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 student is selected at random.
(a) Find the probability that she reads neither Hindi nor English news papers.
(b) If she reads Hindi news paper, find the probability that she reads English news paper.
(c) If she reads English news paper, find the probability that she reads Hindi news paper.

• A. $0.56,0.78,0.76$
• B. $0.20,0.33,0.50$
• C. $0.65,0.45,0.34$
• D. $0.56,0.56,0.65$

Answer 1

The correct option is B $0.20,0.33,0.50$

Let $H$ denote the students who read Hindi newspaper and $E$ denote the students who read English newspaper.

It is given that,

$P\left(H\right)=60$% $=\frac{6}{10}=\frac{3}{5}$

$P\left(E\right)=40$% $=\frac{40}{100}=\frac{2}{5}$

$P(H\cap E)=20$% $=\frac{20}{100}=\frac{1}{5}$

(i) Probability that a student reads Hindi or English newspaper is,
$(H\cup E)\prime =1-P(H\cup E)$

$=1-\{P\left(H\right)+P\left(E\right)-P(H\cap E)\}$

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

$=1-\frac{4}{5}$

$=\frac{1}{5}=0.20$

(ii) Probability that a randomly chosen student reads English newspaper, if she reads Hindi news paper, is given by $P\left(E|H\right)$

$P\left(E|H\right)=\frac{P(E\cap H)}{P\left(H\right)}$

$=\frac{\frac{1}{5}}{\frac{3}{5}}$

$=\frac{1}{3}=0.33$

(iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by $P\left(H|E\right)$

$P\left(H|E\right)=\frac{P(H\cap E)}{P\left(E\right)}$

$=\frac{\frac{1}{5}}{\frac{2}{5}}$

$=\frac{1}{2}=0.50$