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CAT 2016 | Question: 81

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Given that,

  • $n(\text{Cable TV)} = \dfrac{2}{3}$
  • $n(\text{VCR})  = \dfrac{1}{5}$
  • $n(\text{Cable TV $\cap$ VCR})  = \dfrac{1}{10}$

Lets draw the Venn diagram.

Now, the fraction of people having either Cable TV or VCR

$\qquad = n(\text{Cable TV $\cup$ VCR})-n(\text{Cable TV $\cap$ VCR})$

$\qquad = n(\text{Cable TV)+n(VCR)}-n(\text{Cable TV $\cap$ VCR})-n(\text{Cable TV $\cap$ VCR})$

$\qquad = \dfrac{2}{3}+\dfrac{1}{5}-\dfrac{1}{10}-\dfrac{1}{10}$

$\qquad = \dfrac{2}{3}+\dfrac{1}{5}-\dfrac{2}{10}$

$\qquad = \dfrac{2}{3}+\dfrac{1}{5}-\dfrac{1}{5}$

$\qquad = \dfrac{2}{3}$

$$\textbf{(OR)}$$

The fraction of people having either Cable TV or VCR

$\qquad = n(\text{Cable TV})-n(\text{Cable TV $\cap$ VCR})+n(\text{VCR})-n(\text{Cable TV $\cap$ VCR})$

$\qquad = \dfrac{2}{3}-\dfrac{1}{10}+\dfrac{1}{5}-\dfrac{1}{10}$

$\qquad = \dfrac{2}{3}$

Correct Answer $:\text{B}$

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