Let there be three independent events $$E_1, E_2$$ and $$E_3$$. The probability that only $$E_1$$ occurs is $$\alpha$$, only $$E_2$$ occurs is $$\beta$$ and only $$E_3$$ occurs is $$\gamma$$. Let '$$p$$' denote the probability of none of events occurs that satisfies the equations $$(\alpha - 2\beta)p = \alpha\beta$$ and $$(\beta - 3\gamma)p = 2\beta\gamma$$. All the given probabilities are assumed to lie in the interval $$(0, 1)$$. Then, $$\frac{\text{Probability of occurrence of } E_1}{\text{Probability of occurrence of } E_3}$$ is equal to ________.

Let the probabilities of $$E_1, E_2, E_3$$ be $$p_1, p_2, p_3$$. Since the events are independent:

$$\alpha = p_1(1-p_2)(1-p_3)$$, $$\beta = (1-p_1)p_2(1-p_3)$$, $$\gamma = (1-p_1)(1-p_2)p_3$$, and $$p = (1-p_1)(1-p_2)(1-p_3)$$.

Observe that $$\frac{\alpha}{p} = \frac{p_1}{1-p_1}$$, $$\frac{\beta}{p} = \frac{p_2}{1-p_2}$$, $$\frac{\gamma}{p} = \frac{p_3}{1-p_3}$$.

Let $$x = \frac{p_1}{1-p_1}$$, $$y = \frac{p_2}{1-p_2}$$, $$z = \frac{p_3}{1-p_3}$$.

From $$(\alpha - 2\beta)p = \alpha\beta$$, dividing by $$p^2$$: $$x - 2y = xy$$, so $$x(1-y) = 2y$$, giving $$x = \frac{2y}{1-y}$$ $$-(1)$$.

From $$(\beta - 3\gamma)p = 2\beta\gamma$$, dividing by $$p^2$$: $$y - 3z = 2yz$$, so $$y(1-2z) = 3z$$, giving $$y = \frac{3z}{1-2z}$$ $$-(2)$$.

From $$(2)$$: $$1 - y = \frac{1-2z-3z}{1-2z} = \frac{1-5z}{1-2z}$$. Substituting into $$(1)$$: $$x = \frac{2 \cdot \frac{3z}{1-2z}}{\frac{1-5z}{1-2z}} = \frac{6z}{1-5z}$$.

Now $$1+x = \frac{1-5z+6z}{1-5z} = \frac{1+z}{1-5z}$$ and $$1+z = 1+z$$.

Since $$p_1 = \frac{x}{1+x}$$ and $$p_3 = \frac{z}{1+z}$$: $$\frac{p_1}{p_3} = \frac{x}{1+x} \cdot \frac{1+z}{z} = \frac{x(1+z)}{z(1+x)}$$.

$$= \frac{\frac{6z}{1-5z} \cdot (1+z)}{z \cdot \frac{1+z}{1-5z}} = \frac{6z(1+z)(1-5z)}{z(1+z)(1-5z)} = 6$$.

The answer is 6.