{"id":65,"date":"2019-06-18T12:50:20","date_gmt":"2019-06-18T16:50:20","guid":{"rendered":"https:\/\/pressbooks.library.torontomu.ca\/pohmath\/chapter\/sets-and-counting\/"},"modified":"2023-12-17T11:04:47","modified_gmt":"2023-12-17T16:04:47","slug":"sets-and-counting","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.torontomu.ca\/pohmath\/chapter\/sets-and-counting\/","title":{"raw":"5.1. Set Theory and Venn Diagrams","rendered":"5.1. Set Theory and Venn Diagrams"},"content":{"raw":"[Latexpage]\r\n

Set Theory<\/h1>\r\nIn this section, we will familiarize ourselves with set operations and notations, so that we can apply these concepts to both counting and probability problems. We begin by defining some terms.\r\n\r\nA set<\/strong> is a collection of objects, and its members are called the elements<\/strong> of the set. We name the set by using capital letters, and enclose its members in curly brackets. Suppose we need to list the members of the chess club. We use the following set notation.\r\n

C<\/em> ={Ken, Bob, Tran, Shanti, Eric}<\/p>\r\nA set that has no members is called an empty set<\/strong>. The empty set is denoted by the symbol \u00d8.\r\n\r\nTwo sets are equal<\/strong> if they have the same elements.\r\n\r\nA set A<\/em> is a subset<\/strong> of a set B<\/em> if every member of A<\/em> is also a member of B<\/em>.\r\n\r\nSuppose C<\/em> = {Al, Bob, Chris, David, Ed} and A<\/em> = {Bob, David}. Then A<\/em> is a subset of C<\/em>, written as $A\\subseteq C$.\r\n\r\nEvery set is a subset of itself, and the empty set is a subset of every set.\r\n

\r\n\r\nUnion Of Two Sets <\/strong>\r\n\r\nLet A<\/em> and B<\/em> be two sets, then the union of A<\/em> and B<\/em>, written as $A\\cup B$, is the set of all elements that are either in A<\/em> or in B<\/em>, or in both A<\/em> and B<\/em>.\r\n\r\n<\/div>\r\n \r\n
\r\n\r\nIntersection Of Two Sets <\/strong>\r\n\r\nLet A<\/em> and B<\/em> be two sets, then the intersection of A<\/em> and B<\/em>, written as $A\\cap B$, is the set of all elements that are common to both sets A<\/em> and B<\/em>.\r\n\r\n<\/div>\r\n \r\n\r\nA universal set<\/strong> U<\/em> is the set consisting of all elements under consideration.\r\n
\r\n\r\nComplement of a Set <\/strong>\r\n\r\nLet A<\/em> be any set, then the complement of set A<\/em>, written as $A^c$ , is the set consisting of elements in the universal set U<\/em> that are not in A<\/em>.\r\n\r\n<\/div>\r\n \r\n
\r\n\r\nDisjoint Sets <\/strong>\r\n\r\nTwo sets A<\/em> and B<\/em> are called disjoint sets if their intersection is an empty set.\r\n\r\n<\/div>\r\n \r\n
\r\n

Example 5.1.1<\/p>\r\n\r\n<\/header>\r\n

List all the subsets of the set of primary colors {red, yellow, blue}.<\/div>\r\n
Solution <\/strong><\/div>\r\n
The subsets are \u2205, {red}, {yellow}, {blue}, {red, yellow}, {red, blue}, {yellow, blue}, {red, yellow, blue}<\/div>\r\n
Note that the empty set is a subset of every set, and a set is a subset of itself.<\/div>\r\n<\/div>\r\n \r\n
\r\n

Example 5.1.2<\/p>\r\n\r\n<\/header>\r\n

Let F<\/em> = {Aikman, Jackson, Rice, Sanders, Young}, and B<\/em> = {Griffey, Jackson, Sanders, Thomas}. Find the intersection of the sets F<\/em> and B<\/em>.<\/div>\r\n
Solution <\/strong><\/div>\r\n
The intersection of the two sets is the set whose elements belong to both sets. Therefore,<\/div>\r\n
$F\\cap B$ = {Jackson, Sanders}<\/div>\r\n<\/div>\r\n \r\n
\r\n

Example 5.1.3<\/p>\r\n\r\n<\/header>\r\n

Find the union of the sets F<\/em> and B<\/em> given as follows.<\/div>\r\n
F<\/em> = {Aikman, Jackson, Rice, Sanders, Young}<\/div>\r\n
B<\/em> = {Griffey, Jackson, Sanders, Thomas}<\/div>\r\n
Solution <\/strong><\/div>\r\n
The union of two sets is the set whose elements are either in A<\/em> or in B<\/em> or in both A<\/em> and B<\/em>. Therefore<\/div>\r\n
$F\\cup B$ = {Aikman, Griffey, Jackson, Rice, Sanders, Thomas, Young}<\/div>\r\n
Observe that when writing the union of two sets, the repetitions are avoided.<\/div>\r\n<\/div>\r\n \r\n
\r\n

Example 5.1.4<\/p>\r\n\r\n<\/header>\r\n

Let the universal set U<\/em> = {red, orange, yellow, green, blue, indigo, violet}, and P<\/em> = {red, yellow, blue}. Find the complement of P<\/em>.<\/div>\r\n
Solution <\/strong><\/div>\r\n
The complement of a set P<\/em> is the set consisting of elements in the universal set U<\/em> that are not in P<\/em>. Therefore:<\/div>\r\n
$P^c$ = {orange, green, indigo, violet}<\/div>\r\n<\/div>\r\n \r\n\r\nTo achieve a better understanding, let us suppose that the universal set U<\/em> represents the colors of the spectrum, and P<\/em> the primary colors, then $P^c$ represents those colors of the spectrum that are not primary colors.\r\n\r\n \r\n
\r\n

Example 5.1.5<\/p>\r\n\r\n<\/header>\r\n

Let U<\/em> = {red, orange, yellow, green, blue, indigo, violet},<\/div>\r\n
P<\/em> = {red, yellow, blue},<\/div>\r\n
Q<\/em> = {red, green}, and<\/div>\r\n
R<\/em> ={orange, green, indigo}.<\/div>\r\n
Find $({P \\cup Q})^c\\cap R^c$.<\/div>\r\n
Solution <\/strong><\/div>\r\n
We do the problems in steps.<\/div>\r\n
$P\\cup Q$ = {red, yellow, blue, green}<\/div>\r\n
$({P \\cup Q})^c$ = {orange, indigo, violet}<\/div>\r\n
\u00a0$R^c$ = {red, yellow, blue, violet}<\/div>\r\n
$({P \\cup Q})^c\\cap R^c$ = {violet}<\/div>\r\n<\/div>\r\n \r\n

Venn Diagrams<\/h1>\r\nWe now use Venn diagrams to illustrate the relationships between sets. In the late 1800's, an English logician named John Venn developed a method to represent relationship between sets. He represented these relationships using diagrams, which are now known as Venn diagrams. A Venn diagram represents a set as the interior of a circle. Often two or more circles are enclosed in a rectangle where the rectangle represents the universal set. To visualize an intersection or union of a set is easy. In this section, we will mainly use Venn diagrams to sort various populations and count objects.\r\n\r\n \r\n
\r\n

Example 5.1.6<\/p>\r\n\r\n<\/header>\r\n

Suppose a survey of car enthusiasts showed that over a certain time period, 30 drove cars with automatic transmissions, 20 drove cars with standard transmissions, and 12 drove cars of both types. If every one in the survey drove cars with one of these transmissions, how many people participated in the survey?<\/div>\r\n
Solution <\/strong><\/div>\r\n
We will use Venn diagrams to solve this problem.<\/div>\r\n
Let the set A<\/em> represent those car enthusiasts who drove cars with automatic transmissions, and set S<\/em> represent the car enthusiasts who drove the cars with standard transmissions. Now we use Venn diagrams to sort out the information given in this problem.<\/div>\r\n
Since 12 people drove both cars, we place the number 12 in the region common to both sets.<\/div>\r\n
\r\n
\r\n\r\n\r\n\r\n
\\begin{tikzpicture}\r\n[thick]\r\n\\draw (0,0) circle (1) node[below,shift={(-1,1.1)}] {$\\bm{A}$};\r\n\\draw (1.2,0) circle (1) node[below,shift={(1,1.1)}] {$\\bm{S}$};\\node at (.6,0) {12};\\end{tikzpicture}\r\n

(a)<\/p>\r\n<\/td>\r\n

\\begin{tikzpicture}[thick]\r\n\\draw (0,0) circle (1) node[below,shift={(-1,1.1)}] {$\\bm{A}$};\r\n\\draw (1.2,0) circle (1) node[below,shift={(1,1.1)}] {$\\bm{S}$};\\node at (.6,0) {12};\\node at (-.3,0) {18};\r\n\\end{tikzpicture}\r\n

(b)<\/p>\r\n<\/td>\r\n

\\begin{tikzpicture}[thick]\r\n\\draw (0,0) circle (1) node[below,shift={(-1,1.1)}] {$\\bm{A}$};\r\n\\draw (1.2,0) circle (1) node[below,shift={(1,1.1)}] {$\\bm{S}$};\\node at (.6,0) {12};\r\n\\node at (1.5,0) {8};\r\n\\node at (-.3,0) {18};\r\n\\end{tikzpicture}\r\n

(c)<\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\nBecause 30 people drove cars with automatic transmissions, the circle A<\/em> must contain 30 elements. This means x<\/em> + 12 = 30, or x<\/em> = 18. Similarly, since 20 people drove cars with standard transmissions, the circle B<\/em> must contain 20 elements, or y<\/em> +12 = 20 which in turn makes y<\/em> = 8.\r\n\r\nNow that all the information is sorted out, it is easy to read from the diagram that 18 people drove cars with automatic transmissions only, 12 people drove both types of cars, and 8 drove cars with standard transmissions only. Therefore, 18 + 12 + 8 = 38 people took part in the survey.\r\n\r\n<\/div>\r\n<\/div>\r\n \r\n

\r\n

Example 5.1.7<\/p>\r\n\r\n<\/header>\r\n

A survey of 100 people in California indicates that 60 people have visited Disneyland, 15 have visited Knott's Berry Farm, and 6 have visited both. How many people have visited neither place?<\/div>\r\n
Solution <\/strong><\/div>\r\n
Let the set D<\/em> represent the people who have visited Disneyland, and K<\/em> the set of people who have visited Knott's Berry Farm.<\/div>\r\n
\r\n
\r\n\r\n\r\n\r\n
\\begin{tikzpicture}[thick]\r\n\\draw (2.7,-1.5) rectangle (-1.5,1.5) node[below,shift={(3.9,0)}] {$\\bm{U}$};\r\n\\draw (0,0) circle (1) node[below,shift={(-1.2,0.8)}] {$\\bm{D}$};\r\n\\draw (1.2,0) circle (1) node[below,shift={(1.2,0.8)}] {$\\bm{K}$};\\node at (.6,0) {6};\r\n\\node at (1.5,0) {9};\r\n\\node at (-.3,0) {54};\r\n\\node at (2.3,-0.8) {x};\r\n\\end{tikzpicture}\r\n

(a)<\/p>\r\n<\/td>\r\n

\\begin{tikzpicture}[thick]\r\n\\draw (2.7,-1.5) rectangle (-1.5,1.5) node[below,shift={(3.8,0)}] {$\\bm{U}$};\r\n\\draw (0,0) circle (1) node[below,shift={(-1.2,0.8)}] {$\\bm{D}$};\r\n\\draw (1.2,0) circle (1) node[below,shift={(1.2,0.8)}] {$\\bm{K}$};\\node at (.6,0) {6};\r\n\\node at (1.5,0) {9};\r\n\\node at (-.3,0) {54};\r\n\\node at (2.3,-0.8) {31};\r\n\\end{tikzpicture}\r\n

(b)<\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n

We fill the three regions associated with the sets <\/span>D<\/em> and <\/span>K<\/em> in the same manner as before. Since 100 people participated in the survey, the rectangle representing the universal set <\/span>U<\/em> must contain 100 objects. Let <\/span>x<\/em> represent those people in the universal set that are neither in the set <\/span>D<\/em> nor in <\/span>K<\/em>. This means 54 + 6 + 9 + <\/span>x<\/em> = 100, or <\/span>x<\/em> = 31.<\/span><\/p>\r\n

Therefore, there are 31 people in the survey who have visited neither place.<\/span><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n \r\n

\r\n

Example 5.1.8<\/p>\r\n\r\n<\/header>\r\n

A survey of 100 exercise conscious people resulted in the following information:<\/div>\r\n
    \r\n \t
  • 50 jog, 30 swim, and 35 cycle<\/li>\r\n \t
  • 14 jog and swim<\/li>\r\n \t
  • 7 swim and cycle<\/li>\r\n \t
  • 9 jog and cycle<\/li>\r\n \t
  • 3 people take part in all three activities<\/li>\r\n<\/ul>\r\n
    a.<\/strong> How many jog but do not swim or cycle?<\/div>\r\n
    b.<\/strong> How many take part in only one of the activities?<\/div>\r\n
    c.<\/strong> How many do not take part in any of these activities?<\/span><\/div>\r\n
    \r\n\r\nSolution <\/strong>\r\n\r\n<\/div>\r\n
    Let J<\/em> represent the set of people who jog, S<\/em> the set of people who swim, and C<\/em> who cycle. In using Venn diagrams, our ultimate aim is to assign a number to each region. We always begin by first assigning the number to the innermost region and then working our way out.<\/div>\r\n
    \r\n
    \r\n\r\n\r\n\r\n
    \\begin{tikzpicture}[thick]\r\n\\draw (2.7,-2.6) rectangle (-1.5,1.5) node[below,shift={(3.9,0)}] {$\\bm{U}$};\r\n\\draw (0,0) circle (1) node[below,shift={(-1.2,0.8)}] {$\\bm{J}$};\r\n\\draw (1.2,0) circle (1) node[below,shift={(1.2,0.8)}] {$\\bm{S}$};\\draw (0.6,-1.1) circle (1) node[below,shift={(1.1,-0.4)}] {$\\bm{C}$};\\node at (.6,-0.4) {3};\\end{tikzpicture}\r\n

    (a)<\/p>\r\n<\/td>\r\n

    		\\begin{tikzpicture}[thick]\r\n\\draw (2.7,-2.6) rectangle (-1.5,1.5) node[below,shift={(3.9,0)}] {$\\bm{U}$};\r\n\\draw (0,0) circle (1) node[below,shift={(-1.2,0.8)}] {$\\bm{J}$};\r\n\\draw (1.2,0) circle (1) node[below,shift={(1.2,0.8)}] {$\\bm{S}$};\\draw (0.6,-1.1) circle (1) node[below,shift={(1.1,-0.4)}] {$\\bm{C}$};\\node at (.6,-0.4) {3};\r\n\\node at (1.1,-0.7) {4};\r\n\\node at (0.1,-0.7) {6};\r\n\\node at (0.6,0.2) {11};\\end{tikzpicture}\r\n

    (b)<\/p>\r\n<\/td>\r\n

    		\\begin{tikzpicture}[thick]\r\n\\draw (2.7,-2.6) rectangle (-1.5,1.5) node[below,shift={(3.9,0)}] {$\\bm{U}$};\r\n\\draw (0,0) circle (1) node[below,shift={(-1.2,0.8)}] {$\\bm{J}$};\r\n\\draw (1.2,0) circle (1) node[below,shift={(1.2,0.8)}] {$\\bm{S}$};\\draw (0.6,-1.1) circle (1) node[below,shift={(1.1,-0.4)}] {$\\bm{C}$};\\node at (.6,-0.4) {3};\\node at (1.1,-0.7) {4};\r\n\\node at (0.1,-0.7) {6};\r\n\\node at (0.6,0.2) {11};\r\n\\node at (1.5,0) {12};\r\n\\node at (-.3,0) {30};\r\n\\node at (2.3,-2.2) {12};\\node at (0.6,-1.5) {22};\\end{tikzpicture}\r\n

    (c)<\/p>\r\n<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/div>\r\n

    We place a 3 in the innermost region of Figure (a) because it represents the number of people who participate in all three activities. Next we compute <\/span>x<\/em>, <\/span>y<\/em> and <\/span>z<\/em>.<\/span><\/p>\r\n\r\n

      \r\n \t
    • Since 14 people jog and swim, <\/span>x<\/em> +3 = 14, or <\/span>x<\/em> = 11.<\/span><\/li>\r\n \t
    • The fact that 9 people jog and cycle results in y<\/em> + 3 = 9, or <\/span>y<\/em> = 6.<\/span><\/li>\r\n \t
    • Since 7 people swim and cycle, z<\/em> + 3 = 7, or <\/span>z<\/em> = 4.<\/span><\/li>\r\n \t
    • This information is depicted in Figure (b).<\/li>\r\n<\/ul>\r\n<\/div>\r\n
      Now we proceed to find the unknowns m<\/em>, n<\/em> and p<\/em>:<\/div>\r\n
        \r\n \t
      • Since 50 people jog, m + 11 + 6 + 3 = 50, or m = 30.<\/li>\r\n \t
      • 30 people swim, therefore, n + 11 + 4 + 3 = 30, or n = 12.<\/li>\r\n \t
      • 35 people cycle, therefore, p<\/em> + 6 + 4 + 3 = 35, or <\/span>p<\/em> = 22.<\/span><\/li>\r\n \t
      • By adding all the entries in all three sets, we get a sum of 88. Since 100 people were surveyed, the number inside the universal set but outside of all three sets is 100 \u2013 88, or 12.<\/li>\r\n \t
      • In Figure (c), the information is sorted out, and the questions can readily be answered.<\/li>\r\n<\/ul>\r\n
        a.<\/strong> The number of people who jog but do not swim or cycle is 30.<\/div>\r\n
        b.<\/strong> The number who take part in only one of these activities is 30 + 12 + 22 = 64.<\/div>\r\n
        c.<\/strong> The number of people who do not take part in any of these activities is 12.<\/div>\r\n<\/div>\r\n \r\n

        Practice questions<\/h1>\r\n1.<\/strong> Let the Universal set U<\/em> = {a, b, c, d, e, f, g, h, i, j},V<\/em> = {a, e, i, f, h}, and W<\/em> = {a, c, e, g, i}. List the members of the following sets:\r\n

        a.<\/strong> $V\\cup W$<\/p>\r\n

        b.<\/strong> $V^c\\cap W$<\/p>\r\n2. <\/strong>Consider the following sets: A<\/em> = {SARS, H1N1, H5N1, MERS-CoV, COVID-19,\u00a0Influenz<\/span>a, Norovirus<\/span>}, <\/span>B<\/em> = {Listeria, Campylobacter, Salmonella, E. coli O157, Norovirus, Shigella<\/span>}, and <\/span>C<\/em> = {<\/span>SARS, Listeria, Tuberculosis, H5N1, Salmonella, HIV, COVID-19<\/span>}.<\/span> List the members of the following sets:<\/span>\r\n

        a.<\/strong> $A\\cap C$<\/p>\r\n

        b.<\/strong> $({A \\cup B})^c\\cap C$<\/p>\r\n3.<\/strong> A survey of athletes revealed that for their minor aches and pains, 30 used aspirin, 50 used ibuprofen, and 15 used both. All surveyed athletes used at least one of the two painkillers. How many athletes were surveyed?\r\n\r\n4. <\/strong>A study of 150 high school students found that 25 reported having a previous concussion or head injury, 52 reported experiencing mental illness, and 15 reported both outcomes. How many students did not report either outcome?\r\n\r\n5.<\/strong> A survey of 100 students at Toronto Metropolitan University finds that 50 subscribe to Netflix, 40 subscribe to Amazon Prime, and 30 subscribe to Disney+. Of these, 15 subscribe to both Netflix and Amazon Prime, 10 to both Amazon Prime and Disney+, 10 to both Netflix and Disney+, and 5 have all three subscription services. Draw a Venn diagram and determine the following:\r\n

        a.<\/strong> The number of students subscribing to Amazon Prime but not the other two streaming services.<\/p>\r\n

        b.<\/strong> The number of students subscribing to Netflix or Amazon Prime but not Disney+.<\/p>\r\n

        c.<\/strong> The number of students not subscribing to any of these services.<\/p>\r\n \r\n\r\n ","rendered":"

        Set Theory<\/h1>\n

        In this section, we will familiarize ourselves with set operations and notations, so that we can apply these concepts to both counting and probability problems. We begin by defining some terms.<\/p>\n

        A set<\/strong> is a collection of objects, and its members are called the elements<\/strong> of the set. We name the set by using capital letters, and enclose its members in curly brackets. Suppose we need to list the members of the chess club. We use the following set notation.<\/p>\n

        C<\/em> ={Ken, Bob, Tran, Shanti, Eric}<\/p>\n

        A set that has no members is called an empty set<\/strong>. The empty set is denoted by the symbol \u00d8.<\/p>\n

        Two sets are equal<\/strong> if they have the same elements.<\/p>\n

        A set A<\/em> is a subset<\/strong> of a set B<\/em> if every member of A<\/em> is also a member of B<\/em>.<\/p>\n

        Suppose C<\/em> = {Al, Bob, Chris, David, Ed} and A<\/em> = {Bob, David}. Then A<\/em> is a subset of C<\/em>, written as \"A\subseteq C\".<\/p>\n

        Every set is a subset of itself, and the empty set is a subset of every set.<\/p>\n

        \n

        Union Of Two Sets <\/strong><\/p>\n

        Let A<\/em> and B<\/em> be two sets, then the union of A<\/em> and B<\/em>, written as \"A\cup B\", is the set of all elements that are either in A<\/em> or in B<\/em>, or in both A<\/em> and B<\/em>.<\/p>\n<\/div>\n

         <\/p>\n

        \n

        Intersection Of Two Sets <\/strong><\/p>\n

        Let A<\/em> and B<\/em> be two sets, then the intersection of A<\/em> and B<\/em>, written as \"A\cap B\", is the set of all elements that are common to both sets A<\/em> and B<\/em>.<\/p>\n<\/div>\n

         <\/p>\n

        A universal set<\/strong> U<\/em> is the set consisting of all elements under consideration.<\/p>\n

        \n

        Complement of a Set <\/strong><\/p>\n

        Let A<\/em> be any set, then the complement of set A<\/em>, written as \"A^c\" , is the set consisting of elements in the universal set U<\/em> that are not in A<\/em>.<\/p>\n<\/div>\n

         <\/p>\n

        \n

        Disjoint Sets <\/strong><\/p>\n

        Two sets A<\/em> and B<\/em> are called disjoint sets if their intersection is an empty set.<\/p>\n<\/div>\n

         <\/p>\n

        \n
        \n

        Example 5.1.1<\/p>\n<\/header>\n

        List all the subsets of the set of primary colors {red, yellow, blue}.<\/div>\n
        Solution <\/strong><\/div>\n
        The subsets are \u2205, {red}, {yellow}, {blue}, {red, yellow}, {red, blue}, {yellow, blue}, {red, yellow, blue}<\/div>\n
        Note that the empty set is a subset of every set, and a set is a subset of itself.<\/div>\n<\/div>\n

         <\/p>\n

        \n
        \n

        Example 5.1.2<\/p>\n<\/header>\n

        Let F<\/em> = {Aikman, Jackson, Rice, Sanders, Young}, and B<\/em> = {Griffey, Jackson, Sanders, Thomas}. Find the intersection of the sets F<\/em> and B<\/em>.<\/div>\n
        Solution <\/strong><\/div>\n
        The intersection of the two sets is the set whose elements belong to both sets. Therefore,<\/div>\n
        \"F\cap B\" = {Jackson, Sanders}<\/div>\n<\/div>\n

         <\/p>\n

        \n
        \n

        Example 5.1.3<\/p>\n<\/header>\n

        Find the union of the sets F<\/em> and B<\/em> given as follows.<\/div>\n
        F<\/em> = {Aikman, Jackson, Rice, Sanders, Young}<\/div>\n
        B<\/em> = {Griffey, Jackson, Sanders, Thomas}<\/div>\n
        Solution <\/strong><\/div>\n
        The union of two sets is the set whose elements are either in A<\/em> or in B<\/em> or in both A<\/em> and B<\/em>. Therefore<\/div>\n
        \"F\cup B\" = {Aikman, Griffey, Jackson, Rice, Sanders, Thomas, Young}<\/div>\n
        Observe that when writing the union of two sets, the repetitions are avoided.<\/div>\n<\/div>\n

         <\/p>\n

        \n
        \n

        Example 5.1.4<\/p>\n<\/header>\n

        Let the universal set U<\/em> = {red, orange, yellow, green, blue, indigo, violet}, and P<\/em> = {red, yellow, blue}. Find the complement of P<\/em>.<\/div>\n
        Solution <\/strong><\/div>\n
        The complement of a set P<\/em> is the set consisting of elements in the universal set U<\/em> that are not in P<\/em>. Therefore:<\/div>\n
        \"P^c\" = {orange, green, indigo, violet}<\/div>\n<\/div>\n

         <\/p>\n

        To achieve a better understanding, let us suppose that the universal set U<\/em> represents the colors of the spectrum, and P<\/em> the primary colors, then \"P^c\" represents those colors of the spectrum that are not primary colors.<\/p>\n

         <\/p>\n

        \n
        \n

        Example 5.1.5<\/p>\n<\/header>\n

        Let U<\/em> = {red, orange, yellow, green, blue, indigo, violet},<\/div>\n
        P<\/em> = {red, yellow, blue},<\/div>\n
        Q<\/em> = {red, green}, and<\/div>\n
        R<\/em> ={orange, green, indigo}.<\/div>\n
        Find \"({P \cup Q})^c\cap R^c\".<\/div>\n
        Solution <\/strong><\/div>\n
        We do the problems in steps.<\/div>\n
        \"P\cup Q\" = {red, yellow, blue, green}<\/div>\n
        \"({P \cup Q})^c\" = {orange, indigo, violet}<\/div>\n
        \u00a0\"R^c\" = {red, yellow, blue, violet}<\/div>\n
        \"({P \cup Q})^c\cap R^c\" = {violet}<\/div>\n<\/div>\n

         <\/p>\n

        Venn Diagrams<\/h1>\n

        We now use Venn diagrams to illustrate the relationships between sets. In the late 1800’s, an English logician named John Venn developed a method to represent relationship between sets. He represented these relationships using diagrams, which are now known as Venn diagrams. A Venn diagram represents a set as the interior of a circle. Often two or more circles are enclosed in a rectangle where the rectangle represents the universal set. To visualize an intersection or union of a set is easy. In this section, we will mainly use Venn diagrams to sort various populations and count objects.<\/p>\n

         <\/p>\n

        \n
        \n

        Example 5.1.6<\/p>\n<\/header>\n

        Suppose a survey of car enthusiasts showed that over a certain time period, 30 drove cars with automatic transmissions, 20 drove cars with standard transmissions, and 12 drove cars of both types. If every one in the survey drove cars with one of these transmissions, how many people participated in the survey?<\/div>\n
        Solution <\/strong><\/div>\n
        We will use Venn diagrams to solve this problem.<\/div>\n
        Let the set A<\/em> represent those car enthusiasts who drove cars with automatic transmissions, and set S<\/em> represent the car enthusiasts who drove the cars with standard transmissions. Now we use Venn diagrams to sort out the information given in this problem.<\/div>\n
        Since 12 people drove both cars, we place the number 12 in the region common to both sets.<\/div>\n
        \n
        \n\n\n\n
        \n

        \"Rendered<\/p>\n

        (a)<\/p>\n<\/td>\n

        		\n

        \"Rendered<\/p>\n

        (b)<\/p>\n<\/td>\n

        		\n

        \"Rendered<\/p>\n

        (c)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

        Because 30 people drove cars with automatic transmissions, the circle A<\/em> must contain 30 elements. This means x<\/em> + 12 = 30, or x<\/em> = 18. Similarly, since 20 people drove cars with standard transmissions, the circle B<\/em> must contain 20 elements, or y<\/em> +12 = 20 which in turn makes y<\/em> = 8.<\/p>\n

        Now that all the information is sorted out, it is easy to read from the diagram that 18 people drove cars with automatic transmissions only, 12 people drove both types of cars, and 8 drove cars with standard transmissions only. Therefore, 18 + 12 + 8 = 38 people took part in the survey.<\/p>\n<\/div>\n<\/div>\n

         <\/p>\n

        \n
        \n

        Example 5.1.7<\/p>\n<\/header>\n

        A survey of 100 people in California indicates that 60 people have visited Disneyland, 15 have visited Knott’s Berry Farm, and 6 have visited both. How many people have visited neither place?<\/div>\n
        Solution <\/strong><\/div>\n
        Let the set D<\/em> represent the people who have visited Disneyland, and K<\/em> the set of people who have visited Knott’s Berry Farm.<\/div>\n
        \n
        \n\n\n\n
        \n

        \"Rendered<\/p>\n

        (a)<\/p>\n<\/td>\n

        		\n

        \"Rendered<\/p>\n

        (b)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

        We fill the three regions associated with the sets <\/span>D<\/em> and <\/span>K<\/em> in the same manner as before. Since 100 people participated in the survey, the rectangle representing the universal set <\/span>U<\/em> must contain 100 objects. Let <\/span>x<\/em> represent those people in the universal set that are neither in the set <\/span>D<\/em> nor in <\/span>K<\/em>. This means 54 + 6 + 9 + <\/span>x<\/em> = 100, or <\/span>x<\/em> = 31.<\/span><\/p>\n

        Therefore, there are 31 people in the survey who have visited neither place.<\/span><\/p>\n<\/div>\n<\/div>\n

         <\/p>\n

        \n
        \n

        Example 5.1.8<\/p>\n<\/header>\n

        A survey of 100 exercise conscious people resulted in the following information:<\/div>\n
          \n
        • 50 jog, 30 swim, and 35 cycle<\/li>\n
        • 14 jog and swim<\/li>\n
        • 7 swim and cycle<\/li>\n
        • 9 jog and cycle<\/li>\n
        • 3 people take part in all three activities<\/li>\n<\/ul>\n
          a.<\/strong> How many jog but do not swim or cycle?<\/div>\n
          b.<\/strong> How many take part in only one of the activities?<\/div>\n
          c.<\/strong> How many do not take part in any of these activities?<\/span><\/div>\n
          \n

          Solution <\/strong><\/p>\n<\/div>\n

          Let J<\/em> represent the set of people who jog, S<\/em> the set of people who swim, and C<\/em> who cycle. In using Venn diagrams, our ultimate aim is to assign a number to each region. We always begin by first assigning the number to the innermost region and then working our way out.<\/div>\n
          \n
          \n\n\n\n
          \n

          \"Rendered<\/p>\n

          (a)<\/p>\n<\/td>\n

          		\n

          \"Rendered<\/p>\n

          (b)<\/p>\n<\/td>\n

          		\n

          \"Rendered<\/p>\n

          (c)<\/p>\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n

          We place a 3 in the innermost region of Figure (a) because it represents the number of people who participate in all three activities. Next we compute <\/span>x<\/em>, <\/span>y<\/em> and <\/span>z<\/em>.<\/span><\/p>\n

            \n
          • Since 14 people jog and swim, <\/span>x<\/em> +3 = 14, or <\/span>x<\/em> = 11.<\/span><\/li>\n
          • The fact that 9 people jog and cycle results in y<\/em> + 3 = 9, or <\/span>y<\/em> = 6.<\/span><\/li>\n
          • Since 7 people swim and cycle, z<\/em> + 3 = 7, or <\/span>z<\/em> = 4.<\/span><\/li>\n
          • This information is depicted in Figure (b).<\/li>\n<\/ul>\n<\/div>\n
            Now we proceed to find the unknowns m<\/em>, n<\/em> and p<\/em>:<\/div>\n
              \n
            • Since 50 people jog, m + 11 + 6 + 3 = 50, or m = 30.<\/li>\n
            • 30 people swim, therefore, n + 11 + 4 + 3 = 30, or n = 12.<\/li>\n
            • 35 people cycle, therefore, p<\/em> + 6 + 4 + 3 = 35, or <\/span>p<\/em> = 22.<\/span><\/li>\n
            • By adding all the entries in all three sets, we get a sum of 88. Since 100 people were surveyed, the number inside the universal set but outside of all three sets is 100 \u2013 88, or 12.<\/li>\n
            • In Figure (c), the information is sorted out, and the questions can readily be answered.<\/li>\n<\/ul>\n
              a.<\/strong> The number of people who jog but do not swim or cycle is 30.<\/div>\n
              b.<\/strong> The number who take part in only one of these activities is 30 + 12 + 22 = 64.<\/div>\n
              c.<\/strong> The number of people who do not take part in any of these activities is 12.<\/div>\n<\/div>\n

               <\/p>\n

              Practice questions<\/h1>\n

              1.<\/strong> Let the Universal set U<\/em> = {a, b, c, d, e, f, g, h, i, j},V<\/em> = {a, e, i, f, h}, and W<\/em> = {a, c, e, g, i}. List the members of the following sets:<\/p>\n

              a.<\/strong> \"V\cup W\"<\/p>\n

              b.<\/strong> \"V^c\cap W\"<\/p>\n

              2. <\/strong>Consider the following sets: A<\/em> = {SARS, H1N1, H5N1, MERS-CoV, COVID-19,\u00a0Influenz<\/span>a, Norovirus<\/span>}, <\/span>B<\/em> = {Listeria, Campylobacter, Salmonella, E. coli O157, Norovirus, Shigella<\/span>}, and <\/span>C<\/em> = {<\/span>SARS, Listeria, Tuberculosis, H5N1, Salmonella, HIV, COVID-19<\/span>}.<\/span> List the members of the following sets:<\/span><\/p>\n

              a.<\/strong> \"A\cap C\"<\/p>\n

              b.<\/strong> \"({A \cup B})^c\cap C\"<\/p>\n

              3.<\/strong> A survey of athletes revealed that for their minor aches and pains, 30 used aspirin, 50 used ibuprofen, and 15 used both. All surveyed athletes used at least one of the two painkillers. How many athletes were surveyed?<\/p>\n

              4. <\/strong>A study of 150 high school students found that 25 reported having a previous concussion or head injury, 52 reported experiencing mental illness, and 15 reported both outcomes. How many students did not report either outcome?<\/p>\n

              5.<\/strong> A survey of 100 students at Toronto Metropolitan University finds that 50 subscribe to Netflix, 40 subscribe to Amazon Prime, and 30 subscribe to Disney+. Of these, 15 subscribe to both Netflix and Amazon Prime, 10 to both Amazon Prime and Disney+, 10 to both Netflix and Disney+, and 5 have all three subscription services. Draw a Venn diagram and determine the following:<\/p>\n

              a.<\/strong> The number of students subscribing to Amazon Prime but not the other two streaming services.<\/p>\n

              b.<\/strong> The number of students subscribing to Netflix or Amazon Prime but not Disney+.<\/p>\n

              c.<\/strong> The number of students not subscribing to any of these services.<\/p>\n

               <\/p>\n

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