Question

In a town of $10,000$ families it was found that $40\%$ family buy newspaper $A$, $20\%$ families buy newspaper $B$ and $10\%$ families buy newspaper $C$, $5\%$ families buy $A$ and $B$, $3\%$ buy and $C$ and $4\%$ buy $A$ and $C$.

If $2\%$ families buy all the three newspapers, then the number of families which buy $A$ only is

• A. $3100$
• B. $3300$
• C. $2900$
• D. $1400$

Answer 1

The correct option is B

$3300$

Explanation for the correct option:

Step 1: Define all possible event

Total number of families = $10,000$

$$\begin{gathered} n\left(A\right)=40\% \\ n\left(A\right)=\frac{\left[40\times 10000\right]}{100}=4000 \end{gathered}$$

$$\begin{gathered} n\left(B\right)=20\% \\ n\left(B\right)=\frac{\left[20\times 10000\right]}{100}=2000 \end{gathered}$$

$$\begin{gathered} n\left(C\right)=10\% \\ n\left(C\right)=\frac{\left[10\times 10000\right]}{100}1000 \end{gathered}$$

Now,

$$\begin{gathered} n\left(A\cap B\right)=5\% \\ n\left(A\cap B\right)=\frac{\left[5\times 10000\right]}{100}=500 \end{gathered}$$

$$\begin{gathered} n\left(A\cap C\right)=4\% \\ n\left(A\cap C\right)=\frac{\left[4\times 10000\right]}{100}=400 \end{gathered}$$

$$\begin{gathered} n\left(B\cap C\right)=3\% \\ n\left(B\cap C\right)=\frac{\left[3\times 10000\right]}{100}=300 \end{gathered}$$

$$\begin{gathered} n\left(A\cap B\cap C\right)=2\% \\ n\left(A\cap B\cap C\right)=\frac{\left[2\times 10000\right]}{100}=200 \end{gathered}$$

Step 2: Determining the number of families which buy only $A$ newspaper

$$\begin{gathered} n\left(A\cap B^c\cap C^c\right)=n\left[A\cap {\left(B\cup C\right)}^c\right] \\ n\left(A\cap B^c\cap C^c\right)=n\left(A\right)-n\left[A\cap \left(B\cup C\right)\right] \\ n\left(A\cap B^c\cap C^c\right)=n\left(A\right)-n\left\{n\left(A\cap B\right)\cup \left(A\cap C\right)\right\} \\ n\left(A\cap B^c\cap C^c\right)=n\left(A\right)-\left\{n\left(A\cap B\right)+n\left(A\cap C\right)-n\left(A\cap B\cap C\right)\right\} \\ n\left(A\cap B^c\cap C^c\right)=4000-\left\{500+400-200\right\} \\ n\left(A\cap B^c\cap C^c\right)=3300 \end{gathered}$$

Hence, the correct option is B.