7.2 Applications of Venn Diagrams


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1 7.2 Applications of Venn Diagrams In this section, we will see how we can use Venn diagrams to understand or interpret data that may be obtained from real life situations. As in the previous section, we have different regions representing different sets. Let us first consider the following Venn diagrams, where we can see the different regions into which the universal set (represented by the rectangular region) is partitioned by the sets under consideration. In the figure above, we see that a single set, A divides the universal set into 2 regions. Region 1 represents the set of those elements of U which are outside A (that is, the elements of A ). Region 2 represents those elements which are in A. In the above figure, we see that the two sets A and B divide U into 4 regions. The above Venn diagram shows that, if the two sets A and B are disjoint, then they divide the universal set, U into 3 regions. Fall 2010 Page 1 Penn State University
2 If A B, then the sets A and B divide the universal set into 3 regions as shown in the above Venn diagram. Remark. The numbering of the regions in the above Venn diagrams is completely arbitrary. Example 1. Draw Venn diagrams and shade the regions representing each set. (a) A B Solution. Set A contains all the elements outside set A. As labeled in the figure below, A is represented by the regions 1 and 4. Set B is represented by the regions 3 and 4. The intersection of the sets A and B, A B, is given by the region(s) common to A and B. So the set A B is represented by the region 4. (b) A B Solution. Again set A is represented by the regions 1 and 4. The set B is represented by the regions 1 and 2. To find A B, we need to identify the region that represents the set of all elements in A, B, or both. The result, which is shaded in the following figure, includes the regions 1, 2 and 4. Remark. In addition to the fact that region 4 in the Venn diagrams in Example 1 represents A B, notice that region 1 represents A B, region 2 represents A B, and region 3 is A B. Venn diagrams can also be drawn with three sets inside U. These three sets divide the universal set into 8 regions as shown in the following figure. Fall 2010 Page 2 Penn State University
3 Example 2. In a Venn diagram, shade the region that represents A (B C ). Solution. We first need to find B C. From the following figure, we see that B is represented by the regions 3, 4, 6, and 7, and C is represented by the regions 1, 2, 3, and 4. The overlap of these regions (regions 3 and 4) represents B C. Set A is represented by the regions 1, 4, 7, and 8. The union of the set represented by the regions 3 and 4 and the set represented by the regions 1, 4, 7, and 8 is the set represented by the regions 1, 3, 4, 7, and 8. We can now use Venn diagrams to solve some real life problems. Example 3. A researcher collecting data on 100 households finds that 21 have a DVD player; 56 have a videocassette recorder (VCR); and 12 have both. The researcher wants to answer the following questions. (a) How many do not have a VCR? (b) How many have neither a VCR nor a DVD player? (c) How many have a DVD player but not a VCR? Solution. We will use a Venn diagram as shown in the following figure to help sort out the information. We put the number 12 in the region common to both a VCR and a DVD player, because 12 households have both. Of the 21 with a DVD player, = 9 have no Fall 2010 Page 3 Penn State University
4 VCR. So we put 9 in the region for a DVD but no VCR. Similarly, = 44 households have a VCR but not a DVD player, so we put 44 in that region. Finally, the diagram shows that = 35 households have neither a VCR nor a DVD player. Now we can answer the questions: (a) = 44 do not have a VCR. (b) 35 have neither. (c) 9 have a DVD player but not a VCR. Example 4. A survey of 77 freshman business students at a large university produced the following results. 25 of the students read Business Week; 19 read The Wall Street Journal; 27 do not read Fortune; 11 read Business Week but not The Wall Street Journal; 11 read The Wall Street Journal and Fortune; 13 read Business Week and Fortune; 9 read all three. Use this information to answer the following questions: (a) How many students read none of the publications? (b) How many read only Fortune? (c) How many students read Business Week and The Wall Street Journal, but not Fortune? Solution. Since 9 students read all three publications, we begin by placing 9 in the area that belong to all the three regions as shown in the following figure. Of the 13 students who read Business Week and Fortune, 9 also read The Wall Street Journal. Therefore, only 13 9 = 4 read just Business Week and Fortune. We place 4 in the area that is common only to Business Week and Fortune readers. In the same way, we place 11 9 = 2 in the region common only to Fortune and The Wall Street Journal readers. Of the 11 students who read Business Week but not The Wall Street Journal, 4 read Fortune, so we place 11 4 = 7 in the region for those who read only Business Week. The data shows that 25 students read Business Week. However, = 20 readers have already been placed in the region representing Business Week. The balance of this Fall 2010 Page 4 Penn State University
5 region contain only = 5. These students read Business Week and The Wall Street Journal but not Fortune. In the same way, 19 ( ) = 3 students read only The Wall Street Journal. Using the fact that 27 of the 77 students do not read Fortune, we know that 50 do read Fortune. We already have = 15 students in the region. representing Fortune, leaving = 35 who read only Fortune. A total of = 65 students have been placed in the three circles. Since 77 students were surveyed, = 12 students read none of the three publications, and 12 is placed outside all three regions. Now, we can use the figure to answer the questions. (a) There are 12 students who read none of the three publications. (b) There are 35 students who read only Fortune. (c) The overlap of the regions representing readers of Business Week and The Wall Street Journal shows that 5 students read Business Week and The Wall Street Journal but not Fortune. Remark. Note that we do not draw two circles to represent a set and its complement. For example, in the above figure, we did not draw a circle to represent those who read Fortune and another to represent those who do not read Fortune. Such an additional region is not only unnecessary, but also very confusing. Those not in the circle representing a set A are automatically in its complement, A. Example 5. Jeff Friedman is a section chief for an electric utility company. The employees in this section cut down trees, climb poles, and splice wire. Friedman reported the following information to the management of the utility. Of the 100 employees in my section, 45 can cut trees; 50 can climb poles; 57 can splice wire; 22 can climb poles but can t cut trees; 20 can climb poles and splice wire; 25 can cut trees and splice wire; 14 can cut trees and splice wire but can t climb poles; 9 can t do any of the three (management trainees). Fall 2010 Page 5 Penn State University
6 Draw a Venn diagram to represent the data supplied by Friedman. Solution. The data supplied by Friedman can be represented by the following Venn diagram. Now, if we add the numbers from all the regions, we get the total number of employees as = 99, while Friedman claimed to have 100 employees. Hence, management decided that Friedman did not qualify as a section chief, and he was reassigned as a nightshift meter reader. (Moral: He should have taken this course.) Remark. In all the above examples, we started with a piece of information specifying the relationship with all the categories. This is usually the best way to begin solving a problem of this type. Recall that n(a) represents the number of elements in a finite set A. We have the following statement about the number of elements in the union of two sets. This will be useful in our study of probability. Union Rule for Sets For any two sets A and B, n(a B) = n(a) + n(b) n(a B) To prove this statement, let y represent n(a B), x + y represent n(a) and y + z represent n(b), as shown in the following figure. Fall 2010 Page 6 Penn State University
7 Then n(a B) = x + y + z, n(a) + n(b) n(a B) = (x + y) + (y + z) y = x + y + z. So, n(a B) = n(a) + n(b) n(a B). Special Case: When A and B are disjoint sets, then A B = and hence, n(a B) = 0, then we have the following special case of the union rule: n(a B) = n(a) + n(b) if A B =. Example 6. A group of 10 students meet to plan a school function. All are majoring in Accounting or Economics or both. 5 of the students are Economics majors and 7 are majors in Accounting. How many major in both subjects? Solution. Let A represent the set of Accounting majors and E represent the set of Economics majors. So, we have n(a) = 7, n(e) = 5, and n(a E) = 10. We need to find n(a E). Using the union rule, we have n(a E) = n(a) + n(e) n(a E) that is, 10 = n(a E). So, n(a E) = = 2. Thus, the number of students majoring in both Accounting and Economics is 2. Example 7. The following table gives the number of threatened and endangered animal species in the world as of Endangered (E) Threatened (T ) Totals Amphibians and reptiles (A) Arachnids and insects (I) Birds (B) Clams, crustaceans, and snails (C) Fishes (F ) Mammals (M) Totals Using the letters given in the table to denote each set, find the number of species in each of the following sets. (a) E B Solution. The set E B consists of all species that are endangered and are birds. From the table, we see that there are 252 such species that is, n(e B) = 252. (b) E B Fall 2010 Page 7 Penn State University
8 Solution. The set E B consists of all species that are either endangered or are birds. The total number of endangered species, that is, n(e) = 919, and the total number of species of birds, n(b) = 273. Thus, using the union rule, n(e B) = n(e) + n(b) n(e B) = = 940, using the result that n(e B) = 252 from part (a). Thus, n(e B) = 940. (c) (F M) T Solution. We begin with the set F M, which is all species that are fish or mammals. This consists of the four categories with 85, 47, 323, and 33 species. Of this set, we take those that are not threatened, for a total of = 408 species. This is the number of species of fish and mammals that are not threatened. Symbolically, n((f M) T ) = 408. Example 8. Suppose that a group of 150 students have joined at least one of three chat rooms: one on autoracing, one on bicycling, and one for college students. For simplicity, we call these rooms A, B, and C. In addition, 90 students joined room A; 50 students joined room B; 70 students joined room C; 15 students joined rooms A and C; 12 students joined rooms B and C; 10 students joined all three rooms. Determine how many students joined both chat rooms A and B. Solution. Since 10 students joined all three rooms, we begin by placing 10 in the area common to all the three regions as shown in the following figure. Of the 15 students who joined A and C, 10 also joined room B. Thus only = 5 are in the area common only to A and C. Likewise, there are = 2 students who joined only B and C. Since there are already = 17 students in C, there are = 53 students who joined only room C. Now, let x be the number of students who joined only rooms A and B. Then 90 (x ) = 75 x students joined only room A and 50 (x ) = 38 x students who joined only room B. Fall 2010 Page 8 Penn State University
9 Note that since all 150 students joined at least one room, there are no elements in the region outside the three circles. Now that the diagram is filled out, we can determine the value of x by using the fact that the total number of students who joined at least one chat room is 150. Thus, (75 x) x (38 x) = 150. Simplifying, we have 183 x = 150, implying that x = 33. Hence, the number of students who joined both chat rooms A and B is = 43. Fall 2010 Page 9 Penn State University
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