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I am trying to solve the following question;

There are 3 news papers in a town, A, B, and C. 20% of the population read A, 16% read B, and 14% read C.

8% read A and B

5% read A and C

4% read B and C

2% read all

Determine the probability that a person reads none of the news papers.

I've realised I need to calculate what percentage of the population reads news papers.

To do this I've done the following;

A = 20 - 8 - 5 = 7

B = 16 - 8 - 4 = 4

C = 14 - 5 - 4 = 5

Population that reads News papers = 7 + 4 + 5 = 16

Therefore, 84 / 100 = probability a person doesn't read any news papers.

However, my answer is wrong. What is the best way to approach this problem?

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    $\begingroup$ Make a Venn diagram and all will be clear. $\endgroup$
    – Arby
    Mar 4 '17 at 4:23
  • $\begingroup$ Are you familiar with inclusion/exclusion? $\endgroup$ Mar 4 '17 at 4:26
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In these type of questions you should approach by set theory and Venn diagram.

The percentage of people who read any of the newspapers is given by "$A$ or $B$ or $C$."

$A \cup B \cup C = A+B+C-(A \cap B)-(A \cap C)-(B \cap C)+(A \cap B \cap C)$ $\implies A \cup B \cup C = (20+16+14-8-5-4+2)\%=35\%$

Therefore the probability that a person reads none of the newspapers $= 65/100 = 0.65$.

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$P(A\cup B\cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(A \cap C) + P(A \cap B \cap C)$

$P(A\cup B\cup C) = 20 + 16 + 14 - 8 - 5 - 4 + 2$

$P(A\cup B\cup C) = 35$

Person reads no newspaper $= 100 - 35 = 65$

Hope you can proceed further.

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  • $\begingroup$ @Ross Millikan thank u. $\endgroup$
    – Amar
    Mar 4 '17 at 5:02
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It should be clear that you can't have $84\%$ that don't read any papers because $20\%$ read A and that is already too many. Your variable A is intended to be (you should define this) the number that read A and not either B or C. When you did $20-8-5$ you subtracted the people who read all three papers twice, once for reading A and B (the $8$) and once for reading A and C (the $5$), so you need to add them back in once. That means the number that read only A is $20-8-5+2=9$. When you compute the population that reads any paper, you need to add your A,B,C (correctly calculated) but also all the people that read more than one paper. You should draw a Venn diagram to help the computation. The point of the exercise is the inclusion-exclusion principle.

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$$A \cup B \cup C = A+B+C-(A \cap B)-(A \cap C)-(B \cap C)+(A \cap B \cap C)$$ $$\implies A \cup B \cup C = (20+16+14-8-5-4+2)\%=35\%$$ Therefore, required percent $= 65%$

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