In order to find the intersection of chocolate and vanilla, it is easiest to make a Venn Diagram. The outside of the Venn Diagram is 10, and the total of the entire diagram must equal 35. Therefore the two circles of the Venn Diagram including just chocolate, just vanilla and the intersection must equal 25, with the just chocolate plus intersection side equalling 15 and the just vanilla plus intersection side equalling 13.

We know:

(A U B) = A + B – (A ∩ B)

We have found that (A U B) = 25 and we are trying to find (A ∩ B). Plug in A and B

25 = 15 + 13 – (A ∩ B) = 28 – (A ∩ B)

or – (A ∩ B) = –3 

(A ∩ B) = 3

Refer to the following Venn Diagram:

If 25% of the 800 CS students are also mathematics students, the number of students sharing these majors is 800 * 0.25 or 200 students. Furthermore, if this represents one third of the total of math students, we then know:

Math students * 1/3 = 200 or (1/3)M = 200

Solving for M we get 600.  This means that the number of students that are ONLY math students is 400.

Looking at our diagram above, we must be careful not to "double add" the intersection. The easiest way to do this is to take the intersection and add to it the number of CS-only and math-only students: 600 + 200 + 400 = 1200.  This number represents the total number of students that have either a math or CS major (that is, the number of students in the union of the two sets). This leaves 1500 – 1200 or 300 students.

Probability (tabby but not pregnant)

= Prob (tabby) – Prob (tabby and pregnant)

= 10% - 3%

= 7%

A way of solving this problem is by drawing a Venn diagram based on what is known:

There are  customers; within them there are those that purchased something (pants, shirts, or possibly a combination), and those that purchased nothing.

The amount of individuals that purchased something is given as:

The reason that the number of shoppers that bought both is subtracted is so that it is not counted twice when the customers that bought pants and the customers that bought shirts are added together.

Since  bought nothing,

The amount that bought both then is:

When you add both class rosters you get a total of 55 students 

.

You must subtract the 7 that are in both because they are counted twice in the 55 

.  

The total in neither class will be the total students minus the adjusted enrollment in both classes 

.

The easiest way to understand this problem is to draw a Venn Diagram:

S = Total number of students

A = Total students taking ethics

B = Total students taking metaphysics

A - B = Students taking only ethics

B - A = Students taking only metaphysics

A ∪ B = The total students taking either ethics or metaphysics

A ∩ B = The total students taking both ethics and metaphysics

We know there are 15,000 total and that 9,500 are taking neither class.  Therefore, we know that 15,000 - 9,500 = 5,500 are taking at least one of the classes.  Based on our prompt, we know that there are 1,700 taking both and that 2,200 are taking ethics.  To fiind out how many are taking ONLY ethics, we have to subtract off the amount that are taking ethics and metaphysics.  Hence, 2,200 - 1,700 = 500.  Finally, if we know that there are 5,500 taking at at least one of these classes, we want to get rid of that portion taking ethics.  This will leave us with those who are taking at least metaphysics (regardless of whether or not they are taking ethics): 5,500 - 500 = 5,000.

In order to determine how many students are not enrolled in a sport, we must first determine how many students are. The simplest way to do this is to begin by adding the students of both sports together.

43 + 37 = 80

But wait! 9 of those students are play both baseball and basketball. To avoid double counting these students, subtract 9 from the total.  

80 – 9 = 71

Now we know 71 students play sports. If there are 100 students, all that's left to do is subtract.

100 – 71 = 29

29 students do not play basketball or baseball.

You could represent this question in the following Venn Diagram:

We know that the two circles must contain a total of  students.  Now, since there is an extra case of the overlap area when you add together the German and the Latin students, you can say:

This is the simple answer to this question!  You do not need to compute the overlap at all, as you are merely looking for the contents of the two circles.