Determine $P\left(\frac{E}{F}\right)$
Two coins are tossed once, where
E: Tail appears on one coin F: one coin shows head

E: no tail appears F : no head appears

It two coins are tossed once, then the sample space S = {HH, HT, TH, TT}
E = Set of events having tail on one coin = {HT, TH},
F = Set of events having one coin showing head $=HT,THE\cap F=HT,TH\therefore P\left(E\right)=\frac{1}{2},P\left(F\right)=\frac{2}{4}=\frac{1}{2}andP(E\cap F)=\frac{2}{4}=\frac{1}{2}$
$\therefore P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$

IF two coins are tossed once, then the sample space S = {HH, HT, TH, TT}

E = Set of events having no tail = {HH},
F = Set of events having no head $=TTandE\cap F=\varphi P\left(E\right)=\frac{1}{4},\frac{1}{4}andP(E\cap F)=\frac{0}{4}=0.\because P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{0}{\frac{1}{4}}=0$