# 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

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 night-shift 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 auto-racing, 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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