Ratio of Occurrences of Independent Events E1 and E3

Let the probabilities of occurrence of the independent events $E_1, E_2,$ and $E_3$ be denoted by: $P(E_1) = x, \quad P(E_2) = y, \quad P(E_3) = z$

Since $E_1, E_2,$ and $E_3$ are independent events, their complements $E_1^c, E_2^c,$ and $E_3^c$ are also independent.

The probability $p$ that none of the events $E_1, E_2,$ or $E_3$ occur is: $p = P(E_1^c \cap E_2^c \cap E_3^c) = (1 - x)(1 - y)(1 - z)$

The probabilities that only $E_1$, only $E_2$, and only $E_3$ occur are given by: $\alpha = P(E_1 \cap E_2^c \cap E_3^c) = x(1 - y)(1 - z)$ $\beta = P(E_1^c \cap E_2 \cap E_3^c) = (1 - x)y(1 - z)$ $\gamma = P(E_1^c \cap E_2^c \cap E_3) = (1 - x)(1 - y)z$

Expressing $\alpha, \beta,$ and $\gamma$ in terms of $p$: $\alpha = \frac{x}{1 - x} \cdot p \implies \frac{\alpha}{p} = \frac{x}{1 - x}$ $\beta = \frac{y}{1 - y} \cdot p \implies \frac{\beta}{p} = \frac{y}{1 - y}$ $\gamma = \frac{z}{1 - z} \cdot p \implies \frac{\gamma}{p} = \frac{z}{1 - z}$

Now, consider the first given relation: $(\alpha - 2\beta)p = \alpha\beta$

Dividing both sides by $p^2$: $\frac{\alpha}{p} - 2\frac{\beta}{p} = \frac{\alpha}{p} \cdot \frac{\beta}{p}$

Substituting $\frac{\alpha}{p} = \frac{x}{1 - x}$ and $\frac{\beta}{p} = \frac{y}{1 - y}$: $\frac{x}{1 - x} - 2\left(\frac{y}{1 - y}\right) = \left(\frac{x}{1 - x}\right)\left(\frac{y}{1 - y}\right)$

Multiplying through by $(1 - x)(1 - y)$: $x(1 - y) - 2y(1 - x) = xy$ $x - xy - 2y + 2xy = xy$ $x - 2y + xy = xy \implies x = 2y$

Next, consider the second given relation: $(\beta - 3\gamma)p = 2\beta\gamma$

Dividing both sides by $p^2$: $\frac{\beta}{p} - 3\frac{\gamma}{p} = 2 \left(\frac{\beta}{p}\right) \left(\frac{\gamma}{p}\right)$

Substituting $\frac{\beta}{p} = \frac{y}{1 - y}$ and $\frac{\gamma}{p} = \frac{z}{1 - z}$: $\frac{y}{1 - y} - 3\left(\frac{z}{1 - z}\right) = 2 \left(\frac{y}{1 - y}\right)\left(\frac{z}{1 - z}\right)$

Multiplying through by $(1 - y)(1 - z)$: $y(1 - z) - 3z(1 - y) = 2yz$ $y - yz - 3z + 3yz = 2yz$ $y - 3z + 2yz = 2yz \implies y = 3z$

Combining the relations for $x, y,$ and $z$: $x = 2y = 2(3z) = 6z$

We need to calculate: $\frac{\text{Probability of occurrence of } E_1}{\text{Probability of occurrence of } E_3} = \frac{P(E_1)}{P(E_3)} = \frac{x}{z}$

Since $x = 6z$, we get: $\frac{x}{z} = 6$