Question

In a hostel 60% of the students read Hindi news paper, 40% read English news paper and 20% read both Hindi and English news papers. A students is selected at random.
Find the probability that the reads neither Hindi nor English news papers.

Answer 1

Let the no. of students in Hostel$=P$

No. of students who read Hindi$=\frac{60}{100}\times P=\frac{3P}{5}$

No. of students who read English$=\frac{40}{100}\times P=\frac{2P}{5}$

No. of students who read both$=\frac{20}{100}\times P=\frac{P}{5}$

P(Hindi)$=\frac{n\left(Hindi\right)}{total}=\frac{3P}{5\times P}=\frac{3}{5}$

P(English)$=\frac{n\left(English\right)}{total}=\frac{2P}{5\times P}=\frac{2}{5}$

P(Both)$=\frac{n\left(both\right)}{total}=\frac{P}{5P}=\frac{1}{5}$

$P\left(\overline{Hindi\cup English}\right)=1-P(English\cup Hindi)$

$=1-(P\left(English\right)+P(Hindi)-P(both\left)\right)$

$=1-(\frac{2}{5}+\frac{3}{5}-\frac{1}{5})$

$=1-1+\frac{1}{5}=\frac{1}{5}$

$=0.2$