Question

In a hostel \(60\%\) of the students read Hindi newspaper,\(40\%\) read English newspaper and \(20\%\) read both Hindi and English newspapers.A student is selected at random.
(a)Find the probability that she reads neither Hindi nor English newspapers.
(b)If she reads Hindi newspaper,find the probability that she reads English newspaper.
(c) If she reads English newspapers,find the probability she reads Hindi newspaper.

Answer 1

Let H denote the students who read Hindi newspaper and E denote the students who read English newspaper.
It is given that,
\(P(H)=60\%=\frac{6}{10}=\frac{3}{5}\)
\(P(E)=40\%=\frac{40}{100}=\frac{2}{5} \)
\(P(H \cap E)=20\%=\frac{20}{100}=\frac{1}{5}\)
(a)Probability that a student reads neither Hindi or English newspaper is,
\((H\cup E)'=1-P(H\cup E)\)
\(1-[P(H)+P(E)-P(H\cap E)]\)
\(=1-\bigg(\frac{3}{5}+\frac{2}{5}-\frac{1}{5}\bigg)\)
\(=1-\frac{4}{5}=\frac{1}{5}\)
(b)Probability that a randomly chosen student reads English newspaper, if she reads Hindi news paper, is given by \(P (E|H)\).
\(P(E|H)=\frac{P(E∩H)}{P(H)}=\frac{\frac{1}{5}}{\frac{3}{5}}=\frac{1}{3}\)
(c)Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by \(P (H|E).\)
\(P(H|E)=\frac{P(H∩E)}{P(E)}=\frac{\frac{1}{5}}{\frac{2}{5}}=\frac{1}{2}\)