Hey. I sort of used the min/max to solve this one, as an alternative way to the formula, which is what I usually do (no thinking involved...
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To begin with, I used 16 as the total number of students. This, based on the fractions we are given, dials down to:
A = 8
B = 10
C = 12
All three clubls = 3.
Looking at the venn diagram, we add 6 in the middle, as 3/8 are in all three clubs.
Focusing on A, we know that there should be 8 people there. 6 are already in A (in the union of the 3 clubs), so we are left with 2 people. Let's put 1 person only in A (the min possible number, as we know that 1 person is in at least one club) and 1 person in the union of A and B. This automatically means that there are 0 people in the union of A and C.
Moving on to B. There should be 10 people in B. Now, we already have 7 people in B. So, we need 3 more. Again using the min possible number, we add 1 person in B alone, which means that there should be 2 people in the union of B and C.
Finally, we need 12 people for C. We already have 8 people in C, so we still need to add 4 people. As the union of A and C includes 0 people, all 4 remaining people will be in C alone.
The question asks for the fraction of those in exactly 2 clubs (the unions between 2 different clubs), relative to the total number of people. This is 3/16 = 1/8. So,
ANS A
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