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)=100n(A)=48n(B)=46n(C)=55n(AB)=18n(AC)=20n(BC)=23n(ABC)=8. Noting that AB=AcapB Then what I did was: n(AcupBcupC)=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.

Modelfino0g

Modelfino0g

Answered question

2022-09-14

In a survey searching at reading conduct at a college, it become observed that 48% study mag A, 46% read mag B, 55% examine magazine C, 18% examine magazines A and B, 20% examine A and C, 23% examine B and C, ad 8% examine all three. what number read atleast one of the three magazines.
Then what I did was state:
n ( U ) = 100 n ( A ) = 48 n ( B ) = 46 n ( C ) = 55 n ( A B ) = 18 n ( A C ) = 20 n ( B C ) = 23 n ( A B C ) = 8.
Noting that A B = A B
Then what I did was:
n ( A B C ) = n ( A ) + n ( B ) + n ( C ) n ( A C ) n ( A B ) n ( B C ) + n ( A B C )
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.

Answer & Explanation

trestegp0

trestegp0

Beginner2022-09-15Added 12 answers

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 B C | = | A | + | B | + | C | | A C | | A B | | B C | + | A B C |
If we plug in the values you provided, then we have
| A B C | = 48 + 46 + 55 18 20 23 + 8 = 96
Note that the general form of the inclusion-exclusion principle is given by
| i = 1 n A i | = i = 1 n | A i | 1 i < j n | A i A j | + + 1 i < j < k n | A i A j A k | + ( 1 ) n 1 | A 1 A n |

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