If A and B are two events such that $\mathrm{P}(\mathrm{A}\cap \mathrm{B})=0.1$, and P(A|B) and P(B|A) are the roots of the equation 12x² – 7x + 1 = 0, then the value of $\frac{\mathrm{P}(\overline{\mathrm{A}}\cup \overline{\mathrm{B}})}{\mathrm{P}(\overline{\mathrm{A}}\cap \overline{\mathrm{B}})}$ is : 

$\begin{array}{l}12{\mathrm{x}}^2-7\mathrm{x}+1=0\\ \mathrm{x}=\frac{1}{3},\frac{1}{4}\end{array}$
Let $\mathrm{P}\left(\frac{\mathrm{A}}{\mathrm{B}}\right)=\frac{1}{3} \mathrm{and} \mathrm{P}\left(\frac{\mathrm{B}}{\mathrm{A}}\right)=\frac{1}{4}$
$\begin{array}{l}\frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{B}\right)}=\frac{1}{3} \mathrm{and} \frac{\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{\mathrm{P}\left(\mathrm{A}\right)}=\frac{1}{4}\\ \Rightarrow \mathrm{P}\left(\mathrm{B}\right)=0.3\\ \mathrm{and} \mathrm{P}\left(\mathrm{A}\right)=0.4\\ \mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})\\ =0.3+0.4-0.1=0.6\\ \text{ Now }\frac{\mathrm{P}(\overline{\mathrm{A}}\cup \overline{\mathrm{B}})}{\mathrm{P}(\overline{\mathrm{A}}\cap \overline{\mathrm{B}})}=\frac{\mathrm{P}\left(\overline{\mathrm{A}\cap \mathrm{B}}\right)}{\mathrm{P}\left(\overline{\mathrm{A}\cup \mathrm{B}}\right)}\\ =\frac{1-\mathrm{P}(\mathrm{A}\cap \mathrm{B})}{1-\mathrm{P}(\mathrm{A}\cup \mathrm{B})}=\frac{1-0.1}{1-0.6}=\frac{9}{4}\end{array}$