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Step-by-Step Solution
Step 1: Denote the probabilities of the events
Let $P(E) = x$ and $P(F) = y$. We are given that $E$ and $F$ are independent events.
Step 2: Write the condition for both events happening
For two independent events,
$$
P(E \cap F) = P(E) \times P(F).
$$
Given $P(E \cap F) = \frac{1}{12}$, this implies
$$
x \times y = \frac{1}{12}.
$$
Step 3: Write the condition for neither event happening
The probability that neither $E$ nor $F$ occurs is given by
$$
P(E' \cap F') = (1 - x)(1 - y).
$$
Since $P(E' \cap F') = \frac{1}{2}$, we have
$$
(1 - x)(1 - y) = \frac{1}{2}.
$$
Step 4: Expand and simplify the second condition
Expanding
$$
(1 - x)(1 - y) = 1 - x - y + xy = \frac{1}{2}.
$$
Using $xy = \frac{1}{12}$ from Step 2, substitute to get
$$
1 - x - y + \frac{1}{12} = \frac{1}{2}.
$$
Rearrange to find $x + y$:
$$
1 - \frac{1}{2} + \frac{1}{12} = x + y
\;\;\Longrightarrow\;\;
\frac{1}{2} + \frac{1}{12} = x + y
\;\;\Longrightarrow\;\;
x + y = \frac{7}{12}.
$$
Step 5: Solve the system of equations
We now have two equations:
$x + y = \frac{7}{12}$
$xy = \frac{1}{12}$
To solve, set $y = \frac{7}{12} - x$ and substitute into $xy = \frac{1}{12}$:
$$
x \left(\frac{7}{12} - x\right) = \frac{1}{12}.
$$
Multiply through by 12:
$$
12 \times x \left(\frac{7}{12} - x\right) = 1
\;\Longrightarrow\;
7x - 12x^2 = 1.
$$
Rearrange to a quadratic equation:
$$
12x^2 - 7x + 1 = 0.
$$
Solve using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=12, b=-7, c=1$:
$$
x = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 12 \cdot 1}}{2 \cdot 12}
= \frac{7 \pm \sqrt{49 - 48}}{24}
= \frac{7 \pm 1}{24}.
$$
Thus,
$$
x = \frac{8}{24} = \frac{1}{3} \quad \text{or} \quad x = \frac{6}{24} = \frac{1}{4}.
$$
Correspondingly,
$$
y = \frac{7}{12} - x =
\begin{cases}
\frac{7}{12} - \frac{1}{3} = \frac{1}{4}, & \text{if } x = \frac{1}{3}, \\
\frac{7}{12} - \frac{1}{4} = \frac{1}{3}, & \text{if } x = \frac{1}{4}.
\end{cases}
$$
Step 6: Determine the ratio $P(E)/P(F)$
The pairs $(x, y)$ are either $(\tfrac{1}{3}, \tfrac{1}{4})$ or $(\tfrac{1}{4}, \tfrac{1}{3})$.
For $x = \frac{1}{3}$ and $y = \frac{1}{4}$,
$$
\frac{P(E)}{P(F)} = \frac{x}{y} = \frac{\frac{1}{3}}{\frac{1}{4}} = \frac{4}{3}.
$$
Hence,
$$
\frac{P(E)}{P(F)} = \frac{4}{3}.
$$
Answer
The correct value of $\frac{P(E)}{P(F)}$ is $\frac{4}{3}$.
