Two coins are tossed once, where

(i) $E$ : tail appears on one coin, $F$ : one coin shows head

(ii) $E$ : no tail appears, $F$ : no head appears

.: When two coins are tossed, then the sample space $S$ contain 4 equally likely events.

$S = \{ HH,HT,\;TH,TT\}$

(i) Let $E$ : ‘tail appears on one coin’ and $F$ : ‘one coin shows head’

$\Rightarrow$ $E = \{ HT,TH\}$ and $F = \{ TH,HT\}$ $\Rightarrow E \cap F = \{ HT,TH\}$

$\therefore$ $P(E) = \cfrac{2}{4} = \cfrac{1}{2},\;P(F) = \cfrac{2}{4} = \cfrac{1}{2}{\kern 1pt} {\kern 1pt} {\rm{and}}{\kern 1pt} {\kern 1pt} P(E \cap F) = \cfrac{2}{4} = \cfrac{1}{2}$

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

(ii) Let $E$ : ‘no tail appears’ and $F$ : ‘no head appears’

$\Rightarrow E = \{ HH\}$ and $F = \{ TT\}$

$\Rightarrow$ $\Rightarrow E \cap F = \phi$

Hence, $P(E \cap F) = 0$ and $P(F) = 1/4.$

$\therefore$ $P(E/F) = \cfrac{{P(E \cap F)}}{{P(F)}} = \cfrac{{P(\phi )}}{{P(F)}} = \cfrac{0}{{1/4}} = 0$