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 newspaper, find the probability that she reads Hindi newspaper.

.: Let $E$ : `A smdent reads Hindi newspaper'
$F$ : `A student reads English newspaper',

Then $P(E) = \cfrac{{60}}{{100}} = \cfrac{3}{5},P(F) = \cfrac{{40}}{{100}} = \cfrac{2}{5}$ and $P(E \cap F) = \cfrac{{20}}{{100}} = \cfrac{1}{5}$

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(a) Required probability $= P$

(student reads neither Hindi nor English newspaper)

$= P\left( {{E^c} \cap {F^c}} \right) = P\left( {{{(E \cup F)}^c}} \right) = 1 - P(E \cup F)$

$= 1 - \{ P(E) + P(F) - P(E \cap F)\} = 1 - \left( {\cfrac{3}{5} + \cfrac{2}{5} - \cfrac{1}{5}} \right) = \cfrac{1}{5}$

(b) Required probability $= P$

(student reads English newspaper when it is given that she reads Hindi newspaper)

$= P(F/E) = \cfrac{{P(F \cap E)}}{{P(E)}} = \cfrac{{P(E \cap F)}}{{P(E)}} = \cfrac{{1/5}}{{3/5}} = \cfrac{1}{3}$

(c) Required probability $= P$

(student reads Hindi newspaper when it is given that she reads English newspaper)

$= P(E/F) = \cfrac{{P(E \cap F)}}{{P(F)}} = \cfrac{{1/5}}{{2/5}} = \cfrac{1}{2}$

Choose the correct answer in Exercises 17 and 18.