SOLUTION: At a small midwestern college, a survey determined that 61% of freshmen participated in intercollegiate athletics. Further 48% participated in a campus social club. However, 18% d

Question 1153206: At a small midwestern college, a survey determined that 61% of freshmen participated in intercollegiate athletics. Further 48% participated in a campus social club. However, 18% did not participate in either type of activity. What percentage participated in either sports or a social club? From the same survey, what percentage of freshman participated in both sports and a social club?
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
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In my previous post,


    https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1153204.html


I just answered the first question, saying that the answer is the complement of 18% to 100%, i.e.  100% - 18% = 82%.


So, the union of the two sets, athletics and social, is 82% of the entire population.


Each of the components, athletic and social, is 61% and 48%, respectively.


Hence, their intersection is  61% + 48% - 82% = 27%.    ANSWER.

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The major fact you need to know to solve such problems is that

n(A U B) = n(A) + n(B) - n(A ∩ B)

for subsets A, B, (A U B) and (A ∩ B) of a universal set,
where n(X) is the number of elements of subset X.

See the lesson
- Counting elements in sub-sets of a given finite set
in this site.

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