
. f)
. i) Find and explain P

S

T

.
(b)
Simple events: (1) received products in time for holidays; (2) did not receive products in time for holiday; (3) were satisfied with experience; (4) were not satisfied with experience.
Joint events: (1) received products in time for holidays and were satisfied with experience; (2) did not receive products in time for holidays but were satisfied with experience; (3) received products in time for holidays but were not satisfied with experience; (2) did not receive products in time for holidays and were not satisfied with experience.
"Were not satisfied with experience".
P
 

P (satisfied) = 1230/1500 = 0.82
P (received their product in time) = 1324/1500 = 0.883
P (satisfied and received product in time) = 1197/1500 = 0.798
P (satisfied and did not receive product in time) = 33/1500 = 0.022
P (satisfied or received product in time) = (127+1197+33)/1500 = 0.9047
P (not satisfied or did not receive product in time) = (127+143+33)/1500 = 0.202