Question

In a city$20\%$ people read English newspapers,$40\%$ read Hindi newspapers and $5\%$read both newspapers. The percentage of non-reader either paper is

• A. $60\%$
• B. $35\%$
• C. $25\%$
• D. $45\%$

Answer 1

The correct option is D

$45\%$

Explanation for correct answer:

Find the percentage of non-reader of either paper

The probability that people read English newspapers,

$$\begin{gathered} P\left(E\right)=20\% \\ =\frac{1}{5} \end{gathered}$$

The probability that people read Hindi newspapers,

$$\begin{gathered} P\left(H\right)=40\% \\ =\frac{2}{5} \end{gathered}$$

The probability that people read both English and Hindi newspapers,

$$\begin{gathered} P(E\cap H)=5\% \\ =\frac{1}{20} \end{gathered}$$

We need to find$P(E’\cap H’)$

$$\begin{gathered} \mathit{P}\mathbf{(}\mathit{E}\mathbf{’}\mathbf{\cap }\mathit{H}\mathbf{’}\mathbf{)}\mathbf{=}\mathbf{1}\mathbf{-}\mathit{P}\mathbf{(}\mathit{E}\mathbf{\cup }\mathit{H}\mathbf{)}\mathbf{} \\ P(E’\cap H’)=1-\left[P\right(E)+P(H)-P(E\cap H\left)\right] \end{gathered}$$

Now, substitute the values, we get

$$\begin{gathered} P(E’\cap H’)=1-\left[\right(\frac{1}{5})+(\frac{2}{5})-(\frac{1}{20}\left)\right] \\ P(E’\cap H’)=\frac{9}{20} \end{gathered}$$

Thus, the percentage of non-reader of either paper

$$\begin{gathered} =0.45\left(100\right) \\ =45\% \end{gathered}$$

Hence, the correct option is D.