Probability of Pulling17 Red Cards From a Deck With 40 Deck Cards and 60 Blue Cards
1D Set Cardinality
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Some sets have a finite number of elements, while other sets, such as the collection of positive integers, have an infinite number. Tangible sets we encounter in everyday life invariably are finite, although the number of elements might be very large. The set of grains of sand on Bellows Beach has an enormous number of elements, although it is still a finite set. To find an infinite set you generally have to look at sets conjured up in the minds of humans, such as sets of numbers or of points on a line, or of other mathematical entities. The cardinality of a finite set is defined as simply the number of elements in the set. We will say that the cardinality of an infinite set is infinity (written as ∞). However, infinite sets are another topic � in discussing cardinality we restrict our attention to finite sets.
We use the notation |A| to refer to the cardinality of a set A; that is,
|A| = number of elements in A .
As the empty set has no elements, we have |Ø| = 0. For A = {a , b , c} we have |A| = 3. If a set has not too many elements and they are all listed, then of course it is a simple matter to count the elements and determine the set�s cardinality. However, if the elements are numerous and not listed but only described, then it might be quite difficult if not impossible to count them. There is a whole branch of mathematics, called combinatorics, devoted to methods of counting elements in various sets. Later in this book we will discuss some of these methods, but right now we keep things simple and look at only a few problems concerning the number of elements in the unions of sets.
First suppose that two sets A and B are disjoint. Since A ∪ B is constructed by combining the elements of A and B into one set, and because A and B have no elements in common, the number of elements in A ∪ B is just the sum of the number of elements in A and the number of elements in B; thus
|A ∪ B| = |A| + |B| , if A and B are disjoint.
However, if A and B are not disjoint then the situation becomes more complicated, as illustrated in the next example.
example 1
The art club at Waipahu High School has 10 members, the music club has 12 members, and 5 students belong to both of these clubs. How many students will be invited to the joint art club � music club Christmas party? If the art and music clubs were disjoint, then the answer would be 10 + 12 = 22. However, the 5 students who are members of both clubs are counted twice in this addition; thus we must subtract 5 from the total, to arrive at 10 + 12 � 5 = 17 as the number of students in the union of the art and music clubs.
The preceding example demonstrates why the general formula for the cardinality of the union of two sets A and B (when A and B might intersect) is
|A ∪ B| = |A| + |B| � |A ∩ B|.
In adding the number of elements in A to the number in B, we count those elements common to A and B twice; these are the elements in the intersection A ∩ B. To correct for this double-counting we must subtract from our sum the number of elements in A ∩ B.
example 2
| Kings : |  | |||
| Queens : |  | |||
| Jacks : |  | |||
| Numer- |  |  |  |  |
| Spds. | Hts. | Diams. | Clubs | |
In an ordinary deck of playing cards, how many cards are
(a) either red cards or face cards?
(b) either spades or queens?
For the benefit of those not familiar with playing cards, at the right is displayed a standard deck of 52 cards. There are four suits of 13 cards each, called spades (♠), hearts (♥), diamonds (¨), and clubs (♣). The spades and clubs are black, and the hearts and diamonds red. Each suit has ten numerical cards, labeled ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, as well as three face cards, called jack, queen, and king. (The �one� card is called an �ace�; in some games it is considered the highest card in the suit, and in other games the lowest.) Altogether there are 26 black cards, 26 red cards, 12 face cards, and 40 numerical cards.
(a) Let R denote the set of red cards, and F the set of face cards. Then
|R| = 26 , |F| = 12 .
We are interested in the number of cards in the union R ∪ F. It is tempting to add 26 and 12 and conclude that there are 38 cards that are either red or face cards. However, the sets R and F are not disjoint; in fact the intersection R ∩ F contains 6 cards which are both red cards and face cards, and these have been counted twice. Therefore, the number of cards that are either red or face cards is
|R ∪ F| = |R| + |F| � |R ∩ F| = 26 + 12 � 6 = 32 .
(b) Let S denote the spades and Q the queens. Then |S| = 13, |Q| = 4, |S ∩ Q| = 1, and
|S ∪ Q| = |S| + |Q| � |S ∩ Q| = 13 + 4 � 1 = 16 .
(The queen of spades gets counted twice in adding the spades and the queens.)
Any set S is disjoint from its complement ~S, while the union of S and ~S is the universal set U. Consequently, if we add the number of elements in S to the number not in S, we get the total number of elements in U � that is,
|S| + |~S| = |U| .
We illustrate this basic formula with a simple example.
example 3
 | In Susan�s refrigerator there are 9 eggs, of which 3 are rotten. How many of the eggs are good? The universe U, consisting of all the eggs, has 9 members, while the subset R of rotten eggs has 3 members. The good eggs make up the complement, ~R. Therefore,
There are 6 good eggs. |
One can solve counting problems involving unions and intersections of sets also with the help of Venn diagrams. This method is perhaps even preferred, because no formulas are needed and usually in the end we wind up with more information than was initially sought. The next several examples demonstrate the Venn diagram method.
example 4
Among the voters at a neighborhood board meeting there are 10 females, 23 Democrats, and 7 female Democrats. How many voters are either female or Democrat?
First we work the problem with our formula. Let F be the set of females at the meeting and D the set of Democrats. We are given that |F| = 10, |D| = 23, and |F ∩ D| = 7. The number of voters who are female or Democrat then is
|F ∪ D| = |F| + |D| � |F ∩ D| = 10 + 23 � 7 = 26 .
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Next we use a Venn diagram to solve the problem. The universe U consists of all voters at the meeting. We draw circles representing the sets F and D of females and Democrats. We imagine that all the females are standing in the circle labeled F, and all the Democrats in the circle labeled D. Then the female Democrats will be in the intersection of the circles F and D; we place the number 7 in this region to indicate the number of members standing there. Now, since there are 10 women in all, we must place 3 in the remaining region of the F circle. Likewise, since there are 23 Democrats, we must place 16 in the remaining region of the D circle. The picture now tells us everything we need to know. To find the number of members who are female or Democrat, we add the three numbers in the union F ∪ D, getting 3 + 7 + 16 = 26. Note that the number 3 represents the number of females who are not Democrats, while 16 is the number of male Democrats.
example 5
 | A caterer prepared 60 beef tacos for a birthday party. Among these tacos, he made 45 with tomatoes, 30 with both tomatoes and onions, and 5 with neither tomatoes nor onions. How many tacos did he make with (a) tomatoes or unions? (b) onions? (c) onions but not tomatoes? |
| We construct a Venn diagram, where the universe U consists of all the tacos, the set T has the tacos with tomatoes, and set O the tacos with onions. First we fill in the intersection T ∩ O with 30, the number of tacos with both tomatoes and onions. Since there are 45 tacos with tomatoes, we place 15 in the other part of the tomato circle. We place 5 in the region outside both circles, where we have tacos with neither tomatoes nor onions. Now we have filled in three of the four regions. Since all numbers must add up to 60, the total number of tacos, in the last region of the onion circle we need the number 60 � (30 + 15 + 5) = 10. Now we may answer all the questions. The number of tacos with tomatoes or onions is 15 + 30 + 10 = 55. The number with onions is 30 + 10 = 40, and the number with onions but not tomatoes is 10. |  |
Venn diagrams can be used also in problems involving unions of three sets.
example 6
 | Student taking | A dormitory of college freshmen has 110 students. Among these students, 75 are taking English, 52 are taking history, 50 are taking math, 33 are taking English and history, 30 are taking English and math, 22 are taking history and math, 13 are taking English, history, and math. |
We will determine how many students are taking
| (a) English and history, but not math, | (b) neither English, history, nor math, | |
| (c) math, but neither English nor history, | (d) English, but not history, | |
| (e) only one of the three subjects, | (f) exactly two of the three subjects. |
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The problem appears formidable, but a Venn diagram makes it easy. We draw circles E, H, and M, representing students taking English, history, and math, respectively, inside a rectangle representing the universe of all students in the dormitory. Then we fill numbers in the eight regions of the diagram, beginning from the inside and working our way out. The steps are as follows:
1) Write 13 in the red region, for students taking all 3 courses.
2) Write 9 in the pink region, since 22 take history and math.
3) Write 17 in the gray region, since 30 take English and math.
4) Write 20 in the brown region, since 33 take English and history.
5) Write 11 in the blue region, since 50 take math.
6) Write 10 in the green region, since 52 take history.
7) Write 25 in the yellow region, since 75 take English.
8) Add all seven numbers in the circles to get 105; since there are 110 students in all, write 5 in the region outside the three circles.
After the eight regions have numbers, we merely read from the Venn diagram the answers to the posed questions:
(a) Students taking English and history, but not math, are in the brown region; there are 20 of these.
(b) Students taking none of the three courses are outside the three circles; there are 5 of these.
(c) Students taking math, but neither English nor history, are in the blue region; there are 11 of these.
(d) Students taking English, but not history, are in the yellow and gray regions; there are 25 + 17 = 42 of these.
(e) Students taking only one of the three subjects are in the yellow, green, and blue regions; there are 25 + 10 + 11 = 46 of these.
(f) Students taking two of the three subjects are in the pink, gray, and brown regions; there are 9 + 17 + 20 = 46 of these.
EXERCISES 1D
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- In a deck of playing cards, how many cards are
a.
kings or queens?
b.
black cards or hearts?
c.
black cards or numerical cards?
d.
red cards or aces?
e.
diamonds or numerical cards?
f.
face cards or spades?
- A bowl of fruits has 8 apples and 5 oranges. How many fruits are either apples or oranges?
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Determine how many of the pieces contain
| a. none of the three items, | b. only a cherry, | |
| c. a nut and a raisin, but no cherry, | d. at least a raisin, but no cherry, | |
| e. at least a cherry or a nut, | f. neither a nut nor a raisin, | |
| g. exactly one of the three items, | h. at least one of the three items, | |
| i. exactly two of the three items, | j. at least two of the three items. |
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| Calculate how many cups
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Source: http://www2.hawaii.edu/~hile/setsd.htm
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