.: 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.