In a statistical investigation of 1003 families of Calcutta, it was...

Let $\mathrm{R}$ be the set of families having a radio and $T$ the set families having a T.V. then $\mathrm{n}(\mathrm{R} \cup \mathrm{T})=$ The number of families having at least on of the radio and $\mathrm{T} \mathrm{V}=1003-63=940n(\mathrm{R})=794$ and $\mathrm{n}(\mathrm{T})=187$ Let $x$ families have both a radio and a T.V. Then number of families who have only radio $=794-x$ And the number of families who have only T.V. $=187-x$ Erom Venn diagram, $794-x+x-187-x=940\Rightarrow 981-x=940$ or $x=981-940=41$ Hence, the required number of families having both a radio and a T.V.=41