1. Key Concepts and Formulas

This problem involves probabilities of independent events. Understanding how to translate verbal descriptions into mathematical expressions is crucial.

• Independence of Events: If events $E_1, E_2, E_3$ are independent, then the probability of their intersection is the product of their individual probabilities: $P(E_1 \cap E_2 \cap E_3) = P(E_1)P(E_2)P(E_3)$. Furthermore, if events are independent, their complements are also independent.
• Complements of Events: The complement of an event $E_i$, denoted $E_i'$, is the event that $E_i$ does not occur. Its probability is $P(E_i') = 1 - P(E_i)$.
• Probability of "Only Event $E_i$ Occurs": For independent events, this means $E_i$ occurs, and all other events do not occur. For example, $P(\text{only } E_1) = P(E_1 \cap E_2' \cap E_3') = P(E_1)P(E_2')P(E_3')$.
• Probability of "None of the Events Occur": This means all events $E_1, E_2, E_3$ do not occur. For independent events, $P(\text{none}) = P(E_1' \cap E_2' \cap E_3') = P(E_1')P(E_2')P(E_3')$.

2. Step-by-Step Solution

Step 1: Define Probabilities and Translate the Given Information

Let $P(E_1)$, $P(E_2)$, and $P(E_3)$ be the probabilities of events $E_1$, $E_2$, and $E_3$ occurring, respectively. For simplicity, we denote them as $p_1, p_2, p_3$. Since all probabilities are given to lie in the interval $(0, 1)$, we have $0 < p_i < 1$ for $i=1, 2, 3$. Let $P(E_i')$ be the probability that event $E_i$ does not occur, denoted as $q_i$. So, $q_i = 1 - p_i$. Consequently, $0 < q_i < 1$.

Now, let's express the given probabilities in terms of $p_i$ and $q_i$, leveraging the independence of the events:

• The probability that only $E_1$ occurs is $\alpha$: $\alpha = P(E_1 \cap E_2' \cap E_3') = P(E_1)P(E_2')P(E_3') = p_1 q_2 q_3$
• The probability that only $E_2$ occurs is $\beta$: $\beta = P(E_1' \cap E_2 \cap E_3') = P(E_1')P(E_2)P(E_3') = q_1 p_2 q_3$
• The probability that only $E_3$ occurs is $\gamma$: $\gamma = P(E_1' \cap E_2' \cap E_3) = P(E_1')P(E_2')P(E_3) = q_1 q_2 p_3$
• The probability that none of the events occur is $p$: $p = P(E_1' \cap E_2' \cap E_3') = P(E_1')P(E_2')P(E_3') = q_1 q_2 q_3$ Since all given probabilities are in $(0,1)$, $\alpha, \beta, \gamma, p$ are all non-zero.

Step 2: Simplify Relationships using the Probability 'p'

A common strategy in such problems is to find ratios involving $p$. This simplifies the expressions significantly as common terms $q_1 q_2 q_3$ cancel out.

• To relate $\alpha$ and $p$: $\frac{\alpha}{p} = \frac{p_1 q_2 q_3}{q_1 q_2 q_3} = \frac{p_1}{q_1}$
• To relate $\beta$ and $p$: $\frac{\beta}{p} = \frac{q_1 p_2 q_3}{q_1 q_2 q_3} = \frac{p_2}{q_2}$
• To relate $\gamma$ and $p$: $\frac{\gamma}{p} = \frac{q_1 q_2 p_3}{q_1 q_2 q_3} = \frac{p_3}{q_3}$ These simplifications are valid because $q_1, q_2, q_3 \in (0,1)$, so they are non-zero.

Step 3: Utilize the Given Equations to Find Relationships between $p_1, p_2, p_3$

We are given two equations:

1. $(\alpha - 2\beta)p = \alpha\beta$
2. $(\beta - 3\gamma)p = 2\beta\gamma$

From Equation 1: $(\alpha - 2\beta)p = \alpha\beta$

• Divide by $p^2$ on both sides: This allows us to use the simplified ratios from Step 2. Since $p \in (0,1)$, $p \neq 0$. $\frac{(\alpha - 2\beta)p}{p^2} = \frac{\alpha\beta}{p^2}$ $\frac{\alpha}{p} - 2\frac{\beta}{p} = \left(\frac{\alpha}{p}\right)\left(\frac{\beta}{p}\right)$
• Substitute the simplified ratios: Replace $\frac{\alpha}{p}$ with $\frac{p_1}{q_1}$ and $\frac{\beta}{p}$ with $\frac{p_2}{q_2}$. $\frac{p_1}{q_1} - 2\frac{p_2}{q_2} = \left(\frac{p_1}{q_1}\right)\left(\frac{p_2}{q_2}\right)$
• Clear denominators: Multiply the entire equation by $q_1 q_2$. Since $q_1, q_2 \in (0,1)$, $q_1 q_2 \neq 0$. $p_1 q_2 - 2 p_2 q_1 = p_1 p_2$
• Substitute $q_i = 1-p_i$: Express the equation solely in terms of $p_1$ and $p_2$. $p_1(1-p_2) - 2p_2(1-p_1) = p_1 p_2$
• Expand and simplify: $p_1 - p_1 p_2 - 2p_2 + 2p_1 p_2 = p_1 p_2$ Combine like terms: $p_1 - 2p_2 + p_1 p_2 = p_1 p_2$ Subtract $p_1 p_2$ from both sides: $p_1 - 2p_2 = 0$ This gives our first crucial relationship: $p_1 = 2p_2$

From Equation 2: $(\beta - 3\gamma)p = 2\beta\gamma$

• Divide by $p^2$ on both sides: $\frac{(\beta - 3\gamma)p}{p^2} = \frac{2\beta\gamma}{p^2}$ $\frac{\beta}{p} - 3\frac{\gamma}{p} = 2\left(\frac{\beta}{p}\right)\left(\frac{\gamma}{p}\right)$
• Substitute the simplified ratios: Replace $\frac{\beta}{p}$ with $\frac{p_2}{q_2}$ and $\frac{\gamma}{p}$ with $\frac{p_3}{q_3}$. $\frac{p_2}{q_2} - 3\frac{p_3}{q_3} = 2\left(\frac{p_2}{q_2}\right)\left(\frac{p_3}{q_3}\right)$
• Clear denominators: Multiply the entire equation by $q_2 q_3$. $p_2 q_3 - 3 p_3 q_2 = 2 p_2 p_3$
• Substitute $q_i = 1-p_i$: $p_2(1-p_3) - 3p_3(1-p_2) = 2 p_2 p_3$
• Expand and simplify: $p_2 - p_2 p_3 - 3p_3 + 3p_2 p_3 = 2 p_2 p_3$ Combine like terms: $p_2 - 3p_3 + 2p_2 p_3 = 2 p_2 p_3$ Subtract $2p_2 p_3$ from both sides: $p_2 - 3p_3 = 0$ This gives our second crucial relationship: $p_2 = 3p_3$

Step 4: Calculate the Required Ratio

The problem asks for the ratio $\frac{Probability\ of\ occurrence\ of\ E_{1}}{Probability\ of\ occurrence\ of\ E_{3}} = \frac{p_1}{p_3}$. We have derived the following two relationships:

1. $p_1 = 2p_2$
2. $p_2 = 3p_3$

To find $\frac{p_1}{p_3}$, we can substitute the expression for $p_2$ from the second relationship into the first relationship: $p_1 = 2(3p_3)$ $p_1 = 6p_3$ Now, divide both sides by $p_3$ (which is non-zero as $p_3 \in (0,1)$): $\frac{p_1}{p_3} = 6$

3. Common Mistakes & Tips

• Incorrectly assuming dependence: Always remember to use the product rule for probabilities of independent events ($P(A \cap B) = P(A)P(B)$).
• Algebraic errors: Be careful with expanding terms and combining like terms, especially when substituting $q_i = 1-p_i$. A common mistake is not distributing negative signs properly.
• Dividing by zero: Ensure that any probability or expression you divide by is non-zero. The problem statement "All the given probabilities are assumed to lie in the interval (0, 1)" is key to ensuring $p_i, q_i, \alpha, \beta, \gamma, p$ are all non-zero.

4. Summary

By first defining the probabilities of individual events and their complements, we translated the given descriptive probabilities ($\alpha, \beta, \gamma, p$) into mathematical expressions based on the independence of events. We then used the ratio of these probabilities to simplify the given equations. Solving the simplified equations yielded two direct relationships: $p_1 = 2p_2$ and $p_2 = 3p_3$. Combining these relationships, we found that $p_1 = 6p_3$, leading to the final required ratio of $\frac{p_1}{p_3} = 6$.

5. Final Answer

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