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*similar to problem 3 from section 1.3 of your text

1.

In a universal set U, assume that n(U) = 100, n(A') = 42, n(B) = 55, and n(A ∪ B) = 94. Find n(A ∩ B).

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Many problems in finite will require you to draw on your knowledge of the relationships between things in order to identify information you need that isn't explicitly stated. In this problem, for example, we're asked to calculate n(A ∩ B). We know from earlier in the chapter that A ∩ B appears in the equation for calculating the union of two sets: n(A) + n(B) - n(A ∩ B) = n(A ∪ B). If we know the value of every other set in that equation, we can solve for the value of (A ∩ B).

So we know n(B) = 55 and we know n(A ∪ B) = 94, but how do we solve for n(A) when we're given n(A')? Well, recall that any set and its complement will total U when added together. So if n(A) + n(A') = n(U), then we have n(A) + 42 = 100. Subtract 42 from both sides of the equation, and we get n(A) = 58.

With that, we can now plug each value into the union equation and solve for the intersection. We have 58 + 55 - n(A ∩ B) = 94. That reduces to 113 - n(A ∩ B) = 94. Subtract 113 from both sides and divide by -1 to eliminate the negative signs, and we get n(A ∩ B) = 19.

*similar to problem 6 from section 1.3 of your text

2.

In a universal set U, assume that A and B are disjoint subsets and that n(U) = 100, n(A') = 81, n(B) = 24. Find n(A' ∩ B).

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This problem is interesting because it's actually a bit of a trick question. If A and B are disjoint subsets, then they have no intersection and, thus, no shared elements in common. And since the set A' ∩ B refers to all the elements in B that are also not in A, that's just B -- because every element in B is, by definition, also not in A since we know the two sets have no elements in common.

So n(A' ∩ B) = n(B), which equals 24.

*similar to problem 15 from section 1.3 of your text

3.

Assume 159 students who were denied direct admission to the Kelley School of Business chose to pursue a different degree instead. Of these students, 74 expressed interest in a Media Studies degree, 62 expressed interest in an English degree, and 24 expressed interest in both of these alternatives. How many were interested in neither alternative degree?

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Let's designate the students who expressed interest in Media Studies as n(M) = 74, the students who expressed interest in English as n(E) = 62, and the students who expressed interest in both as n(M ∩ E) = 24.

Because we want to know the number of students interested in neither degree, we need to calculate that section of the Venn diagram outside the two circles (the union of the two subsets) but within the universal set, which, as you may recall from prior questions, we designate n(M ∪ E)'.

So how do we find that? Well, we should first recognize that if we're looking for n(M ∪ E)' and we know the value of n(U), then if we could find n(M ∪ E), it would have to equal n(U) when added to n(M ∪ E)' -- the sum of two complementary sets will always equal the universal set. Now we need to ask ourselves, "Do I have all the information needed to find n(M ∪ E)?" And the answer to that is yes.

We know that n(M ∪ E) = n(M) + n(E) - n(M ∩ E), and we're given each of those values. Plugging them into the equation, we have n(M ∪ E) = 74 + 62 - 24, which equals 112.

Now we can plug that new information into the equation n(M ∪ E) + n(M ∪ E)' = n(U), giving us 112 + n(M ∪ E)' = 159. By subtracting 112 from both sides, we can determine that n(M ∪ E)' -- the number of students interested in neither alternative degree -- is 47.

*similar to problem 17 from section 1.3 of your text

4.

In a universal set U, assume that n(U) = 100, n(A) = 39, and n(B’) = 66. Assume further that there are 27 elements of U in A which are not in B. Find the number of elements in B which are not in A.

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The primary function of this question seems to be training you to represent in mathematical terms what's been stated in plain English. In other words, how do you represent "elements of U in A which are not in B" mathematically? Well, it's those elements of A (A) that are simultaneously (∩) -- or that it has in common with being -- not in B (B'), which we designate A ∩ B'. Conversely, we designate the "number of elements in B which are not in A" as n(A' ∩ B). So now that we have everything notated, let's set our equations to solve for what we're missing.

Set B is divided into two parts: its intersection with A (A ∩ B) and what it doesn't have in common with A (A' ∩ B). When you add the number of elements in each part together, you get the total amount in B. So to figure out the number of elements B doesn't share with A (A' ∩ B), we need to know the amount in B as a whole, as well as in its intersection with A.

Because we're told n(U) = 100 and n(B') = 66, we can quickly plug those values into the equation n(B) + n(B') = n(U) and determine that n(B) = 34. To find n(A ∩ B), we'll need to use the same logic on set A that we're using on set B: its two parts -- what it has in common with B (A ∩ B) and what it doesn't (A ∩ B') -- will total the whole of A when added together.

Knowing n(A) = 39 and n(A ∩ B') = 27, we can plug those values into the equation n(A ∩ B') + n(A ∩ B) = n(A), giving us 27 + n(A ∩ B) = 39. Subtracting 27 from both sides reveals that n(A ∩ B) = 12.

Lastly, we plug that number into the same equation for B -- n(A' ∩ B) + n(A ∩ B) = n(B) -- giving us n(A' ∩ B) + 12 = 34, and n(A' ∩ B) = 22.

*similar to problem 25 from section 1.3 of your text

5.

Assume 194 students were asked to complete a survey about their first-year college experience. Of these students, 97 responded positively to the campus climate,  102 responded positively to social activities, and 78 responded positively to academics. Also, 65 responded positively to both climate and social activities, 41 to climate and academics, 17 to social activities and academics, and 25 to none of the items. How many responded positively to all three?

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If this were a question with only two subsets instead of three, we would simply use the equation n(A ∪ B) = n(A) + n(B) - n(A ∩ B) to solve for the intersection of A and B, provided that we're given the values of all the other sets in that equation. Here, that remains true, only by adding a third subset, we'll have to use the logic behind that equation twice. Here's why:

We're told that 25 students responded positively to none of the three subsets, which represents the section outside the three circles of your Venn diagram but within the universal set. Assuming we notate the campus climate set as C, the social activities set as S, and the academics set as A, then we have n(C ∪ S ∪ A)' = 25, and that means the union of the three sets -- n(C ∪ S ∪ A) -- equals 169 because it will total n(U) = 194 when added to its complement.

So, using the logic of the union formula I described at the beginning, if we add the number in each of the three sets -- n(C) + n(S) + n(A) -- then subtract from their union, we discover that there are 108 students who fall somewhere in the intersections of these three sets. Keep in mind, this does not represent n(C ∩ S ∩ A) because that refers to the intersection of all three sets, and it might be possible that some of these 108 students fall in the intersection of two of the sets but not the third. Therefore, we'll need to use the logic of the union formula a second time.

Because we know 108 students fall somewhere in the intersections of these three sets, when we add those intersections together -- n(C ∩ S) + n(C ∩ A) + n(S ∩ A) -- we'll see that their sum (123) exceeds the amount of students we know to be in these intersections by 15 (123 - 108 = 15). That tell us there are 15 students who responded positively to all three aspects of their first-year experience.

*similar to problem 27 from section 1.3 of your text

6.

In a universal set U, assume n(U) = 100, n(A) = 30, n(A ∩ C) = 15, n(B) = 33, n((A ∪ C)') = 39, and ((A ∪ C) ∩ B) = Ø. Find n(B'), n(A ∩ B), and n(A' ∩ C).

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Like so many problems in this class, this one is partially a trick and the way the information is presented makes it seem far more complicated than it actually is. Without even thinking, start with what you're first asked to find: n(B'). This is an easy one because we know n(B) + n(B') = n(U), and since we can easily substitute 100 for n(U) and 33 for n(B), we know n(B') must be 67. One down, two to go!

Finding n(A ∩ B) is also an easy one because it's a trick. We're told ((A ∪ C) ∩ B) = Ø. And while it may look weird, all that's saying is that not a single element in B is shared by either A or C (since the two sets have been joined as a union). In other words, B is completely by itself, all alone (#relatable). So if B doesn't intersect with the union of A and C, it sure as shit doesn't intersect with just A. The answer to n(A ∩ B) is also Ø.

Finding n(A' ∩ C) is the only one that requires a little bit more math. We now know that only A and C overlap (in fact, we're told their intersection is 15), so we can treat this as we have any other problem with only two sets. We know n(A) = 30, so all we need to complete  our union formula is n(C) and n(A ∪ C). Fortunately, we're told the complement of n(A ∪ C), which we can just subtract from n(U), giving us n(A ∪ C) = 61 (if I lost you there, go back and review my more in depth explanations of this rule in Questions 1-5 until you can move as quickly as we have here). From there, we can plug our known values into the union formula to solve for n(C), which we learn to be 46.

Lastly, we determine n(A' ∩ C) by subtracting n(A ∩ C) from n(C) because the value of what C has in common with A and what C doesn't have in common with A will total the whole of C when added together. Therefore, 46 - 15 = n(A' ∩ C), which equals 31.

7.

Let X, Y, Z ⊂ U. If n(U) = 80, n(X) = 27, n(Y) = 22, n(Z) = 41, n(X ∩ Y) = 10, n(X ∩ Z) = 12, n(Y ∩ Z) = 13, and n(X ∩ Y ∩ Z) = 4, find n(X' ∩ Y' ∩ Z), n(X’ ∩ (Y' ∪ Z)), n(Y’), and n(X ∪ Y' ∪ Z').

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This problem gives you all the information you need to solve, so really, it's just an exercise in understanding what each section of the Venn diagram is called and their mathematical relationship to other sections.

To begin, I want to suggest that when you read something like n(X ∩ Y ∩ Z) in your head, don't read it as, "X intersect Y intersect Z." Instead, I recommend reading it as, "elements that are in X and and Z" because to exist in their intersection, the elements must be in all three sets simultaneously. And when you read n(X ∪ Y ∪ Z), read it as, "elements that are in X or or Z" because the union of those three sets joins together all the elements in each, so any one particular element may have come from any of those sets -- X, Y, or Z.

Hopefully that makes solving for n(X' ∩ Y' ∩ Z) less intimidating because it's simply elements that are not in X and not in Y and are in Z. When you draw your Venn diagram, you'll see that Z is divided into four sections: one that overlaps with both X and Y (the center), one that overlaps with just X, one that overlaps with just Y, and one that doesn't overlap with either . Of those four sections, only one includes elements that are neither in X nor Y, and that's the last section, the one by itself.

To solve for that section, put a 4 in the center (X ∩ Y ∩ Z) and then subtract it from n(X ∩ Z). The 8 that remain go in the section of Z that only overlaps with X but not Y, notated (X ∩ Y' ∩ Z). Repeat this process for n(Y ∩ Z), placing a 9 in the section of Z that only overlaps with Y but not X, notated (X' ∩ Y ∩ Z). Now that you've filled in three of the four sections of Z, you can add them together (8 + 4 + 9 = 21) and subtract from the total number of elements in Z (41), giving you 20 elements in the section by itself, notated (X' ∩ Y' ∩ Z).

Now let's pick up the pace. We're then asked to solve for n(X' ∩ (Y' ∪ Z)), which refers to all the elements that are either not in Y or are in Z and not in X. Looking at our Venn diagram, the only sections for which that's true are the two sections of Z that don't overlap with X (one of which overlaps with Y, but that's okay because we want elements in Z or not in Y) and the section outside the sets but within the universal set. We already know that the section of Z by itself contains 20 elements and the section that overlaps with Y but not X contains 9 elements. By using the method described above to fill in the remaining sections of the Venn diagram, we can add them all together and conclude that n(X ∪ Y ∪ Z) = 59. Because n(U) = 80, that must mean there are 21 elements in the universal set that aren't in any subset. Adding those to the elements from Z, we get n(X' ∩ (Y' ∪ Z)) = 50.

To find n(Y') simply subtract n(Y) from n(U): 80 - 22 = 58.

Finally, you could waste a lot of time trying to count up all the elements in n(X ∪ Y' ∪ Z'), or you could use the complement rule I described in Question 3 of Episode 1.2. By reversing the complement of each set and flipping the union signs, we know that the complement to (X ∪ Y' ∪ Z') is (X' ∩ Y ∩ Z) and that, when added, the value of each will total the universal set. Since we've already found (X' ∩ Y ∩ Z) -- the section of Z that overlaps with Y but not X -- we can subtract it from 80 and find that n(X ∪ Y' ∪ Z') = 71.

8.

Let A, B, and C be subsets of a universal set U with A and B disjoint, n(U) = 100, n(A) = 29, n(B) = 35, n(A ∪ B ∪ C) = 81, and n((A ∪ B) ∩ C) = 21. Find n(C).

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Because A and B don't overlap, this problem is actually easier than it looks. Recall from our knowledge of the union formula that if we had the number of elements in each set together and subtract their intersection, we'll get their union. If we know the union but not one of the other sets, we can work the equation backward and solve, which is exactly what we'll do here.

The trick to solving this problem really lies in recognizing what n((A ∪ B) ∩ C) refers to, which are the elements that are in either A or B and in C. In other words, it's the union of the two intersections with C, which you might think of as ((A ∩ C) ∪ (A ∩ B)). It doesn't matter how many of those 21 elements are in (A ∩ B) vs (A ∩ C), only that our union formula can be assembled thusly: n(A) + n(B) + n(C) - n((A ∪ B) ∩ C) = n(A ∪ B ∪ C). Plugging in our known values, we have 29 + 35 + n(C) - 21= 81. By reducing and adding 21 to both sides, we have 64 + n(C) = 102. After subtracting 64, we learn n(C) = 28.

9.

Let A, B, and C be distinct subsets of a universal set U with A ⊂ B ⊂ C. Also suppose n(A) = 2 and n(C) = 5. In how many different ways can you select B such that n(B) = 3? n(B) = 4?

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This problem sounds a bit strange to describe, but it's actually extremely easy to visualize. Set A is the smallest circle, drawn inside a slightly larger circle for set B, drawn inside an even larger circle for set C.

Now, set C has 5 elements, and we know 2 of those elements are in set A. Set B can either have 3 or 4 elements (because if it had all 5, it would be Set C, and if it had 2 it would either be Set A or Set A wouldn't be a subset of B).

So since set A is using 2 of the 5 elements, that means there are 3 elements left in C we can choose from to allocate to set B. In order to make n(B) = 3, we would need to assign 1 of those 3 elements left in C to set B. How many ways are there to do that? Well, 3 because you could either choose the first, second, or third.

In order to make n(B) = 4, we would need to assign 2 of those 3 elements left in C to set B. And there are also 3 ways to do that because we could either choose the first and second elements, the first and third elements, or the second and third elements - in other words, 3 unique pairs.

10.

Let A and B be subsets of a universal set U, such that n(U) = 100, n(A) = 40, and n(A' ∪ B) = 75. Find (A ∩ B) and n(A ∩ B').

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The easiest way to solve this problem is by first solving for n(A ∩ B'), which we can do by remembering the complement rule. The complement of n(A' ∪ B) is n(A ∩ B') -- we know this because we flipped the union sign to intersection and reversed the complement symbol of each subset. Because we know n(A' ∪ B) + n(A ∩ B') = n(U), we can plug in our known values and solve for n(A ∩ B'), which is 25.

We should then recognize that n(A ∩ B) and n(A ∩ B') are the two pieces of set A that will total n(A) when added together. Again, we plug in our known values and get n(A ∩ B) + 25 = 40. Subtracting 25 from both sides, we learn n(A ∩ B) = 15.

11.

On the floor of your residence hall, 2/5 of students plan to major in business, 3/10 plan to major in computer science, and 1/8 plan to major in both. What share of students plan to major in exactly one of either business or computer science?

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There are a couple of ways to solve this problem, but in both cases, we need to first visualize the problem as a Venn diagram. In one circle, we have the share of students planning to major in business; in the other, the share planning to major in computer science; and in the middle, the share planning to do both.

The first method is to find the union of the two majors (thus, the total number of students in the problem). To find the union, per the formula described in the tutorial for this topic, we add together the share in each major, then subtract the number in their intersection. This gives us 2/5 + 3/10 - 1/8 = 23/40.

Next, we subtract the number in the intersection again because the problem asks for only those who are majoring in exactly one major (and those in the intersection are majoring in two). Therefore, 23/40 - 1/8 = 9/20.

The other way to solve this would be to draw our Venn diagram, which has three sections (the intersection, the section for only business majors, and the section for only computer science majors), then fill in each section and add together the two sections of students with only one major.

The intersection, as we know, is 1/8. And to find the sections for students with only one major, we must take the share of students in each major and subtract the intersection. For students in business, we have 2/5 - 1/8 = 11/40. For students in computer science, we have 3/10 - 1/8 = 7/40.

Then, after adding together 11/40 and 7/40, we get 18/40, which reduces to 9/20, same as above.

12.

In your graduating high school class, 149 students took AP Calculus, 102 students took AP History, and 54 students took both. How many students took exactly one of these two courses?

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This problem is solved exactly like Question 11 above, so please refer to my answer there for a full explanation.

In short, though, you should identify the number of students who took only AP Calc by subtracting the intersection from the total in that set (149 - 54 = 95), then add it to the number of students who took only AP History, which you find the same way as with AP Calc (102 - 54 = 48). Thus, the number of students who took exactly one of these courses is 95 + 48 = 143.

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