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 $\frac{\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})}{\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})}$ is :

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Aicharya · Accepted Answer

$P(A \cap B)=0.1, P(A \mid B)$ and $P(B \mid A)$ are the roots of the equation $12 x^2-7 x+1=0$
$\begin{aligned}
& \Rightarrow \quad P(A \mid B) P(B \mid A)=\frac{1}{12} \\ & \Rightarrow \quad \frac{P(A \cap B)}{P(B)} \times \frac{P(A \cap B)}{P(A)}=\frac{1}{12} \\ & \Rightarrow \quad P(A) P(B)=12(0.1)^2 \\ & =0.12
\end{aligned}$
Also, $P(A \mid B)+P(B \mid A)=\frac{7}{12}$
$\begin{aligned}
\Rightarrow & P(A \cap B)\left(\frac{1}{P(B)}+\frac{1}{P(A)}\right)=\frac{7}{12} \\ \Rightarrow & P(A)+P(B)=\frac{7}{12} \times \frac{0.12}{0.1} \\ \Rightarrow & P(A)+P(B)=0.7 \\ & \frac{P(\bar{A} \cup \bar{B})}{P(\bar{A} \cap \bar{B})}=\frac{P(\overline{A \cap B})}{P(\overline{A U B})} \\ & =\frac{1-P(A \cap B)}{1-P(A \cup B)} \\ & =\frac{1-0.1}{1-(0.7-0.1)}=\frac{0.9}{0.4}=\frac{9}{4}
\end{aligned}$ ~

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…

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