The formula for intersection is P(a or b) = P(a) + P(b) – P(a and b).

Now, 30 students out of 100 swim, so P(swim) = 30/100 = 3/10.

40 students run out of 100, so P(run) = 40/100 = 4/10. Notice how we are keeping 10 as the common denominator even though we could simplify this further. Keeping all of the fractions similar will make the addition and subtraction easier later on. 

Finally, 20 students swim AND run, so P(swim AND run) = 20/100 = 2/10. (Again, we keep this as 2/10 instead of 1/5 so that we can combine the three fractions more easily.)

P(swim OR run) = P(swim) + P(run) – P(swim and run)

                       = 3/10 + 4/10 – 2/10 = 5/10 = 1/2.

The intersection of two sets contains every element that is present in both sets, so  is the correct answer.

The idea is to draw a Venn Diagram and find the intersection. We have one circle of 70 and another with 40. When we add the two circles plus the 10 students who joined neither, we should get 100 students. However, when adding the two circles, we are adding the intersections twice, therefore we need to subtract the intersection once.

We get , which means the intersection is 20.

If  students are enrolled in sciences, but there are only  students then we must find how many overlap in the subjects they take.

To do this we can subtract  from .

Therefore,  of those "enrollments" must be doubles.

Those  students take both Chemistry and Biology. 

Draw a Venn diagram with three subsets:  Math, Science, and English.  Start in the center with students that like all three subjects.  Next, look at students that liked two subjects.  Be sure to subtract out the ones already counted in the middle.  Then, look at the students that only like one subject.  Be sure to subtract out the students already accounted for.  Once all of the subsets are filled, look at those students who don’t like any of these subjects.  To find the students who don’t like any of these subjects add all of the students who like at least one subject from the total number of students surveyed, which is 50. 

M = math

S = science

E = English

M∩S∩E = 3        

M∩S = 5 (but 3 are already accounted for) so 2 for M and S ONLY

M∩E = 7 (but 3 are already accounted for) so 4 for M and E ONLY

S∩E = 8 (but 3 are already accounted for) so 5 for S and E ONLY

M = 14 (but 3 + 2 + 4 are already accounted for) so 5 for M ONLY

S = 20 (but 3 + 2 + 5 are already accounted for) so 10 for S ONLY

E = 28 (but 3 + 4 + 5 are already accounted for) so 16 for E ONLY

Therefore, the students already accounted for is 3 + 2 +4 + 5 + 5 + 10 + 16 = 45 students

So, those students who don’t like any of these subjects are 50 – 45 = 5 students

A = {2, 4, 6, 8, 10, 12}

B = {3, 6, 9, 12, 15, 18}

The intersection of a set means that the elements are in both sets: A∩B = {6, 12}

This problem can be solved two ways, with a formula or with reason.

Using the formula, the intersection of the Venn diagram for which classes students take is:

By using reason, it is clear that 60 + 70 is greater than 100 by 30. It is assumed that this extra 30 students come from students who were counted twice because they took both classes.

The statement says all seniors take statistics so if you are a senior you are in statistics automatically.  It also said some baseball team members are seniors which means at least some teammates must be in statistics.

 represents the intersection of the two sets. In other words, we want all the elements that appear in both sets. The elements that appear in both sets are 2, 4, 6, and 10.

Therefore, 

First, subtract the students that are in neither class; 75 – 30 = 45 students.

Thus, 45 students are enrolled in Chemistry, Physics, or both.  Of these 45 students, we know 40 are in Chemistry, so that leaves 5 students who are enrolled in Physics only; with 15 total students in Physics, that means 10 must be in Chemistry as well. So 10 students are in both Physics and Chemistry.