Key Concepts and Formulas

• Independence of Events: Two events $E$ and $F$ are independent if the probability of their simultaneous occurrence is the product of their individual probabilities: $P(E \cap F) = P(E)P(F)$
• Complement of an Event: The probability of an event $E$ not happening (denoted $E'$) is $P(E') = 1 - P(E)$. A crucial property for independent events is that if $E$ and $F$ are independent, then their complements $E'$ and $F'$ are also independent. Therefore: $P(E' \cap F') = P(E')P(F')$
• Algebraic Relation for Sum and Product: If $x$ and $y$ are two quantities, and their sum $(x+y)$ and product $(xy)$ are known, they can be found by solving the quadratic equation $t^2 - (x+y)t + xy = 0$.

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Step-by-Step Solution

Step 1: Define Variables and Translate Given Conditions into Equations

To simplify the notation, let's assign variables to the probabilities of events $E$ and $F$:

• Let $P(E) = x$
• Let $P(F) = y$

We are given two pieces of information, which we will translate into mathematical equations:

• Condition 1: "The probability that both E and F happen is $\frac{1}{12}$."

  • This means $P(E \cap F) = \frac{1}{12}$.
  • Why this step? This directly converts the first statement into a standard probability expression.
  • Since $E$ and $F$ are independent events, we apply the definition of independence: $P(E \cap F) = P(E)P(F)$
  • Substituting our variables and the given value: $xy = \frac{1}{12} \quad \text{... (Equation 1)}$

• Condition 2: "The probability that neither E nor F happens is $\frac{1}{2}$."

  • The phrase "neither E nor F happens" means that event $E$ does not happen ($E'$) AND event $F$ does not happen ($F'$).
  • This translates to $P(E' \cap F') = \frac{1}{2}$.
  • Why this step? It's crucial to correctly interpret "neither A nor B" as the intersection of their complements, $A' \cap B'$.
  • Since $E$ and $F$ are independent, their complements $E'$ and $F'$ are also independent. Therefore: $P(E' \cap F') = P(E')P(F')$
  • Using the complement rule, $P(E') = 1 - P(E) = 1 - x$ and $P(F') = 1 - P(F) = 1 - y$.
  • Substituting these into the equation for $P(E' \cap F')$: $(1 - x)(1 - y) = \frac{1}{2} \quad \text{... (Equation 2)}$
  • Why this step? This expresses the second condition in terms of our variables $x$ and $y$, allowing us to form a system of equations.

Step 2: Formulate a System of Equations for $x$ and $y$

Now, we will expand Equation 2 and substitute Equation 1 to find the sum $(x+y)$:

• Expand the left side of Equation 2: $1 - y - x + xy = \frac{1}{2}$
• Rearrange the terms to group $(x+y)$: $1 - (x + y) + xy = \frac{1}{2}$ Why this step? Our goal is to find the individual values of $x$ and $y$. We currently have an equation for their product ($xy$) and another equation involving both their sum ($x+y$) and product. By simplifying and substituting, we can derive a clean equation for the sum.
• Substitute the value of $xy$ from Equation 1 ($xy = \frac{1}{12}$) into this expanded equation: $1 - (x + y) + \frac{1}{12} = \frac{1}{2}$
• Isolate the term $(x+y)$ to solve for the sum: $x + y = 1 + \frac{1}{12} - \frac{1}{2}$
• To combine the terms on the right side, find a common denominator, which is 12: $x + y = \frac{12}{12} + \frac{1}{12} - \frac{6}{12}$ $x + y = \frac{12 + 1 - 6}{12}$ $x + y = \frac{7}{12} \quad \text{... (Equation 3)}$

We now have a system of two fundamental equations for $x$ and $y$:

1. $xy = \frac{1}{12}$
2. $x+y = \frac{7}{12}$

Step 3: Determine Individual Probabilities ($x$ and $y$)

We can find the individual values of $x$ and $y$ (which are $P(E)$ and $P(F)$) using the sum and product we just found.

• Why this step? Knowing the sum and product of two variables is a classic algebraic setup that allows us to find the individual values.
• Consider $x$ and $y$ as the roots of a quadratic equation. The general form for a quadratic equation with roots $r_1$ and $r_2$ is $t^2 - (r_1+r_2)t + r_1r_2 = 0$.
• Substituting $x+y = \frac{7}{12}$ and $xy = \frac{1}{12}$: $t^2 - \frac{7}{12}t + \frac{1}{12} = 0$
• To eliminate the fractions, multiply the entire equation by 12: $12t^2 - 7t + 1 = 0$
• Now, factor the quadratic equation: $12t^2 - 4t - 3t + 1 = 0$ $4t(3t - 1) - 1(3t - 1) = 0$ $(4t - 1)(3t - 1) = 0$
• This gives two possible values for $t$: $4t - 1 = 0 \Rightarrow t = \frac{1}{4}$ $3t - 1 = 0 \Rightarrow t = \frac{1}{3}$
• Thus, the probabilities $P(E)$ and $P(F)$ are $\frac{1}{4}$ and $\frac{1}{3}$ in some order. That is, either $P(E) = \frac{1}{3}$ and $P(F) = \frac{1}{4}$, or $P(E) = \frac{1}{4}$ and $P(F) = \frac{1}{3}$.

Step 4: Calculate the Required Ratio $\frac{P(E)}{P(F)}$

The question asks for "a value of" $\frac{P(E)}{P(F)}$, meaning we need to find one of the possible ratios that matches the options.

• Why this step? This is the final step to answer the specific question posed in the problem.

Based on the probabilities found in Step 3, we have two possibilities for the ratio:

• Possibility 1: If $P(E) = \frac{1}{3}$ and $P(F) = \frac{1}{4}$ $\frac{P(E)}{P(F)} = \frac{1/3}{1/4} = \frac{1}{3} \times \frac{4}{1} = \frac{4}{3}$

• Possibility 2: If $P(E) = \frac{1}{4}$ and $P(F) = \frac{1}{3}$ $\frac{P(E)}{P(F)} = \frac{1/4}{1/3} = \frac{1}{4} \times \frac{3}{1} = \frac{3}{4}$

Comparing these values with the given options: (A) $\frac{4}{3}$ (B) $\frac{3}{2}$ (C) $\frac{1}{3}$ (D) $\frac{5}{12}$

The value $\frac{4}{3}$ matches option (A).

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Common Mistakes & Tips

• Misinterpreting "Neither E nor F": A common error is to confuse $P(E' \cap F')$ with $P(E' \cup F')$. Remember that "neither A nor B" means "not A AND not B".
• Forgetting Independence of Complements: If $E$ and $F$ are independent, it's a crucial property that their complements $E'$ and $F'$ are also independent. This allows $P(E' \cap F') = P(E')P(F')$.
• Algebraic Errors: Be careful when expanding $(1-x)(1-y)$ and when solving the quadratic equation or manipulating fractions. A small arithmetic mistake can lead to an incorrect answer.

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Summary

This problem effectively tests the understanding of independent events and their properties, including the independence of their complements. By assigning variables $P(E)=x$ and $P(F)=y$, we translated the given verbal conditions into a system of two algebraic equations involving $xy$ and $x+y$. Solving this system, we found that the probabilities $P(E)$ and $P(F)$ must be $\frac{1}{3}$ and $\frac{1}{4}$ (in either order). Finally, calculating the ratio $\frac{P(E)}{P(F)}$ yielded two possible values, $\frac{4}{3}$ and $\frac{3}{4}$, with $\frac{4}{3}$ being one of the provided options.

The final answer is $\boxed{\text{4/3}}$, which corresponds to option (A).