Question 1198177: During one year in a school, 5/8 of the students had measles, 1/2 had chickenpox and 1/8 had neither. What fraction of the school had both measles and chickenpox?
Found 2 solutions by math_tutor2020, MathLover1:
Answer by math_tutor2020(3817)   (Show Source):
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Answer: 1/4
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Explanation:
Consider a very small school with 8 students total.
1/8 of the students had neither condition.
In other words, 1 student (of the 8 total) had neither measles nor chickenpox.
This must mean 8-1 = 7 students had either condition, or both.
5/8 of the students had measles, aka 5 out of the 8 had measles
1/2 = 4/8 of the students had chickenpox, aka 4 out of the 8 got it
In short:
5 students have measles
4 students have chickenpox
In this zoomed-in context, it's not clear if there is any overlap between these two groups.
Naively we calculate that
5+4 = 9 students have either condition
But clearly we exceeded the number who have either (7) and even the total school population (8).
This overcounting error is due to the fact we double-counted at least one individual who has both conditions.
So there must be overlap going on.
I recommend drawing out a Venn Diagram.
A = number who have either or both = 7
B = number who have measles = 5
C = number who have chickenpox = 4
D = number who have both = x
A = B+C - D
7 = 5+4 - x
7 = 9-x
7+x = 9
x = 9-7
x = 2 students, of the 8 total, have both conditions.
2/8 = 1/4 of the students have both measles and chickenpox.
Answer by MathLover1(20850)  (Show Source):
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