The question is:
In a survey looking at reading habits at a university, it was found that 48% read magazine A, 46% read magazine B, 55% read magazine C, 18% read magazines A and B, 20% read A and C, 23% read B and C, ad 8% read all three. What percentage read atleast one of the 3 magazines.
Then what I did was state: $$ n(U) = 100 \\ n(A) = 48 \\ n(B) = 46 \\ n(C) = 55 \\ n(AB) = 18 \\ n(AC)= 20 \\ n(BC) = 23 \\ n(ABC) = 8.$$ Noting that $AB = A\cap B $
Then what I did was: $$ n(A\cup B\cup C) = n(A) + n(B) +n(C) - n(AC) - n(AB) - n(BC) + n(ABC) $$ when I did this I got that 96% read either A, B, or C and I was just wondering if I had made an error because it seemed a bit high when I first did it.
Yes. You are correct, and your approach is correct. The inclusion-exclusion principle is a method used to calculate the number of elements in the union of finite sets. For three sets, the inclusion-exclusion principle is given by
$$|A\cup B\cup C| = |A| + |B| + |C| - |A \cap C| - |A \cap B| - |B \cap C| + |A \cap B \cap C|$$
If we plug in the values you provided, then we have
$$|A\cup B\cup C| = 48+46+55-18-20-23+8 = 96$$
This is exactly your answer.
Note that the general form of the inclusion-exclusion principle is given by
$$\left|\bigcup _{i=1}^{n}A_{i}\right|={} \sum _{i=1}^{n}|A_{i}|-\sum _{1\leq i<j\leq n}|A_{i}\cap A_{j}|+\cdots {}\cdots +\sum _{1\leq i<j<k\leq n}|A_{i}\cap A_{j}\cap A_{k}|-\cdots +(-1)^{n-1}\left|A_{1}\cap \cdots \cap A_{n}\right|$$