1. Key Concepts and Formulas

• Independent Events: If events $A_1, A_2, \ldots, A_n$ are independent, then the probability of their intersection is the product of their individual probabilities: $P(A_1 \cap A_2 \cap \ldots \cap A_n) = P(A_1)P(A_2)\ldots P(A_n)$. Importantly, if events are independent, their complements are also independent. For example, $B_1, B_2^c, B_3^c$ are independent.
• Complement of an Event: The probability of an event's complement $A^c$ is $P(A^c) = 1 - P(A)$.
• Probabilities in (0, 1): All probabilities are strictly between 0 and 1, meaning no event has a probability of 0 or 1. This ensures that we can safely divide by any probability term without risking division by zero.

2. Step-by-Step Solution

Step 1: Define variables for event probabilities and their complements. Let $P(B_1) = x$, $P(B_2) = y$, and $P(B_3) = z$. Since $P(B_i) \in (0, 1)$, we have $0 < x, y, z < 1$. Consequently, the probabilities of their complements are:

• $P(B_1^c) = 1-x$
• $P(B_2^c) = 1-y$
• $P(B_3^c) = 1-z$ These are also in $(0, 1)$.

Step 2: Express $\alpha, \beta, \gamma, p$ in terms of $x, y, z$. We use the definition of independent events and their complements:

• $\alpha$ (only $B_1$ occurs): $B_1$ occurs, $B_2$ does not, and $B_3$ does not. $\alpha = P(B_1 \cap B_2^c \cap B_3^c) = P(B_1)P(B_2^c)P(B_3^c) = x(1-y)(1-z)$
• $\beta$ (only $B_2$ occurs): $\beta = P(B_1^c \cap B_2 \cap B_3^c) = P(B_1^c)P(B_2)P(B_3^c) = (1-x)y(1-z)$
• $\gamma$ (only $B_3$ occurs): $\gamma = P(B_1^c \cap B_2^c \cap B_3) = P(B_1^c)P(B_2^c)P(B_3) = (1-x)(1-y)z$
• $p$ (none of the events occurs): $p = P(B_1^c \cap B_2^c \cap B_3^c) = P(B_1^c)P(B_2^c)P(B_3^c) = (1-x)(1-y)(1-z)$ All $\alpha, \beta, \gamma, p$ are in $(0, 1)$, so they are non-zero.

Step 3: Simplify the first given equation to find a relationship between $x$ and $y$. The first equation is $(\alpha - 2\beta)p = \alpha\beta$. Since $\alpha, \beta, p$ are non-zero, we can divide the entire equation by $\alpha\beta p$: $\frac{(\alpha - 2\beta)p}{\alpha\beta p} = \frac{\alpha\beta}{\alpha\beta p}$ $\frac{1}{\beta} - \frac{2}{\alpha} = \frac{1}{p}$ Now, substitute the expressions from Step 2: $\frac{1}{(1-x)y(1-z)} - \frac{2}{x(1-y)(1-z)} = \frac{1}{(1-x)(1-y)(1-z)}$ Multiply both sides by the common factor $(1-x)(1-y)(1-z)$ (which is $p$, and $p \neq 0$): $\frac{(1-x)(1-y)(1-z)}{(1-x)y(1-z)} - \frac{2(1-x)(1-y)(1-z)}{x(1-y)(1-z)} = \frac{(1-x)(1-y)(1-z)}{(1-x)(1-y)(1-z)}$ Simplify the terms by canceling common factors: $\frac{1-y}{y} - \frac{2(1-x)}{x} = 1$ Split the fractions: $\left(\frac{1}{y} - 1\right) - 2\left(\frac{1}{x} - 1\right) = 1$ $\frac{1}{y} - 1 - \frac{2}{x} + 2 = 1$ $\frac{1}{y} - \frac{2}{x} + 1 = 1$ Subtract 1 from both sides: $\frac{1}{y} - \frac{2}{x} = 0$ $\frac{1}{y} = \frac{2}{x}$ Cross-multiply to get the relationship: $x = 2y$

Step 4: Simplify the second given equation to find a relationship between $y$ and $z$. The second equation is $(\beta - 3\gamma)p = 2\beta\gamma$. Since $\beta, \gamma, p$ are non-zero, divide the entire equation by $\beta\gamma p$: $\frac{(\beta - 3\gamma)p}{\beta\gamma p} = \frac{2\beta\gamma}{\beta\gamma p}$ $\frac{1}{\gamma} - \frac{3}{\beta} = \frac{2}{p}$ Now, substitute the expressions from Step 2: $\frac{1}{(1-x)(1-y)z} - \frac{3}{(1-x)y(1-z)} = \frac{2}{(1-x)(1-y)(1-z)}$ Multiply both sides by the common factor $(1-x)(1-y)(1-z)$ (which is $p$, and $p \neq 0$): $\frac{(1-x)(1-y)(1-z)}{(1-x)(1-y)z} - \frac{3(1-x)(1-y)(1-z)}{(1-x)y(1-z)} = \frac{2(1-x)(1-y)(1-z)}{(1-x)(1-y)(1-z)}$ Simplify the terms by canceling common factors: $\frac{1-z}{z} - \frac{3(1-y)}{y} = 2$ Split the fractions: $\left(\frac{1}{z} - 1\right) - 3\left(\frac{1}{y} - 1\right) = 2$ $\frac{1}{z} - 1 - \frac{3}{y} + 3 = 2$ $\frac{1}{z} - \frac{3}{y} + 2 = 2$ Subtract 2 from both sides: $\frac{1}{z} - \frac{3}{y} = 0$ $\frac{1}{z} = \frac{3}{y}$ Cross-multiply to get the relationship: $y = 3z$

Step 5: Calculate the required ratio $\frac{P(B_1)}{P(B_3)}$. We have found two relationships:

1. $x = 2y$
2. $y = 3z$ Substitute the expression for $y$ from the second relationship into the first relationship: $x = 2(3z)$ $x = 6z$ Now, find the ratio $\frac{P(B_1)}{P(B_3)} = \frac{x}{z}$: $\frac{x}{z} = 6$

3. Common Mistakes & Tips

• Division by Zero: Always ensure that terms you are dividing by are non-zero. The problem statement "All probabilities are assumed to lie in the interval (0, 1)" is crucial here, as it guarantees $x, y, z, (1-x), (1-y), (1-z)$ are all non-zero. This in turn makes $\alpha, \beta, \gamma, p$ non-zero, allowing for safe division.
• Algebraic Errors: Be careful when expanding and combining terms. The method of simplifying $\frac{1}{A} - \frac{k}{B} = C$ by multiplying by the common denominator $ABC$ and then canceling terms can be error-prone. A safer way, as used in this solution, is to keep the fractions and simplify them as $\left(\frac{1}{A}-1\right)$ etc.
• Understanding Independence: Remember that if events $B_i$ are independent, then any combination of $B_i$ and $B_i^c$ are also independent. This is key to writing $\alpha, \beta, \gamma, p$ correctly.

4. Summary

This problem required us to translate a scenario involving independent events into algebraic equations using their probabilities. By defining $P(B_1)=x$, $P(B_2)=y$, and $P(B_3)=z$, we expressed the given probabilities $\alpha, \beta, \gamma, p$ in terms of $x, y, z$. We then systematically simplified the two given equations, leveraging the fact that all probabilities are in $(0,1)$ to allow for division by non-zero terms. This led to two key relationships: $x=2y$ and $y=3z$. Combining these relationships, we found that $x=6z$, from which the desired ratio $\frac{P(B_1)}{P(B_3)}$ was determined.

5. Final Answer

The final answer is $\boxed{6}$.