Question

In a town of $10,000$ families it was found that $40\%$ families buy newspaper $A$, $20\%$ families buy newspaper $B$ and $10$ families buy news-paper $C$. $5\%$ families buy $A$ and $B$, $3\%$ buy $B$ and $C$ and $4\%$ buy $A$ and $C$. If $2\%$ families buy all the three newspapers, find the number of families which buy (i) $A$ only (ii) $B$ only (iii) none of $A$, $B$ and $C$.

Answer 1

Ans. $3300,1400,4000.$
Here we have
$N=10000,n\left(A\right)=40\%of10,000=4000,$
$n\left(B\right)=20\%of10000=2000$
$n\left(C\right)=10\%of10000=1000$
$n(A\cap B)=5\%of10000=500$
$n(B\cap C)=3\%of10000=300$
$n(C\cap A)=4\%of10000=400$
$n(A\cap B)=5\%of10000=500$
$n(B\cap C)=3\%of10000=300$
$n(C\cap A)=4\%of10000=400$
$n(A\cap B\cap C)=2\%of10000=200$
To find
(i) $n(A\cap \overline{B}\cap \overline{C})$
(ii) $n(\overline{A}\cap B\cap \overline{C})$
(iii) $n(\overline{A}\cap \overline{B}\cap \overline{C})$
We have
$n(A\cap \overline{B}\cap \overline{C})=[A\cap \overline{(B\cup C)}]$
$=n\left(A\right)-n[A\cap (B\cup C)]$
$=n\left(A\right)-n\{n(A\cap B)\cup (A\cap C)\}$
$=n\left(A\right)-[n(A\cap B)+n(A\cap C)-n(A\cap B)\cap (A\cap C)]$
$=n\left(A\right)-\{n(A\cap B)+n(A\cap C)-n(A\cap B\cap C)\}$
You may use the formula $\left[6\right]$ on $P.22$ directly
$=4000-500-400+200=3300$
$n(\overline{A}\cap B\cap \overline{C})=n[B\cap \left(\overline{A\cup C}\right)]$
$=n\left(B\right)-n[B\cap (A\cup C)]$
$=n\left(B\right)-n[(B\cap A)\cup (B\cap C)]$
$=n\left(B\right)-[n(A\cap B)+n(B\cap C)-n(A\cap B)\cap (B\cap C)]$
$=n\left(B\right)-[n(A\cap B)+n(B\cap C)-n(A\cap B\cap C)]$
$=2000-500-300+200=1400$
and $n(\overline{A}\cap B\cap \overline{C})=n\left(\overline{A\cup B\cup C}\right)$
$=N-n(A\cup B\cup C)$
$=N-\{n\left(A\right)+n\left(B\right)+n\left(C\right)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)\}$$=10000-4000-2000-1000+500+300+400-200$
$=4000.$