Chapter 6: Section 6-2
Applications of Venn Diagrams
D. S. Malik
Creighton University, Omaha, NE
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6: Section 6-2 Applications of Venn Diagrams
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Suppose that the universal set, U, under consideration is …nite.
Suppose that U, A, and B are sets such that
U = f1, 2, 3, 4, 5, 6, 7, 8g,
A = f1, 3, 7g and B = f2, 5, 6, 8g.
Then
n (A) = 3,
n (B ) = 4,
A \ B = ∅, n (A \ B ) = 0,
A [ B = f1, 2, 3, 5, 6, 7, 8g, and n (A [ B ) = 7.
n (A [ B ) = 7 = 3 + 4
0 = n (A) + n (B )
n (A \ B ).
A0 = f2, 4, 5, 6, 8g.
n (A0 ) = 5 = 8
3 = n (U )
n (A).
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Example
Let U = Set of English alphabet. Let A = fa, e, i, o, u g and
B = fa, b, e, g , i, k, m, u, y , z g. Then
A [ B = fa, b, e, g , i, k, m, o, u, y , z g and A \ B = fa, e, i, u g.
Thus,
n (A [ B ) = 11, n (A) = 5, n (B ) = 10, and n (A \ B ) = 4.
Note that
n (A [ B ) = 11 = 5 + 10
4 = n (A) + n (B )
D. S. Malik Creighton University, Omaha, NEChapter
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n (A \ B ).
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Theorem
Let A and B be subsets of a …nite set U.
(i) n (∅) = 0
(ii) n (U ) = n (A) + n (A0 ), n (A) = n (U ) n (A0 ), and
n (A0 ) = n (U ) n (A ).
(iii) If A \ B = ∅, then
n (A [ B ) = n (A) + n (B ).
(iv)
n (A [ B ) = n (A) + n (B )
n (A \ B ).
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Example
Let A and B be subsets of U such that n (A) = 48, n (B ) = 27, and
n (A \ B ) = 15. Then
n (A [ B ) = n (A) + n (B )
n (A \ B ) = 48 + 27
15 = 60.
Example
Let A and B be subsets of U such that n (U ) = 55, n (A0 ) = 32,
n (B ) = 31, and n (A \ B ) = 9. Then
n (A) = n (U )
n (A0 ) = 55
32 = 23.
Thus,
n (A [ B ) = n (A) + n (B )
n (A \ B ) = 23 + 31
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9 = 47.
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Example
Let A and B be subsets of U such that n (A [ B ) = 74, n (A) = 58,
n (B ) = 47. Let us …nd n (A \ B ).
Suppose that n (A \ B ) = x. Then
)
)
)
)
)
n (A [ B ) = n (A) + n (B )
74 = 58 + 47 x
74 = 105 x
74 105 = x
31 = x
x = 31.
n (A \ B )
Hence,
n (A \ B ) = 31.
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Example
Let U be a universal set with subsets A and B such that n (U ) = 12,
n (A) = 7, n (B ) = 8, and n (A \ B ) = 5. Then
A
B
U
U
2
B
B-A
2
5
A
A B
A-B
5
A
3
B
(b)
(a)
From this …gure,
n (A [ B ) = n (A
B ) + n (B
A) + n (A \ B ) = 2 + 3 + 5 = 10.
Also note that n (U ) = 12 and n (A [ B ) = 10. So the number of elements
that are not in A [ B is 12 10 = 2. That is,
n ((A [ B )0 ) = 2.
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Example
In a survey of 150 people, the following information was gathered.
(i) 110 people use the Internet to pay bills,
(ii) 70 people use regular mail to pay bills, and
(iii) 40 people use both the Internet and regular mail to pay bills.
We determine the number of people who use either the Internet or regular
mail to pay their bills.
Let U denote the set of people surveyed, I denote the set of people who
use the Internet to pay bills, and M denote the set of people who use
regular mail to pay bills. Then,
n (I ) = 110, n (M ) = 70.
Because 40 people use both the Internet and regular mail to pay bills,
n (I \ M ) = 40.
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Then,
n (I ) = 110, n (M ) = 70.
Because 40 people use both the Internet and regular mail to pay bills,
n (I \ M ) = 40.
Thus,
I
M
I-M
U
(a)
M-I
U
70 40
M
M
10
40
I
I
I
30
M
(b)
From this …gure, we have
n (I [ M ) = n (I
M ) + n (M
I ) + n (I \ M ) = 70 + 30 + 40 = 140.
This implies that the number of people who use the Internet or regular
mail to pay their bills is 140.
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Venn Diagram and Three Subsets
Let A, B, and C be subsets of a universal set U. In general, the Venn
diagram containing three sets is:
A
B
C
U
Reg 5
A
Reg 2
Reg 6
Reg 1
Reg 3
Reg 4
Reg 7
C
B
Reg 8
From this region, it is clear that the region is divided into eight regions.
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Remark
As in the case of a Venn diagram of two sets, in general, when writing the
elements of various regions, we start from the inside out.
First we write the number of elements of Reg 1, which is the number of
elements in A \ B \ C . If the number of elements in A \ B \ C is not
given, then we can assume that the number of elements in A \ B \ C is,
say x.
Next, we write the elements in Reg 2, Reg 3, and Reg 4, taking into
account the number of elements in Reg 1. For example, suppose that
n (A \ B ) = 10 and n (A \ B \ C ) = 4. Then
n (Reg 1) = 4
and
n (Reg 2) = 10
4 = 6.
Finally, we write the number of elements in Reg 5, Reg 6, and Reg 7,
taking into account the number of elements in Reg 1 to Reg 4.
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Example
Let A, B, and C be subsets of a universal set U such that n (U ) = 110,
n (A) = 38, n (B ) = 23, n (C ) = 50, n (A \ B ) = 14, n (B \ C ) = 10,
n (A \ C ) = 12, and n (A \ B \ C ) = 6. Then
U
8
18
5
B
A
6
6
4
34
C
Thus,
n (A [ B [ C ) = 6 + 8 + 6 + 4 + 18 + 5 + 34 = 81
Now n (U ) = 110 and n (A [ B [ C ) = 81. So
n ((A [ B [ C )0 ) = n (U )
n (A [ B [ C ) = 110
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81 = 29.
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Example
In a survey of 100 students, at a local university, the following information
is collected. 55 students play basketball, 39 play soccer, and 47 play golf.
22 play basketball and soccer, 15 play soccer and golf, and 18 play
basketball and golf. 8 students play all three sports. Let
B = set of students who play basketball,
S = set of students who play soccer, and
G = set of students who play golf.
Then from the given information, we have
n (B ) = 55,
n (S ) = 39,
n (G ) = 47,
n (B \ S ) = 22,
n (S \ G ) = 15, n (B \ G ) = 18,
n (B \ S \ G ) = 8.
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Then from the given information, we have
n (B ) = 55,
n (S ) = 39,
n (G ) = 47,
n (B \ S ) = 22,
n (S \ G ) = 15, n (B \ G ) = 18,
n (B \ S \ G ) = 8.
Thus, we have the Venn diagram shown in the following …gure:
U
B
Reg 5
23
Reg 2
14
Reg 6
10
S
Reg 1
Reg 3 8 Reg 4
10
7
G
Reg 7
22
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Example
1
The number of students who play basketball and soccer, but not golf
is the number of players in Reg 2. Thus, the number of students who
play basketball and soccer, but not golf is 14.
2
The number of students who play basketball and golf, but not soccer
is 10.
3
The number of students who play soccer and golf, but not basketball
is 7.
4
The number of students who play basketball, but neither soccer nor
golf is the number of players in Reg 5. Thus, the number of students
who play basketball, but neither soccer nor golf is 23. This also means
that the number of students who play only basketball is 23.
5
The number of students who play soccer, but neither basketball nor
golf is 10. This also means that the number of students who play
only soccer is 10.
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Example
6. The number of students who play golf, but neither basketball nor
soccer is 22. This also means that the number of students who play
only golf is 22.
7. The number of students who play at least one of the sports is
n (B [ S [ G ). So we add the numbers in Reg 1 to Reg 7. Thus,
n (B [ S [ G ) = 8 + 14 + 10 + 7 + 23 + 10 + 22 = 94.
Thus, the number of students who play at least one of the sports is
94.
8. The number of students who do not play any of the three sports is
the number of students in Reg 8. Now,
n (Reg 8) = 100
94 = 6.
Hence, there are 6 students who do not play any of the three sports.
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Exercise: During July, a car dealer sold 100 cars. Of these, 83 cars had
either leather seats or DVD players, 48 cars had leather
seats, and 56 cars had DVD players.
1
2
3
How many cars sold had both leather seats and a DVD
player?
How many cars sold had leather seats but not DVD
players?
How many cars sold had neither leather seats nor DVD
players?
Solution: Let L denote the set of cars that had leather seats and D
denote the set of cars that had DVD players. Then,
n (L [ D ) = 83, n (L) = 48, and n (D ) = 56.
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Solution: Let L denote the set of cars that had leather seats and D
denote the set of cars that had DVD players. Then,
n (L [ D ) = 83, n (L) = 48, and n (D ) = 56.
Suppose that n (L \ D ) = x.Then
n (L
D ) = n (L)
n (L \ D ) = 48
x,
n (D
L) = n (D )
n (L \ D ) = 56
x.
and
We thus have the following …gure:
L-D
L
D
D-L
U
48 - x
L
x
56 - x
D
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Solution:
L-D
L
D
D-L
U
48 - x
x
L
56 - x
D
From this …gure,
n (L [ D ) = 48
) 83 = 104 x
) x = 104 83
) x = 21.
x + x + 56
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x
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Solution:
U
27
21
L
1
2
D
The number of cars with both leather seats and a DVD
player is
n (L \ D ) = x = 21.
The number of cars with leather seats but not DVD
players is
n (L
3
35
D ) = 48
x = 48
21 = 27.
The number of cars with neither leather seats nor DVD
players is
n ((L [ D )0 ) = n (U )
n (L [ D ) = 100
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83 = 17.
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Exercise: A campus survey of Internet use by 200 students revealed
the following information.
110 students use the Internet to download music …les.; 90
students use the Internet to download video …les; 60 students
use the Internet to download research papers to write term
papers; 55 students use the Internet to download music and
video …les; 37 students use the Internet to download music
…les and research papers; 25 students use the Internet to
download video …les and research papers; and 160 students
use the Internet to download music …les or video …les or
research papers.
1 How many students use the Internet for all three
activities?
2 How many students use the Internet for activities other
than either of these three activities?
How many students use the Internet to download music
or video …les but not research papers?
3 How many students use the Internet to download only
music …les?
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Solution: Let M be the set of students who use the Internet to
download music …les, V be set of students who use the
Internet to download music …les, and R be the set of
students who use the Internet to download research papers.
Then,
n (M ) = 110, n (V ) = 90, n (R ) = 60,
n (M \ V ) = 55, n (M \ R ) = 37, n (V \ R ) = 25,
n (M [ V [ R ) = 160.
Also n (U ) = 200. Let us suppose that n (M \ V \ R ) = x.
Then
U
18 + x
55 - x
M
10 + x
V
x
37 - x
-2 + x
25 - x
R
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Solution: Because n (M [ V [ R ) = 160, we have
18 + x + 55 x + 10 + x + x + 37
) x + 143 = 160
) x = 17.
x + 25
x
2+x
Hence, n (M \ V \ R ) = 17. So 17 students use the Internet
for all three activities.
U
35
27
38
M
V
17
20
8
15
R
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The number of students who use the Internet for activities other than
either of these three activities is:
n ((M [ V [ R )0 ) = n (U )
n (M [ V [ R ) = 200
160 = 40.
From the Venn diagram, the number of students who use the Internet
to download music or video …les but not research papers is
35 + 38 + 27 = 100.
From the Venn diagram, 35 students use the Internet to download
only music …les.
From the Venn diagram, 15 students use the Internet to download
only research papers.
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