If $A$ and $B$ are two events such that $P(A \cap B) = 0.1$, and $P(A|B)$ and $P(B|A)$ are the roots of the equation $12x^2 - 7x + 1 = 0$, then the value of $\dfrac{P(\bar{A} \cup \bar{B})}{P(\bar{A} \cap \bar{B})}$ is:
Step-by-Step Solution
Key Concept: Use Vieta's formulas on $12x^2-7x+1=0$ to find $P(A|B) \cdot P(B|A) = 1/12$ and $P(A|B) + P(B|A) = 7/12$. Derive $P(A)$, $P(B)$, $P(A\cup B)$, then use $\bar{A}\cup\bar{B} = \overline{A\cap B}$ and $\bar{A}\cap\bar{B} = \overline{A\cup B}$.
$P(A|B)\cdot P(B|A) = \frac{P(A\cap B)^2}{P(A)P(B)} = \frac{1}{12}$, so $P(A)P(B) = 0.12$. Also $P(A)+P(B) = \frac{7}{12} \times \frac{P(A\cap B)}{...} = 0.7$. $P(A\cup B) = 0.6$. $\frac{P(\bar{A}\cup\bar{B})}{P(\bar{A}\cap\bar{B})} = \frac{1-P(A\cap B)}{1-P(A\cup B)} = \frac{0.9}{0.4} = \frac{9}{4}$.
Correct Answer: $\frac{9}{4}$