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 $:\text{B}$