JEE MAIN - Mathematics (2025 - 22nd January Evening Shift - No. 8)
If $A$ and $B$ are two events such that $P(A \cap B)=0.1$, and $P(A \mid B)$ and $P(B \mid A)$ are the roots of the equation $12 x^2-7 x+1=0$, then the value of $\frac{P(\bar{A} \cup \bar{B})}{P(\bar{A} \cap \bar{B})}$ is :
$\frac{4}{3}$
$\frac{7}{4}$
$\frac{9}{4}$
$\frac{5}{3}$
Explanation
$$\begin{aligned}
& 12 \mathrm{x}^2-7 \mathrm{x}+1=0 \\
& \mathrm{x}=\frac{1}{3}, \frac{1}{4} \\
& \text { Let } \mathrm{P}\left(\frac{\mathrm{~A}}{\mathrm{~B}}\right)=\frac{1}{3} \& \mathrm{P}\left(\frac{\mathrm{~B}}{\mathrm{~A}}\right)=\frac{1}{4} \\
& \frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~B})}=\frac{1}{3} \& \frac{\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{\mathrm{P}(\mathrm{~A})}=\frac{1}{4} \\
& \Rightarrow \mathrm{P}(\mathrm{~B})=0.3 \\
& \& \mathrm{P}(\mathrm{~A})=0.4 \\
& \mathrm{P}(\mathrm{~A} \cup \mathrm{~B})=\mathrm{P}(\mathrm{~A})+\mathrm{P}(\mathrm{~B})-\mathrm{P}(\mathrm{~A} \cap \mathrm{~B}) \\
& =0.3+0.4-0.1=0.6 \\
& \mathrm{Now} \frac{\mathrm{P}(\overline{\mathrm{~A}} \cup \overline{\mathrm{~B}})}{\mathrm{P}(\overline{\mathrm{~A}} \cap \overline{\mathrm{~B}})}=\frac{\mathrm{P}(\overline{\mathrm{~A} \cap \mathrm{~B}})}{\mathrm{P}(\overline{\mathrm{~A} \cup \mathrm{~B}})} \\
& =\frac{1-\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})}{1-\mathrm{P}(\mathrm{~A} \cup \mathrm{~B})}=\frac{1-0.1}{1-0.6}=\frac{9}{4}
\end{aligned}$$
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