Let E   &   F be two independent events. The probability that E   &   F happen…

$P\left(E\cap F\right)=P\left(E\right). P\left(F\right)=\frac{1}{12}$

$\Rightarrow xy=\frac{1}{12}$

$\Rightarrow P\left(\stackrel{-}{E}\cap \stackrel{-}{F}\right)=P\left(\stackrel{-}{E}\right) . P\left(\stackrel{-}{F}\right)=\frac{1}{2}$

$\Rightarrow \left(1-P\left(E\right)\right) \left(1-P\left(F\right)\right)=\frac{1}{2}$

$\Rightarrow 1-x-y+xy=\frac{1}{2}$

$\Rightarrow 1-x-y+\frac{1}{12}=\frac{1}{2}$

$\Rightarrow 1-x-y=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}$

$\Rightarrow x+y=\frac{7}{12}$

$\Rightarrow x+\frac{1}{12x}=\frac{7}{12}$

$\Rightarrow \Rightarrow \frac{12x^2+1}{12x}=\frac{7}{12}$

$\Rightarrow 12x^2-7x+1=0$

$\Rightarrow 12x^2-4x-3x+1=0$

$\Rightarrow 4x\left(3x-1\right)-1\left(3x-1\right)=0$

$\Rightarrow x=\frac{1}{3}, x=\frac{1}{4}$

$\Rightarrow y=\frac{1}{4}, y=\frac{1}{3}$

$\Rightarrow \frac{x}{y}=\frac{\frac{1}{3}}{\frac{1}{4}} \& \frac{\frac{1}{4}}{\frac{1}{3}}$

$\Rightarrow \frac{x}{y}=\frac{4}{3} \& \frac{{\displaystyle 3}}{{\displaystyle 4}}$