If and are two events such that P(A ∩ 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 $$\frac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B}}$$ is:

Given that $$A$$ and $$B$$ are two events with $$P(A \cap B) = 0.1$$, and $$P(A \mid B)$$ and $$P(B \mid A)$$ are the roots of the equation $$12x^2 - 7x + 1 = 0$$. We need to find $$\frac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})}$$.

First, solve the quadratic equation $$12x^2 - 7x + 1 = 0$$ to find the roots. Using the quadratic formula:

$$x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 12 \cdot 1}}{2 \cdot 12} = \frac{7 \pm \sqrt{49 - 48}}{24} = \frac{7 \pm \sqrt{1}}{24} = \frac{7 \pm 1}{24}$$

So, the roots are $$x = \frac{8}{24} = \frac{1}{3}$$ and $$x = \frac{6}{24} = \frac{1}{4}$$. Therefore, $$P(A \mid B)$$ and $$P(B \mid A)$$ are $$\frac{1}{3}$$ and $$\frac{1}{4}$$, in some order.

Recall the conditional probability formulas:

$$P(A \mid B) = \frac{P(A \cap B)}{P(B)} \quad \text{and} \quad P(B \mid A) = \frac{P(A \cap B)}{P(A)}$$

Given $$P(A \cap B) = 0.1$$, we consider the two possible cases.

Case 1: $$P(A \mid B) = \frac{1}{3}$$ and $$P(B \mid A) = \frac{1}{4}$$

Then,

$$P(B) = \frac{P(A \cap B)}{P(A \mid B)} = \frac{0.1}{\frac{1}{3}} = 0.1 \times 3 = 0.3$$

$$P(A) = \frac{P(A \cap B)}{P(B \mid A)} = \frac{0.1}{\frac{1}{4}} = 0.1 \times 4 = 0.4$$

Case 2: $$P(A \mid B) = \frac{1}{4}$$ and $$P(B \mid A) = \frac{1}{3}$$

Then,

$$P(B) = \frac{P(A \cap B)}{P(A \mid B)} = \frac{0.1}{\frac{1}{4}} = 0.1 \times 4 = 0.4$$

$$P(A) = \frac{P(A \cap B)}{P(B \mid A)} = \frac{0.1}{\frac{1}{3}} = 0.1 \times 3 = 0.3$$

In both cases, $$P(A) + P(B) = 0.4 + 0.3 = 0.7$$ or $$0.3 + 0.4 = 0.7$$, same sum.

Now, find $$P(A \cup B)$$ using the formula:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

In Case 1: $$P(A \cup B) = 0.4 + 0.3 - 0.1 = 0.6$$

In Case 2: $$P(A \cup B) = 0.3 + 0.4 - 0.1 = 0.6$$

So, $$P(A \cup B) = 0.6$$ in both cases.

Using De Morgan's laws:

$$\overline{A} \cup \overline{B} = (A \cap B)^c \quad \text{and} \quad \overline{A} \cap \overline{B} = (A \cup B)^c$$

Therefore,

$$P(\overline{A} \cup \overline{B}) = P((A \cap B)^c) = 1 - P(A \cap B) = 1 - 0.1 = 0.9$$

$$P(\overline{A} \cap \overline{B}) = P((A \cup B)^c) = 1 - P(A \cup B) = 1 - 0.6 = 0.4$$

The ratio is:

$$\frac{P(\overline{A} \cup \overline{B})}{P(\overline{A} \cap \overline{B})} = \frac{0.9}{0.4} = \frac{9}{10} \times \frac{5}{2} = \frac{9}{4}$$

Using fractions for clarity:

$$P(A \cap B) = \frac{1}{10}$$

Case 1: $$P(A) = \frac{2}{5}$$, $$P(B) = \frac{3}{10}$$

$$P(A \cup B) = \frac{2}{5} + \frac{3}{10} - \frac{1}{10} = \frac{4}{10} + \frac{3}{10} - \frac{1}{10} = \frac{6}{10} = \frac{3}{5}$$

$$P(\overline{A} \cup \overline{B}) = 1 - \frac{1}{10} = \frac{9}{10}$$

$$P(\overline{A} \cap \overline{B}) = 1 - \frac{3}{5} = \frac{2}{5}$$

Ratio $$= \frac{\frac{9}{10}}{\frac{2}{5}} = \frac{9}{10} \times \frac{5}{2} = \frac{45}{20} = \frac{9}{4}$$

Same result in Case 2.

Thus, the ratio is $$\frac{9}{4}$$, which corresponds to option D.