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! .
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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